Pappus Configurations in Finite Hall Affine Planes

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PAPPUS CONFIGURATIONS IN FINITE HALL AFFINE PLANES

Lorinda A.M. Leshock

A dissertation submitted to the Faculty of the University of Delaware in partial fulﬁllment of the requirements for the degree of Doctor of Philosophy in Mathematics

Summer 2020

Lorinda A.M. Leshock

Approved: Louis Rossi, Ph.D. Chair of the Department of Mathematical Sciences

Approved: John A. Pelesko, Ph.D. Dean of the College of Arts and Sciences

Approved: Douglas J. Doren, Ph.D. Interim Vice Provost for Graduate and Professional Education and Dean of the Graduate College I certify that I have read this dissertation and that in my opinion it meets the academic and professional standard required by the University as a dissertation for the degree of Doctor of Philosophy.

Signed: Felix Lazebnik, Ph.D. Professor in charge of dissertation

I certify that I have read this dissertation and that in my opinion it meets the academic and professional standard required by the University as a dissertation for the degree of Doctor of Philosophy.

Signed: Robert Coulter, Ph.D. Member of dissertation committee

Signed: Jason Williford, Ph.D. Member of dissertation committee

Signed: Qing Xiang, Ph.D. Member of dissertation committee ACKNOWLEDGEMENTS

This thesis would not have existed without the guidance of Felix Lazebnik. I appreciate his ingenuity, fortitude, and graciousness as an advisor and a teacher. Learning to do mathematics with him has been an honor. I am thankful to Eric Moorhouse for the discussion of Hall planes, and in particular, for his help in understanding some properties of the collineation groups of Hall planes. Many thanks go to my committee members, Robert Coulter, Jason Williford, and Qing Xiang for their helpful comments and questions. I am grateful to the professors of the classes I took; their lectures and coursework nourished my mind. The staff of the Department of Mathematical Sciences was especially efﬁcient and friendly. I feel fortunate to have met many graduate students who faced challenges with dignity and shared humor during our time at the University of Delaware. Finally, I want to acknowledge my family. Because of their love and support, I had the opportunity to begin and the stamina to complete this endeavor. They encouraged me every step along the way.

iv TABLE OF CONTENTS

LIST OF FIGURES ...... vii ABSTRACT ...... viii

Chapter

1 INTRODUCTION ...... 1

1.1 Afﬁne Planes ...... 2 1.2 Conﬁgurations, Classical Theorems, Coordinatization ...... 5 1.3 Structure of the Dissertation ...... 9

2 FINITE HALL PLANES ...... 11

2.1 Analytic Construction ...... 11

2.1.1 Algebraic System ...... 11 2.1.2 Construction of Hall planes ...... 15 2.1.3 Symmetries ...... 17 2.1.4 Selected Properties ...... 25 2.1.5 Action of Collineation Group of AH on Pairs of Lines...... 26 2.1.6 The Way Mathematica is Used ...... 28

3 MAIN RESULTS ...... 30

3.1 Proof of Theorem 3.0.1...... 34

3.1.1 Case 1: BF/BF ...... 34

3.1.1.1 Intersecting `1, `2 ...... 34 3.1.1.2 Parallel `1, `2 ...... 41

3.1.2 Case 2: NBF/NBF ...... 44

3.1.2.1 Intersecting `1, `2 ...... 44

v 3.1.2.2 Parallel `1, `2 ...... 54

3.2 Proof of Theorem 3.0.2...... 61

3.2.1 Case 3a: NBF/BF ...... 61

3.3 Proof of Theorem 3.0.3...... 70

3.3.1 Case3b: BF/NBF ...... 70

4 SUPPLEMENT ...... 76

4.1 Numerical Search for Pappus Conﬁgurations ...... 76 4.2 Alternative Methods for Constructing Hall planes ...... 77 4.3 Possible Future Work ...... 79

5 CODE SAMPLES ...... 80

5.1 Magma ...... 80 5.2 Mathematica ...... 81

BIBLIOGRAPHY ...... 83

vi LIST OF FIGURES

1.1 Two diagrams of the plane on four points ...... 4

1.2 Two diagrams of the Desargues conﬁguration ...... 7

1.3 Three diagrams of the Pappus conﬁguration ...... 8

vii ABSTRACT

In the classical projective planes, both Desargues’s theorem and Pappus’s theorem hold. According to a result of Ostrom, the Desargues configuration can also be found in every finite projective plane on at least twenty-one points, classical or not. In fact, Ostrom’s argument shows that the number of Desargues configurations in every finite plane is actually quite large. The result is also true in the finite projective plane on thirteen points. The existence of Pappus configurations in every non-classical finite affine or projective plane is unknown. We study whether the Pappus configuration is present in such planes. In particular, we endeavor to prove that in finite Hall affine planes, the following strong version for the existence of Pappus configurations holds: For every pair of lines `1, `2 and every triple of points on `1 and every choice of a single point on `2, a pair of points on `2 can be found to complete a Pappus configuration. This statement is not proven in every case. When it is not, weaker versions for existence are shown. Hall planes are not Pappian, yet this work implies that the number of Pappus configurations in Hall planes is actually quite large.

viii Chapter 1

INTRODUCTION

Geometry was developed in many cultures to describe the world. In 300 BCE, Euclid’s Elements utilized an axiomatic deductive structure to organize mathematical knowledge, see Byer, Lazebnik, Smeltzer [3, p. 9]. By changing an axiom from Euclidean geometry in the 1820’s and 1830’s, two mathematicians, Lobachevsky and Bolyai independently developed non-Euclidean geometries. These mathematicians were contemporaries of Gauss who developed a geometry by changing the same axiom but chose not to publish it. Earlier speculations that such a geometry is possible are credited to Ibn al-Haytham (11th century), Khayyám (12th century), al-Tu¯ s¯ı (13th century), and Saccheri (18th century), see Rosenfeld and Yuschkevich [27]. Descartes formalized the connection between geometry and algebra by using coordinates and algebraic equations in the 17th century, see Brannan, Esplen, and Gray [2, p. 1]. Let us deﬁne two lines in the same plane to be parallel if they have no common point or they coincide. A problem determining whether two lines in the same plane are parallel may be solved geometrically, for example, by using alternate interior angles or analytically, by ﬁnding the equations of the lines and trying to solve the corresponding system of linear equations. This second approach to problem solving, the coordinate method, will be the one we favor in this dissertation. In 1899, Hilbert published the Foundations of Geometry. It is a succinct and groundbreaking text. In this book, Hilbert adjusted the axioms of Euclidean geometry with the goal of creating a simple and complete set of independent axioms from which a geometry can be built. As a part of Hilbert’s endeavor, his abstraction of the concepts of points, lines, and planes extended the scope of geometry. Furthermore, it promoted the use of the axiomatic method in mathematics.

1 1.1 Afﬁne Planes Consider a set P, a point set, with elements called points and another set L, a line set, with elements called lines that are subsets of the point set. We say that a point P ∈ P is on the line ` ∈ L if P ∈ `. In this situation, we may also say that the line ` is on the point P, that the line ` contains the point P, or that the point P and the line ` are incident. If a set of points is a subset of a line, then the points from the set are called collinear. If a point P is on every line from a set of lines, we say that the lines from the set are concurrent at P, or intersect at P. An afﬁne plane is an ordered triple A = (P, L, I) where I ⊆ (P × L) ∪ (L × P) is a symmetric binary relation on P ∪ L satisfying the following three axioms.

A1. For every two distinct points, there exists a unique line containing them.

A2. For every line ` and every point P not on `, there exists a unique line m on P such that m and ` are parallel.

A3. There exist three noncollinear points.

The concept of an affine plane reminds us of high school geometry. It is a generalization of the Euclidean plane with respect to only the notion of incidence. In the usual Euclidean affine plane we can introduce a coordinate system. Points are represented as ordered pairs of real numbers. Lines are described by linear equations with real coefficients. By using this approach, the Euclidean affine plane can be easily generalized to affine planes over other fields, in particular, over finite fields. The following interesting question was implicit in Hilbert’s Foundations and echoed by Artin [1, p. 51]:

Given a plane geometry whose objects are the elements of two sets, the set of points and the set of lines; assume that certain axioms of geometric nature are true. Is it possible to ﬁnd a ﬁeld k such that the points of our geometry can be described by coordinates from k and the lines by linear equations?

2 For affine planes, this question was answered by Hilbert himself, and the answer was negative. Hilbert’s work inspired a search, not for a field k, but for a more general algebraic system {S, +, ·}, see Cerroni [5, p. 322-327]. Here, + and · denote two binary operations on the set S, often referred to as "addition" and "multiplication". For most of the planes considered, more general algebraic systems were found. In this work, we restrict ourselves to a particular infinite family of finite affine planes and the algebraic systems that coordinatize them. In 1907, Veblen and Wedderburn [28, pp.382-383] constructed a finite algebraic system of nine elements, {S, +, ·}, which is a particular example of what is now called a Veblen-Wedderburn system. They used it to build an analytical model of a projective plane on 91 points. Moreover, they could argue that this projective plane cannot be coordinatized by a field. This work was generalized in 1943 by Hall [10, pp. 273-276], the eponym for Hall planes. He built the projective planes using an infinite class of Veblen-Wedderburn systems. These particular systems are usually called Hall systems or quasifields, which will be discussed in the Section 2.1.1.

Unless otherwise stated, the word plane will refer to an afﬁne plane.

Let us discuss an example of a finite plane. It’s easy to argue that a plane must have at least four points. We illustrate three ways to account for incidence in a finite plane: by maintaining a list of elements in a set system, using a diagram, and analytically. A diagram of a plane, A, is a figure in the Euclidean plane where points of A are represented by points of the plane, with lines represented as continuous curves passing through the points. We concentrate only on those points of the curves that represent the points of A.

Example

P = {1, 2, 3, 4}

L = {{1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}}

3 The point 2 is on line {1, 2}. The line {2, 4} is on the point 4. The points 1 and 2 are collinear. The lines {1, 2} and {2, 4} are concurrent. The lines {1, 3} and {2, 4} are parallel.

A diagram of a plane is not unique.

2 2 • •

3 3 • •

1 1 • •4 • •4

Figure 1.1: Two diagrams of the plane on four points

The coordinate method replaces each point label with a pair of coordinates and the list of points on each line with an equation that is satisﬁed by the pair of coordinates of each point on that line.

Here is the analytic model of the same plane. Let F = F2 be the finite field with two elements. We define points and lines as follows.

P = {(x, y) : x, y ∈ F} = {(0, 0), (0, 1), (1, 0), (1, 1)}

For arbitrary ﬁxed m, k ∈ F, the set {(x, xm + k) : x ∈ F}, represents a line with "slope" m and "y-intercept" (0, k). Clearly, y = xm + k is an equation of the line. For an arbitrary ﬁxed c ∈ F, the set {(c, y) : y ∈ F}, represents a "vertical" line with equation x = c. Counting all possible m, k and c, there are 22 = 4 non-vertical lines and and 2 vertical lines. These six lines form the set L of lines of our plane.

The following propositions and theorems along with their proofs may be found in the exposition by Moorhouse [20, pp. 9-11].

Proposition 1.1.1. Parallelism is an equivalence relation on the set of lines.

4 Proposition 1.1.2. Given any two lines `1, `2, there is a bijection between the points on `1 and the points on `2. Therefore, the set of points on these lines have the same cardinality.

Theorem 1.1.3. The number of points in any ﬁnite plane is n2 for some integer n ≥ 2. Every class of parallel lines consists of n lines, and there are n + 1 such classes. The lines from the class partition the set of all points of the plane. Every line is on n points and every point is on n + 1 lines. In particular, A has exactly n2 + n lines.

The integer n in Theorem 1.1.3 is called the order of the plane.

1.2 Configurations, Classical Theorems, Coordinatization A configuration in a plane A is a subset of points together with a subset of lines. Imposing some additional assumptions on A, Hilbert [13, pp. 23-70] uses configurations to develop geometric analogs to the operations of addition and multiplication.

He deﬁnes a segment as a pair of points on a line. In a plane, say A1, he deﬁnes an addition and multiplication of segments, and shows that these operations are commutative and associative. To prove these properties for multiplication, he uses a version of Pappus’s Theorem (which he refers to as Pascal’s Theorem). Hilbert also shows that the distributive law of multiplication over addition holds. He uses this algebra of segments to coordinatize A1 and create equations of lines. Later,

Hilbert introduces another plane A2 to develop an algebra of segments based on Desargues’s Theorem. He proceeds to determine that in a plane where Desargues’s Theorem holds, he can create what he calls a desarguesian number system which has all of the properties of a ﬁeld except commutative multiplication. Now such systems are called division rings. Conversely, he also demonstrates that a division ring can be used to create a plane in which Desargues’s Theorem holds. He determines the exact conditions under which Pappus’s Theorem holds in a Desarguesian plane and shows it is equivalent to multiplication being commutative in a desarguesian number system. Indeed, Hilbert points out that in the algebra of segments used in A1, he used Pappus’s Theorem to prove the commutativity of multiplication.

5 The planes that are coordinatized by ﬁelds are called classical planes with notation AG(2, F), where F is the ﬁeld. To discuss planes, both classical and nonclassical, it is helpful to have some additional vocabulary and notation. If P and Q are two distinct points on line `, then we write ` = PQ. Suppose two distinct lines m and n share a common point, R. Then we write R = m ∧ n, and if they are parallel, we write m k n. Here is Desargues’s Theorem, which we mentioned before.

Theorem 1.2.1 (Desargues’s Theorem). For every triple of concurrent lines `1, `2, `3 and any pair of triangles 4A1 A2 A3, 4B1B2B3 with vertices on those lines, A1, B1 ∈ `1,

A2, B2 ∈ `2 and A3, B3 ∈ `3, the intersection points of the lines through the pairs of corresponding sides of the triangles, P = A1 A2 ∧ B1B2, Q = A1 A3 ∧ B1B3, and R =

A2 A3 ∧ B2B3 are collinear.

The theorem concerns the generic configuration on the left in Figure 1.2. One can easily reformulate Desargues’s Theorem in cases where some lines in the configuration are parallel. For example, the following restatement refers to the non-generic configuration on the right in Figure 1.2.

Theorem 1.2.2 (Desargues’s Theorem (non-generic version 1)). For every triple of parallel lines `1, `2, `3 and any pair of triangles 4A1 A2 A3, 4B1B2B3 with vertices on those lines, A1, B1 ∈ `1, A2, B2 ∈ `2 and A3, B3 ∈ `3, if A1 A2 k B1B2, and A1 A3 k B1B3, then

A2 A3 k B2B3.

There exist several other versions of the conﬁguration where the triple of lines or some sides of the triangles are parallel. The theorem can be restated for them, as well. Desargues’s Theorem does not hold in every plane [13, pp. 47-51]. The planes where the theorem does not hold are called non-Desarguesian planes. Many classes of non-Desarguesian planes have been constructed. Shortly after Hilbert provided an example of a non-Desarguesian plane, a simpler example was constructed by Moulton [23, p. 193]. Hilbert understood that if Desargues’s Theorem holds in A,

6 R• • A1•

A2 Q• A1 • • •A3 •A3 • P A2• • B1• B1 •B2

•B3 •B3 B2•

Figure 1.2: Two diagrams of the Desargues conﬁguration then the coordinatizing system is guaranteed to have all of the properties of a ﬁeld except commutative multiplication. As mentioned before, the algebraic system is called a division ring. For example, the real quaternions form a famous division ring. Pappus’s Theorem also played a fundamental role in Hilbert’s work.

Theorem 1.2.3 (Pappus’s Theorem). For every pair of lines `1, `2 and every triple of points on each line, A1, B1, C1 ∈ `1 and A2, B2, C2 ∈ `2, the intersection points A3 =

B1C2 ∧ B2C1,B3 = A1C2 ∧ A2C1, and C3 = A1B2 ∧ A2B1 are collinear.

The theorem concerns the generic conﬁguration on the left in Figure 1.3. The theorem is restated for the corresponding non-generic versions of the conﬁguration.

Theorem 1.2.4 (Pappus’s Theorem (non-generic version 1)). For every pair of lines `1, `2 and every triple of points on each line, A1, B1, C1 ∈ `1 and A2, B2, C2 ∈ `2, if B1C2 k B2C1 and A1C2 k A2C1, then A1B2 k A2B1.

Theorem 1.2.5 (Pappus’s Theorem (non-generic version 2)). For every pair of lines `1, `2 and every triple of points on each line, A1, B1, C1 ∈ `1 and A2, B2, C2 ∈ `2, if B1C2 k B2C1 and the other pairs of lines intersect in points B3 = A1C2 ∧ A2C1, and C3 = A1B2 ∧ A2B1, then the line B3C3 k B1C2.

7 A C1 1 • A1 B1 C1 • B1 • • • • A1 B 1C1 • • • • • A3 C3 C3• B3 B3 • • • • • • • • C2 B2 A2 • • • A2 B2 C2 A2 C2 B2

Figure 1.3: Three diagrams of the Pappus conﬁguration

The planes where Pappus’s Theorem holds are called Pappian planes, and otherwise, non-Pappian planes. Hilbert understood that if Desargues’s and Pappus’s Theorems hold in a plane, then the coordinatizing system is guaranteed to have commutative multiplication, in which case, the coordinatizing system is a field. It turns out that Desargues’s Theorem and Pappus’s Theorem in planes are not independent: Pappus’s Theorem implies Desargues’s Theorem. This result was first obtained by Hessenberg [12]. One missing case in his proof was closed by Cronheim [6]. For an alternative reference, see Cameron [4, pp. 1-24]. Due to the existence of quaternions, Desargues’s Theorem doesn’t imply Pappus’s Theorem. According to the famous theorem of Wedderburn [19], every finite division ring is a field. Wedderburn provided multiple proofs in his paper. Almost twenty years after it was published, Artin found a nontrivial hole in the first proof. The other proofs were correct and cite a result by Dickson, see Kantor [18, p. 103]. Many additional proofs of Wedderburn’s Theorem exist. A sketch of Witt’s proof of Wedderburn’s Theorem can be found in Dummit and Foote [7, p. 536]. An implication of Wedderburn’s theorem is that, in finite planes, Desargues’s Theorem implies Pappus’s Theorem. Here is another natural question related to Desargues’s Theorem and Desargues configurations.

Suppose that Desargues’s Theorem does not hold in a plane. Is there still at least one Desargues conﬁguration in the plane?

8 In infinite planes, the Desargues configuration doesn’t have to exist! Hall [10, pp. 236- 240] proved this in 1943 with his "free plane" construction. For another reference, see Hartshorne [11, pp. 7-10]. However, the Desargues configuration exists in every finite plane of order greater than or equal to four. We know this due to an elegant combinatorial proof by Ostrom [25, p. 398] in 1957. It actually shows that there are many Desargues configurations in every finite plane of order greater than or equal to four. This result is also true in planes of order three. His paper dealt with projective planes, but the conclusion for affine planes easily follows. Inspired by Ostrom’s result, Lazebnik asked the following question.

Suppose that Pappus’s Theorem does not hold in a plane. Is there still at least one Pappus conﬁguration in the plane?

By Hall’s "free plane" construction, we know that there exist infinite planes where the Pappus configuration does not exist. Note that it is easy to show the existence of Desargues and Pappus configurations in any plane with a sufficient number of points whose algebraic system contains a field, such as a quasifield or a semifield (a division ring with non-associative multiplication). Indeed, restricting the coordinates of the initial points in these configurations to the elements of the field, we can check that the configuration exists using analytic geometry. Whether for any plane, there exists a coordinatizing algebraic system that contains a field is not known.

1.3 Structure of the Dissertation In this work, we use the coordinate method to prove that Hall planes contain many Pappus configurations. The main problem is to find the strongest conditions under which we are certain to find a Pappus configuration on any pair of lines in Hall planes. It should be noted that every Hall plane contains a classical subplane, so the existence of at least one Pappus configuration is guaranteed, see Corollary 2.1.11. Most of our remarks about Hall planes refer to those of order greater than nine. The Hall plane of order nine exhibits "exceptional" properties (see Hughes [14,

9 pp. 936-938]), and we will consider it separately. In Chapter2, we explore Hall planes in detail. First, we describe the method of constructing a Hall plane analytically, starting with introducing Hall systems. Then, we explore the collineation group in so much as it is used to obtain our results. Some additional properties of Hall planes are presented. Next, we gather what we know to carve the main problem into cases. Finally, we discuss the way that Mathematica is used in our proofs. We state our best conjecture to start Chapter3 and then move on to prove what we can in Theorems 3.0.1, 3.0.2, and 3.0.3. We present the best results we could obtain in each case, falling short of proving the conjecture with the strongest conditions that we believe to be true and thereby foregoing the statement of a theorem. However, there is some level of success in minimizing the number of constraints on the parameter sequences by providing multiple solutions covering different parameter sequences for each pair of lines. We end this work with what was actually done at the beginning of the research. In Chapters4 and5, we present some data obtained at the start of this project when we tested a variety of planes to study the occurrence of Pappus conﬁgurations in non- Desarguesian planes. Next, we provide summaries of a few construction techniques that are not used in our solution. Then, some samples of code are provided for the reader who would like to verify our claims or attempt to advance our results.

10 Chapter 2

FINITE HALL PLANES

Finite Hall planes may be constructed in a variety of ways, a few of which are summarized in Chapter4. Since we use the analytic construction to solve the main problem of this work, we will present it in detail. Many of our arguments are based on symbolic algebraic computations. Most of them are straightforward, and some are extensive. In all cases, the details of these computations are suppressed. Most of them were done by using Mathematica, and those which were done manually were veriﬁed by Mathematica.

2.1 Analytic Construction We will construct Hall systems to build Hall planes. It will be shown that the set of coordinates and the associated equations satisfy the axioms of a plane. We will present several symmetries of the plane in analytic form. After a selection of the properties of Hall planes is explored, we begin to solve the main problem.

2.1.1 Algebraic System

This exposition follows Hall [8, p. 211-215] and [10, pp. 253-273]. Let F = Fq represent the finite field of prime power order q and H = {(a1, a2) : a1, a2 ∈ F}. We call F the basefield of H. Clearly, |H| = q2. It is convenient to have multiple notations for elements of H. Let the bold letter a denote the ordered pair, (a1, a2) ∈ H. We will identify an element b = (b1, 0) of H with the element b1 of F and write, for simplicity, b ∈ F, and we will write b = (b1, b2) 6∈ F when b2 6= 0. The Hall system is a two dimensional (right and left) vector space H over F equipped with a certain multiplications of vectors. The addition in our system

11 {H, +, ·} is the usual addition in F2. To define multiplication in {H, +, ·}, we use the operations from the basefield and a quadratic polynomial f (x) = x2 − rx − s with r, s ∈ F which is irreducible over F, and we refer to it as the defining polynomial of the system. For a, b ∈ H,

A. a + b = (a1, a2) + (b1, b2) = (a1 + b1, a2 + b2)

M1. ab = (a1, a2) · (b1, b2) = (a1b1, a2b1), if b ∈ F

−1 M2. ab = (a1, a2) · (b1, b2) = (a1b1 − a2b2 f (b1), a1b2 − a2b1 + a2r), if b 6∈ F

According to [8, p. 214], the multiplication defined in M1. and M2. is equivalent to the multiplication defined by the following three rules. For a, b, c ∈ H, and coefficients of the defining polynomial of the Hall system, r, s ∈ F,

M1∗. ab = ba, if b ∈ F

M2∗. (a + b)c = ac + bc

M3∗. c2 = rc + s, if c 6∈ F

The equivalence of M1. and M1∗. is an artifact of the properties of basefield multiplication. Direct computation is sufficient to show that M2. implies M2∗. and M3∗. Conversely, M2∗. and M3∗. may be viewed as the motivation for and are used in the derivation of M2. For details, see Hughes and Piper [16, pp. 183-187]. For Hall systems over basefields with q > 3, multiplication is neither commutative nor associative; it is right distributive but not left distributive over addition. In Section 2.1.4, we will point out how some properties of Hall systems with q = 2 and q = 3 (with defining polynomial f (x) = x2 − 2) differ those listed in the previous sentence. Clearly, H is a group under addition with identity element 0 = (0, 0). It’s also easy to see that the identity element for multiplication in H (both left and right) is 1 = (1, 0).

12 We will use the following facts concerning solving equations in {H, +, ·} [10, pp.253-254]. They are often presented as properties of Veblen-Wedderburn systems or quasiﬁelds, of which the Hall system is a particular example.

E1. For x, y, z ∈ H, if y 6= 0, then the equation xy = z can be solved for x uniquely.

E2. For x, y, z ∈ H, if x 6= 0, then the equation xy = z can be solved for y uniquely.

E3. If w, x, y, z ∈ H and x 6= y, then wx = wy + z has a unique solution w. Proof. We begin with E1. If y 6= 0 and y ∈ F, then x = z1 , z2 . Otherwise, if y1 y1 y 6= 0 and y 6∈ F, then a short computation (based on M2.) gives the unique solution ry z −y y z −sz −ry z +y2z − + x = − 2 1 1 2 1 2 1 2 1 2 , − y2z1 y1z2 . The denominators in the fractions sy2 s are not zero due to the irreducibility of the deﬁning polynomial over F. For E2., if x 6= 0 and z = kx for some k ∈ F, then it is easy to see that y = (k, 0). If x 6= 0 and z 6= kx for any k ∈ F, a short computation (based on M2.) gives the sx2+rx z −z2 unique solution y = sx1x2+rx2z1−z1z2 , 2 2 2 2 . Obviously, the denominators in x2z1−x1z2 x2z1−x1z2 the fractions are not zero. We turn our attention to E3. As the Hall system does not satisfy the left distributive law, c(a + b) = ca + cb, corresponding to M2∗., this statement cannot be reduced to E1. or E2. immediately. First, we express the elements of the Hall system using components. By varying whether x and y are in the baseﬁeld, we can determine which multiplication rule to apply, M1. or M2., then solve for w.

• x, y ∈ F, x 6= y: That there is a unique solution for w may be seen by applying the multiplication rule M1., and E2.

wx = wy + z

x1w = y1w + z

(x1 − y1)w = z

13 • x ∈ F, y 6∈ F: We expand the products wx and wy using the multiplication

rules M1. and M2., respectively, then solve for the components w1 and w2 of w uniquely.

wx = wy + z 2 −ry2z1+x1y2z1+y1y2z1+sz2+ry1z2−y1z2 y2z1+x1z2−y1z2 w = − 2 , − 2 (s+rx1−x1)y2 s+rx1−x1

The denominators of the components are not zero since y 6∈ F and due to the irreducibility of the deﬁning polynomial over F.

• x 6∈ F, y ∈ F is completely similar to the previous case.

• x, y 6∈ F, x 6= y: We expand the products wx and wy using the multiplication rule M2., then solve for the components of w uniquely.

wx = wy + z 2 2 −sy2z2+x1y2z2−x1y2(x2z1+rz2)+x2(y1y2z1+sz2+ry1z2−y1z2) w = 2 , s(x2−y2) +(x2y1−x1y2)(−x2y1+r(x2−y2)+x1y2) −x2y2(x2z1−y2z1+(−x1+y1)z2) 2 s(x2−y2) +(x2y1−x1y2)(−x2y1+x1y2+r(x2−y2))

2 2 If x2 = y2, then the denominators become (x2y1 − x1y2) = (x2(x1 − y1)) . As

x 6∈ F, x2 6= 0. As, x 6= y, x1 6= y1. − If x 6= y , then the denominator can be written as −(x − y )2 f x2y1 x1y2 , 2 2 2 2 x2−y2 therefore it is nonzero due to the irreducibility of the deﬁning polynomial f over F.

As we mentioned before, a Hall system is an example of a more general algebraic structure sometimes called a Veblen-Wedderburn system, or a quasiﬁeld [8, p. 213].

14 2.1.2 Construction of Hall planes The point set of a Hall plane is P = {(x, y) : x, y ∈ H}. For arbitrary fixed m, k ∈ H, the sets {(x, xm + k) : x ∈ H}, represent "non-vertical" lines with equations of the form y = xm + k. For arbitrary fixed k ∈ H, the sets {(x, k) : x ∈ H}, represent "horizontal" lines with equations of the form y = k. For arbitrary fixed c ∈ H, the sets {(c, y) : y ∈ H}, represent "vertical" lines with equations of the form x = c. These q4 + q2 lines form the set L of lines in a Hall plane. The non-vertical lines y = xm + k will be referred to as type 1 lines or type 2 lines depending on the "slope" m; type 1 lines have m ∈ F and type 2 lines have m 6∈ F. We define a symmetric incidence relation I on the set of all points and lines as follows, for a point P and line `, PI` and ÌP if P ∈ `.

2 Proposition 2.1.1. The incidence system AH = (P, L, I) is a plane of order q .

In keeping with the spirit of this work, we present a direct analytical proof. We would like to point out that historically, the demonstration that AH is indeed a plane followed a slightly different path. It was a corollary of a more general approach, via ternary rings, linear ternary rings, translation planes, Veblen-Wedderburn systems, and often the notion of projective plane was involved. See [8, pp. 211-215], [9, pp. 353- 366], and [16, pp. 127-134,157,158,183-187] for details. To some extent, these types of arguments, from the general to the speciﬁc, elucidate important information about some sets of symmetries of Hall planes which are inherited from a larger class of planes. For us that discussion, based on the deﬁnition of AH, occurs in Section 2.1.3.

Proof. We will verify each of the three axioms of a plane. Let P = (xP, yP) and

Q = (xQ, yQ) be distinct points, and ∆x = (∆x1, ∆x2) = xP − xQ, ∆y = (∆y1, ∆y2) = yP − yQ. To prove A1., we consider all possible pairs of distinct points P, Q and split our argument into cases based on the values of ∆x, ∆y to show that there exists a unique line PQ.

• If ∆x = 0, then ∆y 6= 0 and PQ = {(xP, y) : y ∈ H}.

15 • If ∆y = 0, then ∆x 6= 0 and PQ = {(x, yP) : x ∈ H}.

• For ∆x∆y 6= 0, we solve the system of equations,

yP = xPm + k

yQ = xQm + k

for m, k ∈ H. By M2∗. and E2., ∆y = ∆xm has a unique solution m0 ∈ H. 0 0 0 0 Hence, k = yP − xPm = yQ − xQm and PQ = {(x, xm + yP − xPm ) : x ∈ H}.

To prove A2., we split our argument into cases based on whether ` is a vertical or non-vertical line. We take an arbitrary point P not on line `, then show that there exists a unique line m on P that does not intersect `.

• Take ` : x = c and P with xP 6= c. Then the line m : x = xP has no points in common with ` since all points on ` are of the form (c, y) and all points on m

are of the form (xP, y) for some y ∈ H. Also, any line of the form y = xm + k that passes through P intersects ` at (c, cm + k), so m is the unique line through P that is parallel to `.

• Take ` : y = xm + k and P with yP 6= xPm + k. Then the line m : y =

xm + (yP − xPm) has no points in common with ` since all points on ` are of the

form (x, xm + k) and all points on m are of the form (x, xm + (yP − xPm)). Also,

any line of the form x = xP that passes through P intersects ` at (xP, xPm + k), so m is the unique line through P that is parallel to `.

To prove A3., we show that the three points, P0((0, 0), (0, 0)), Q0((1, 0), (0, 0)), R0((0, 0), (1, 0)), are not collinear. Notice that points P0 and Q0 are on the line y = 0, points P0 and R0 are on the line x = 0, and points Q0 and R0 are on the line

(y1, y2) = (x1, x2)(−1, 0) + (1, 0). Hence, we have three noncollinear points for a plane coordinatized by a Hall system over any baseﬁeld.

16 Since the axioms of a plane are satisﬁed, AH constructed from {H, +, ·} is a plane.

2.1.3 Symmetries For a plane A = (P, L, I), a bijection φ : P → P which preserves collinearity is called a collineation. A collineation φ induces an action on the set of lines L in an obvious way: for ` ∈ L, `φ = {Pφ : P ∈ `}. A point P (line `) is fixed under φ if Pφ = P (`φ = `). We say that a line ` is fixed pointwise when every point on ` is fixed. We extend this notion to parallel classes of lines. The parallel class of lines is fixed under a collineation if the collineation permutes all lines in the class, and it’s fixed linewise if every line from the class is fixed. A translation of a plane A is a collineation of A such that the parallel classes are fixed and there is some parallel class that is fixed linewise. An example of a translation of the classical plane AG(2, F) is τa,b : (x, y) → (x + a, y + b) where F is a field and a, b, x, y ∈ F such that a, b are fixed. When a 6= 0, the parallel class fixed by −1 τa,b is the class containing the line {(x, xa b + k) : x ∈ F} for some k ∈ F. When a = 0, the parallel class fixed by τa,b is the class containing the line {(c, y) : y ∈ F} for some c ∈ F. If the group of translations is transitive on the points of A, the plane A is called a translation plane, see [26, p. 457]. It is known that translation planes are exactly the planes which can be coordinatized by quasifields, of which the Hall system is a particular example [9, pp. 362-363]. It’s obvious using the translations τa,b that classical and Hall planes are translation planes (in the case of the Hall plane, a, b are in the corresponding quasifield). In 1959, Hughes gave an analytic description of the entire collineation group of Hall planes, see [14, pp. 924-933] and his related work [15, pp. 980-981]. Hughes presents the collineation group of Hall planes as generated by six subgroups. Three of those subgroups are relevant to the solution of the main problem in this dissertation and are presented below. Hughes uses different conventions than Hall did in his

17 construction of Hall planes, e.g., multiplication is left not right distributive over addition and a non-vertical equation is of the form mx + y = k rather than y = xm + k. These differences cause changes in: the multiplication rule M2., the component that must be zero for an ordered pair to be identified with a basefield element (including the multiplicative identity), and the analytic formulas for the collineations. We have adapted the analytic formulas Hughes presented for the collineation subgroups to be consistent with Hall’s methodology. We remind the reader that the defining polynomial of the Hall plane is f (x) = x2 − rx − s where r, s ∈ F. In this subsection, the vertical line {(c, y) : y ∈ H} will be denoted as [c] and the non-vertical line {(x, xm + k) : x ∈ H} will be denoted as [m, k], for arbitrary fixed c, m, k ∈ H, as Kallaher does [17, p. 141].

Now, we discuss the three groups of collineations of AH mentioned above which are closely related to our investigation.

We begin with the translations of AH. Let a, b ∈ H and τ = τ(a, b) be a map τ : P → P given by

(x, y)τ = (x + a, y + b).

Let us deﬁne the action of τ on L in the following way,

[c]τ = [c + a];

[m, k]τ = [m, k − am + b].

Such a map τ is called a translation of AH.

Let TR = {τ = τ(a, b) : a, b ∈ H} be the set of all translations of AH = (P, L, I).

Proposition 2.1.2. The following holds for AH.

1. Every translation from TR is a collineation of AH.

18 2.TR is a group under composition, and it is isomorphic to the additive group of the vector space H2 over F.

3.TR is sharply transitive on P, i.e., it is transitive, and the stabilizer of any point is the identity map τ(0, 0).

4.TR preserves the classes of parallel lines and acts transitively on the set of lines in every parallel class.

Proof. The veriﬁcation of Part 1 is straightforward from our deﬁnition of τ. Parts 2, 3, and 4 are obvious.

Next, we introduce the autotopisms of AH. Let S be a 2 × 2 nonsingular matrix over F. For x ∈ H, let S act on H in the following way:

xS = (x1, x2)S.

Deﬁne a = 1S−1, where 1 = (1, 0) is the multiplicative identity of H. Hence, aS = 1. Let σ = σ(S) be a map σ : P → P given by

(x, y)σ = (xS, yS).

We extend the deﬁnition of σ on L in the following way,

[c]σ = [cS];

[m, k]σ = [(am)S, kS].

Such a map σ is called an autotopism of AH. Let ATP = {σ = σ(S) : S ∈ GL(2, F)} be the set of all autotopisms of

AH = (P, L, I).

Proposition 2.1.3. The following holds for AH.

1. Every autotopism from ATP is a collineation of AH.

19 2. ATP is a group under composition, and it is isomorphic to GL(2, F).

3. ATP ﬁxes the parallel classes of type 1 lines, acts transitively on the parallel classes of type 2 lines, and it has two orbits on the set of vertical lines (one orbit consists of the line [0]).

Proof. The veriﬁcation of Part 1 is obvious for vertical lines. For non-vertical lines, we use the identity (xm)S = (xS)((am)S):

(x, y) ∈ [m, k]

y = xm + k ⇔

yS = (xm + k)S ⇔

yS = (xm)S + kS ⇔

yS = (xS)((am)S) + kS ⇔

(x, y)σ ∈ [m, k]σ

Part 2 is true by the deﬁnition of ATP. Part 3 relies on Part 2 For vertical lines, it is clear that there are two orbits and one of them is [0]. To see why ATP ﬁxes the parallel classes of type 1 lines, recall that −1 0 a = 1S and if m ∈ F then (am)S = m1(aS). By E1., for a given m, m 6∈ F, there exists a unique a ∈ H such that am = m0, therefore ATP is transitive on the parallel classes of type 2 lines.

Finally, we present the linear collineations.   −ar + b as Let a, b ∈ F not both zero, and define the matrix L =  , where a b r, s are the coefficients of the defining polynomial f (x) = x2 − rx − s of H. Clearly, L ∈ GL(2, F). For (x, y) ∈ H2, let L act on H2 in the following way:

(x, y)L = ((−ar + b)x + ay, asx + by).

20 Let λ = λ(L) = λ(a, b) be a map λ : P → P given by

(x, y)λ = (x, y)L.

We extend the deﬁnition of λ on L in the following way,

 [bc], if a = 0 [ ] = c λ h i  b b2−abr−a2s  a , 0 , − a c , if a 6= 0 If m ∈ F,  [m, bk], if a = 0   [m, k]λ = b [ak], if a 6= 0 and m1 = r − a  h 2 2 i  as+m1b , 0 , b −abr−a s k , if a 6= 0 and m 6= r − b  am1−ar+b am1−ar+b 1 a If m 6∈ F,

[m, k]λ = [m, −akm + bk]

Such a map λ is called a linear map of AH. Let LNR = {λ = λ(L) = λ(a, b) : a, b ∈ F, (a, b) 6= (0, 0)} be the set of all linear maps of AH = (P, L, I). We will denote the set of lines that are vertical or have slope from the basefield as BF and the set of lines with slope not from the basefield as NBF.

Proposition 2.1.4. The following holds for AH.

1. Every linear map from LNR is a collineation of AH.

2.LNR is a group under composition, and it is isomorphic to the multiplicative group of F[ ] = [ ] 2 = − + × the quadratic extension ﬁeld α Fq α where α rα s, and so to Fq2 .

3.LNR ﬁxes the parallel classes of lines in NBF and is transitive on the parallel classes of lines in BF.

21 Proof. The veriﬁcation of Part 1 is obvious for some mappings. The mappings that use less obvious manipulations are provided below. If a 6= 0 and (x, y) ∈ [c], then

x = c ⇔

b2 b2 asx + by = −br + a x + by − a − br − as c ⇔

b b2−abr−a2s asx + by = ((−ar + b)x + ay) a − a c ⇔ (x, y)λ ∈ [c]λ.

b If m ∈ F, a 6= 0, m1 6= r − a , and (x, y) ∈ [m, k], then

y = xm + k ⇔

2 2 2 2 2 2 (b − abr − a s)y = m1(b − abr − a s)x + (b − abr − a s)k ⇔ ( + ) 2 2 asx + by = ((−ar + b)x + ay) as m1b + b −abr−a s k ⇔ am1−ar+b am1−ar+b (x, y)λ ∈ [m, k]λ.

If m 6∈ F and (x, y) ∈ [m, k], then

y = xm + k ⇔

−aym + by = −arxm − akm − asx + bxm + bk ⇔

asx + by = ((−ar + b)x + ay)m − akm + bk ⇔

(x, y)λ ∈ [m, k]λ

To justify Part 2, since f (x) = x2 − rx − s is irreducible over F and quadratic, it has no roots in F. Hence, f (−x) = x2 + rx − s is irreducible over F. Joining its root α creates the quadratic extension ﬁeld F[α] = {aα + b : a, b ∈ F} where α2 = −rα + s. It is obvious that LNR is isomorphic to the multiplicative group of all of our matrices L. It is a straightforward veriﬁcation that the latter is isomorphic to F[α]× via the

22 map:   −ar + b as   7→ aα + b. a b

To prove Part 3, it is enough to look at the extended deﬁnition of λ on L, and in some cases, set up equations and solve for the parameters a, b ∈ F. Using LNR, one may map a given vertical line to any vertical line or to some type 1 line from any parallel class. Similarly, a type 1 line may be mapped to any vertical line, or to any type 1 line in its own parallel class, or to some type 1 line in a different parallel class. A type 2 line may be mapped to any type 2 line in the same parallel class.

Proposition 2.1.5. The group generated by TR and ATP acts transitively on all type 2 lines and on all vertical lines.

Proof. By Proposition 2.1.3, ATP acts transitively on the parallel classes of type 2 lines and has two orbits on the set of vertical lines, one of which is [0]. By Proposition 2.1.2, TR acts transtively on the lines in every parallel class.

Proposition 2.1.6. The group generated by TR and LNR acts transitively on all BF lines.

Proof. By Part 3 of Proposition 2.1.4, LNR acts transitively on the parallel classes of lines in BF. By Proposition 2.1.2, TR acts transtively on the lines in every parallel class.

Based on the above propositions, it is clear that there are at most two orbits of lines in AH. The sets BF and NBF partition L and are precisely these orbits. This was implicitly shown in [14, pp. 924-927]. The next proposition describes the action of certain collineations in the stabilizer of the origin on lines through the origin.

Proposition 2.1.7. For any line through the origin, there exists a group of collineations which ﬁxes the line, ﬁxes the origin, and acts transitively on the other points of the line.

Proof. We consider the following three cases: x = 0, y = xm with m ∈ F, and y = xm with m 6∈ F.

23 • Take ` : x = 0. The line ` is on the origin. By Part 3 of Proposition 2.1.3, the action of ATP ﬁxes the line `. Since every point on the line ` is of the form (0, y) for some y ∈ H, Part 2 of Proposition 2.1.3 implies that ATP is transitive on the points of ` other than the origin.

• Take ` : y = xm, with m ∈ F. The line ` is on the origin. By Part 3 of Proposition 2.1.3, the action of ATP fixes the parallel class of line `. Since the origin is fixed by elements of ATP, the line ` is fixed. Since every point of the line ` is of the form (x, xm) for m ∈ F, Part 2 of Proposition 2.1.3 implies that ATP is transitive on the points on ` other than the origin.

• Take ` : y = xm, with m 6∈ F. The line ` is on the origin. By Part 3 of Proposition 2.1.4, the action of LNR fixes the parallel class of line `. By the definition of the map λ on points, LNR fixes the origin. Hence, it fixes the line `.

Let (v, vm) and (w, wm) be two distinct points of `, neither is the origin. Let us show that there exists a, b ∈ F with (a, b) 6= (0, 0) such that λ = λ(a, b) ∈ LNR maps (v, vm) to (w, wm). This is equivalent to solving the equation for a, b.

(−ar + b)v + avm = w

Since H is a two dimensional vector space over F, and v and vm are linearly

independent over F because m 6∈ F, there exist c1, c2 ∈ F such that c1v +

c2vm = w. Setting −ar + b = c1 and a = c2, we get b = c1 + c2r. Clearly, (a, b) 6= (0, 0) as otherwise, w = 0. Hence, the transitivity statement is proven.

Corollary 2.1.8. For any pair of lines {`1, `2} through the origin, both from the BF orbit, or both from the NBF orbit, there exists a group of collineations which ﬁxes both lines, ﬁxes the origin, and acts transitively on the points on `1 or on `2 other than the origin.

24 2.1.4 Selected Properties The following properties of Hall planes are notable for their idiosyncrasy.

Proposition 2.1.9. For baseﬁeld F2, the Hall system is isomorphic to F4. Hence, the plane on 16 points coordinatized by a Hall system is a classical plane [16, pp. 185-186].

Proof. Since a Hall system H1 with four elements is a two dimensional vector space over F2, the rule for addition in H1 is the rule for addition in F4. There is only one 2 monic irreducible quadratic polynomial over F2, namely f (x) = x + x + 1, so H1 is the only Hall system with four elements. It is easy to verify that application of the two rules for multiplication, M1. and M2., in H1 produce the multiplication table for

F4. It follows that the plane coordinatized by H1 is isomorphic to the classical plane

AG(2, F4).

Proposition 2.1.10. Embedded in every Hall plane with baseﬁeld F = Fq for prime power q is a classical subplane on q2 points.

∼ Proof. Consider the subset of the Hall system, HF = {(a, 0) : a ∈ F} = F. Only the multiplication rule M1. applies to all elements of this subset. Furthermore, M1. simpliﬁes to (a, 0) · (b, 0) = (ab, 0). Hence, both the addition and multiplication tables of HF are isomorphic to those of F. It follows that (HF, +, ·) coordinatizes 2 a plane AH = (P, L, I) which is isomorphic to the classical plane AG(2, F) on q points.

Now, we can justify the existence of many Pappus conﬁgurations in AH mentioned in the introduction.

Corollary 2.1.11. The Pappus conﬁguration exists in every Hall plane of order at least 3.

2 Proposition 2.1.12. For baseﬁeld F3 and deﬁning polynomial f (x) = x − 2, the Hall system has associative multiplication [16, pp. 185-186].

Proof. For some elements of H, the result is obvious. In general, to establish the associativity of H, we used the rules of multiplication via components.

25 Theorem 2.1.13. For Hall systems over the same basefield, distinct defining polynomials coordinatize the same Hall projective plane. The collineation group of the finite Hall affine plane is transitive on points. The Hall affine plane coordinatized over F3 with defining polynomial f (x) = x2 − 2 is doubly transitive on points [14, pp. 936-938].

2.1.5 Action of Collineation Group of AH on Pairs of Lines. The collineation group of Hall planes acts on the pairs of distinct lines. The goal of this subsection is to describe some special pairs of lines on each orbit of the action. As the collineation group acts on Pappus conﬁgurations, our proofs of the main theorems can be restricted to Pappus conﬁgurations on these special pairs of lines that represent orbits of this action.

Lemma 2.1.14. Any pair of intersecting lines can be translated to a pair of lines that meet at the origin. Any pair of parallel lines may be translated so that one of the lines contains the origin.

Proof. Follows from Proposition 2.1.2.

The proof of the main result will consist of three cases. Since we have two orbits of lines, BF and NBF, the three cases correspond to choosing two lines from BF, or from NBF, or one line from each orbit.

Proposition 2.1.15 (BF/BF). Any pair of intersecting lines from the BF orbit can be mapped to a pair of lines (`1, `2) where `1 : y = xm for some m ∈ F and `2 : x = 0. Any pair of parallel lines from the BF orbit can be mapped to a pair of horizontal lines (`1, `2) where `1 : y = 0 and `2 : y = k for some k ∈ H.

Proof. For a given pair of intersecting lines (`, n) from the BF orbit, we may use LNR to map ` to a vertical line `0: x = c for some c ∈ H. The same map will carry n to a non-vertical line n0: y = xm + k for some m ∈ F and k ∈ H. By Lemma 2.1.14, we can translate their intersection point to the origin. Now, we have the desired lines

`1 : y = xm for some m ∈ F and `2 : x = 0.

26 For a given pair of parallel lines (`, n) from the BF orbit, we may use LNR to map ` to the horizontal line `1 : y = 0. This same map will carry line n to a horizontal line `2 : y = k for some k ∈ H.

Proposition 2.1.16 (NBF/NBF). Any pair of intersecting lines from the NBF orbit can be mapped to a pair of lines (`1, `2) where `1 : y = xm for some m 6∈ F and `2 : y = x(0, 1).

Any pair of parallel lines from the NBF orbit can be mapped to a pair of lines (`1, `2) where

`1 : y = x(0, 1) and `2 : y = x(0, 1) + k for some k ∈ H.

Proof. For a given pair of intersecting lines (`, n) from the NBF orbit, we may use ATP to map ` to a line `0: y = x(0, 1) + k for some k ∈ H. The same map will carry n to a line n0: y = xm + c for some m, c ∈ H with m 6∈ F. By Lemma 2.1.14, we can translate their intersection point to the origin. Now, we have the desired lines

`1 : y = xm for some m 6∈ F and `2 : y = x(0, 1). For a given pair of parallel lines (`, n) from the NBF orbit, we may use ATP to map ` to a line `0: y = x(0, 1) + c for some c ∈ H. The same map will carry n to a line n0: y = x(0, 1) + d for some d ∈ H. By Lemma 2.1.14, we can choose a point on line ` to translate to the origin. Now, we have the desired lines `1 : y = x(0, 1), `2 : y = x(0, 1) + k for some k ∈ H.

Note that if lines in a pair come from different orbits, they cannot be parallel.

Proposition 2.1.17 (BF/NBF). Any pair of intersecting lines with the ﬁrst from BF and the second from NBF can be mapped to `1 : x = 0, `2 : y = x(0, 1).

Proof. For a given pair of intersecting lines (`, n) with ` from the BF orbit and n from the NBF orbit, we may use LNR to map ` to a vertical line `0: x = c for some c ∈ H. The same map will carry n to a line n0: y = xm + d for some m, d ∈ H with m 6∈ F. Next, we may use ATP to map n0 to a line n00: y = x(0, 1) + d0 for some d0 ∈ H. That map will carry `0 to a line `00: x = c0 for some c0 ∈ H. By Lemma 2.1.14, we can translate their intersection point to the origin. Now, we have the desired pair of intersecting lines `1 : x = 0 and `2 : y = x(0, 1).

27 These three cases will be analyzed to determine the strongest conditions for the existence of Pappus conﬁgurations in Hall planes in Chapter3.

2.1.6 The Way Mathematica is Used As we study Hall planes analytically, the existence of Pappus configurations follows from showing that certain systems of equations have solutions. Using Mathematica in numerical mode over finite fields never created sub- stantial difficulties and the computations supported our conjectures. When Mathematica was used in symbolic mode the situation was quite different. It turned out that instead of trying to compute symbolically over finite fields, it was beneficial to assume that all constants in our systems come from the algebraic closure of the rational numbers Q (the default in Mathematica), and then interpret the obtained results as output from symbolic computations over a finite field F = Fq for = a−b ∈ Q some prime power q. For example, if we encounter x t2−rt−s where a, b, r, s, t we must pay special attention to those values of t where t2 − rt − s is zero. However, if we assume that a, b, r, s, t ∈ F, we know that the denominator is not zero due to the irreducibility of the defining polynomial f (x) = x2 − rx − s of H. Another situation that arises is when we obtain a complicated expression involving a square root in our solution over Q. That imposes a restriction on the values of the prime power q for which the square root does not exist over F. Here is a more important difficulty. Suppose we are over a finite field and are interested in proving that for an arbitrary pair of lines `1 and `2, and three arbitrary points A1, B1, C1 on `1, there exist points A2, B2, C2 on `2, leading to a Pappus configuration. We refer to it as a 3 + 0 version of the main result. First the two lines must be given, generally, each by two parameters from H, or, as we represent elements H by ordered pairs of elements of F, by four parameters from

F. To represent the three points on `1 requires three parameters from H, or six parameters from F. Hence, we already have counted ten parameters from F just to represent `1, `2 and the points A1, B1, C1. The ﬁrst coordinate in H from each point

28 A2, B2, C2 on `2 is considered as unknown, so we have six unknowns in F. Proving that points A3, B3, C3 are collinear, leads to showing that the "collinearity equation" with ten parameters and six unknowns in F has a solution. The collineation group of

AH can be used to reduce the number of parameters to as few as six in F in some cases. If we wish to establish a stronger 3 + 1 version of the main result, the number of parameters grows and the number of unknowns decreases, which makes it harder to determine if the corresponding equation has a solution.

And here is the main difficulty. As the multiplication in AH is given by the pair of "independent" rules, namely M1. and M2., it is very sensitive to the values of the original parameters. This sensitivity begins manifesting itself as early as when we try to write the equations of lines A1B2, A2B1, A1C2, A2C1, B1C2, and B2C1. Then it continues when we try to determine the coordinates of points A3, B3, C3 and check their collinearity. Also, some pairs of lines appearing during the solution can be parallel, and some lines can be vertical, type 1, or type 2. Due to all of this we experienced a very fast growth in the number of logical branchings in our programs, which was of the magnitude of thousands of cases, each coming with long lists of conditions on the parameters. When we could not describe all Pappus configurations for a given set of two lines and the sets of initial points on them, we tried to find only some of the configurations. This was done by specializing some of the unknowns in terms of the parameters or with values from F and then trying to solve a simplified collinearity equation for other unknowns. In the beginning of the research, there were trials and errors in finding the specializations. With experience, we were able to develop a variety of efficiencies in finding the appropriate specializations. Many examples of solving this problem with specializations will be provided in Chapter3.

29 Chapter 3

MAIN RESULTS

The following conjecture may be referred to as the 3 + 1 conjecture and is stated below. Based on the numerical evidence, we believe it to be true for all Hall planes of order greater than nine. Our goal in this work is to prove or get the best result concerning this conjecture.

Conjecture 3.0.1 (3 + 1 Conjecture). Given a Hall plane, for every pair of lines `1, `2 and every three points on `1, and one point on `2, two more points can be found on `2 so that the six points deﬁne a Pappus conﬁguration.

We succeeded in proving this conjecture in many cases, but not in all. In places where we could not prove the 3 + 1 Conjecture, we tried to prove a weaker conjecture which we refer to as the 3 + 0 Conjecture.

Conjecture 3.0.2 (3 + 0 Conjecture). Given a Hall plane, for every pair of lines `1, `2 and every three points on `1, three more points can be found on `2 so that the six points deﬁne a Pappus conﬁguration.

Let us consider the 3 + 1 Conjecture. It is clear that for fixed lines `1 and `2, 8 the number of ways of choosing three points on `1 and one point on `2 is ∼ q where q → ∞. It will become obvious from our arguments that the number of these choices when we cannot prove the 3 + 1 Conjecture is o(q8). Therefore, we will say the conjecture is proven asymptotically. Similar statements are true for other conjectures. There was a case where we fell short of proving even the 3 + 0 Conjecture. If a line is fixed, and we want to choose a point on it, it is sufficient to choose either the x-coordinate or the y-coordinate (in the case of a vertical line) of the point. Since

30 each coordinate of a point is an element of H, it has two components from F. If we ﬁx one of the components, we say that we have chosen 0.5 of a point. Hence, when we say we have chosen 2.5 points on a line, we mean two points and one component of the third point are chosen arbitrarily. This leads us to the following conjecture. Whether this is stronger or weaker or neither than the 3 + 0 conjecture is debatable. Nevertheless, here it is.

Conjecture 3.0.3 (2.5 + 1 Conjecture). Given a Hall plane, for every pair of lines `1, `2 and every 2.5 points on `1 and one point on `2, the second component of the 0.5 point on `1 and two more points on `2 can be found so that the six points deﬁne a Pappus conﬁguration.

Here are our results.

Theorem 3.0.1. The conjecture 3 + 1 holds asymptotically in the following cases.

1. Two lines `1, `2 are from BF.

2. Two lines `1, `2 are from NBF.

Theorem 3.0.2. The conjecture 3 + 0 holds if line `1 is from NBF and `2 is from BF.

Theorem 3.0.3. The conjecture 2.5 + 1 holds asymptotically if line `1 is from BF and `2 is from NBF.

Let us comment on the proofs of these theorems. The components of the points on lines `1 and `2 will be represented by parameters from {α, β, γ, δ, e, ζ, η, θ, π, φ} or unknowns from {e, g, h, j, t, v, w, z}. The parameters ρ, ς ∈ F will always denote the coefficients of the defining polynomial f (x) = x2 − ρx − ς of H. Let (κ, λ) ∈ H always denote the x-intercept of a vertical line or the y-intercept of a non-vertical line. Let (µ, ψ) ∈ H always denote the slope of a line. Figure 1.3 illustrates three cases of possible Pappus configurations in a plane. The names of points and lines in the Pappus configuration which we are trying to find in each case refer to those diagrams.

31 In all proofs of these theorems, we use the following logic. Using the basic subroutines of finding a line on two points and a point on two lines, see Section 5.2, we applied a variety of methods to determine unknowns in terms of our parameters that led to Pappus configurations. As mentioned before, these methods included using basic Mathematica commands Solve[] and Reduce[] which were not sufficient to directly determine the unknowns. We had to find particular solutions by specializing some unknowns to simple expressions in terms of parameters or to values from the basefield. After the values (numeric or symbolic) of the unknowns were determined, the case was essentially solved. What we present below are verifications that our unknowns were found correctly and the Pappus configurations exist. Sometimes the values of the unknowns are substituted at the very beginning of our verification and sometimes at later stages. The decision is made to minimize the length of the formulae in this presentation. As we demonstrate the computations leading to each Pappus configuration, we use a sequence of arbitrary fixed parameters. We must list any instances when a parameter cannot take on a certain value in the basefield or in relation to other parameters without causing division by zero in our computations or invalidating our assumptions about the line type we claim to be using. We call these instances "constraints" on the parameter sequence due to our solution. As we verify our solutions, we often encounter the following situations where it appears that we should place additional constraints on our parameter sequence or that our parameter sequence is not completely arbitrary.

• If we allow any of the points of line `1 or of line `2 to coincide, then the Pappus conﬁguration trivially exists. We call this a degenerate case. These parameter sequences automatically satisfy our conjecture, so we do not include them in our arguments.

• In some cases, we must verify that type 2 lines do not have slope in the baseﬁeld. Many times this veriﬁcation results in an equation with a solution involving a square root. Since that imposes a restriction on the values of the prime power q

32 for which the square root does not exist over F, we do not place any additional constraints on our set of valid parameter sequences. We will call this a Hall radical situation.

• If an expression involving only parameters and basefield elements is in a denominator, then we do not always add constraints on our parameter sequence to avoid dividing by zero. When solving the related equation to determine which parameter sequences to avoid (or which constraints to place on our parameter sequence) we can ignore solutions that indicate that a type 2 line has slope in the basefield or that a field element satisfies the defining polynomial of H since these are not possible in a Hall system, though they may be in Q.

• The expressions we work with can be viewed as polynomials or quotients of polynomials in a particular variable. After a solution in terms of parameters and values from the basefield is substituted into the unknowns in any of our verifications, if there are any "free" unknowns, we assume that they can take on any value in the basefield unless otherwise stated. The implication of this is powerful. For example, if our computations lead to an expression involving unknowns in a denominator, we may consider the expression to be a polynomial in that unknown. We may have to constrain the values that the unknown can take on (in terms of parameters and basefield elements) to avoid dividing by zero. However, since polynomials have finite degree, over a basefield of large enough order, there must exist elements that are not roots of the polynomial. Because of this, we do not constrain parameter sequences due to expressions that contain unknowns. We call this a free unknown situation. This is one reason it is easier to prove the 3 + 0 Conjecture than the 3 + 1 Conjecture. In each case of the proof, we will assume that we are working over a basefield of large enough order. To make it easy to distinguish between parameters and unknowns, we denote parameters with Greek letters and unknowns with Latin letters.

33 In each proof, we show the outcome of all computations small enough to ﬁt on a single page. To see outcomes of computations longer than a page, we refer the reader to the Mathematica code in Section 5.2.

3.1 Proof of Theorem 3.0.1 3.1.1 Case 1: BF/BF For case 1, we consider any pair of lines from BF. This pair can be intersecting or parallel.

3.1.1.1 Intersecting `1, `2 We construct the Pappus conﬁguration on the left in Figure 1.3 to prove the 3 + 1 Conjecture asymptotically. We will verify two solutions which are valid for all

(µ, 0) ∈ H (arbitrary slope of `1) and almost all parameter sequences (α, β, γ, δ, e, ζ, η, θ). The constraints on the parameter sequences covered by Solution 1 and Solution 2 are listed below.

1. α 6= γ and (γ, δ) 6= c(η, θ) and (e, ζ) 6= k(η, θ) for any c, k ∈ F

2. β 6= δ and (γ, δ) 6= c(η, θ) and (e, ζ) 6= k(η, θ) for any c, k ∈ F

These constraints were determined after we made the substitutions for t, v, w, z based on Solution 1 and Solution 2, respectively.

Proof. We begin our veriﬁcation by using Proposition 2.1.15 to ﬁx the lines `1 and

`2 to y = xµ for some µ ∈ F and x = 0, respectively. We will construct a Pappus conﬁguration using the following three distinct points on each line.

34 `1 : (y1, y2) = (x1, x2)(µ, 0)

A1 : ((α, β), (αµ, βµ))

B1 : ((γ, δ), (γµ, δµ))

C1 : ((e, ζ), (eµ, ζµ))

`2 : (x1, x2) = (0, 0)

A2 : ((0, 0), (η, θ))

B2 : ((0, 0), (t, v))

C2 : ((0, 0), (w, z))

First, we build type 2 lines between the appropriate points on `1 and `2 to check for the existence of a Pappus conﬁguration. The formulae for these lines are more complicated after expressing t, v, w, z in terms of the parameters, so those substitutions have not been made here.

• Line A1B2

t(v+(−µ+ρ)β)+α(µ2β−ςβ−µ(v+ρβ)) v2+(−2µ+ρ)vβ+(µ2−µρ−ς)β2 (y1, y2) = (x1, x2) −vα+tβ , −vα+tβ + (t, v)

We do not need to recalculate to obtain the formulae for the remaining lines

AiBj, AiCk, BjCk, i, j, k ∈ {1, 2}. We just permute the points from the line A1B2 [3, pp. 155-169].

• Line A2B1

◦ A1 7→ B1 =⇒ (α, β) 7→ (γ, δ)

◦ B2 7→ A2 =⇒ (t, v) 7→ (η, θ)

35 • Line A1C2

◦ B2 7→ C2 =⇒ (t, v) 7→ (w, z)

• Line A2C1

◦ A1 7→ C1 =⇒ (α, β) 7→ (e, ζ)

◦ B2 7→ A2 =⇒ (t, v) 7→ (η, θ, )

• Line B1C2

◦ A1 7→ B1 =⇒ (α, β) 7→ (γ, δ)

◦ B2 7→ C2 =⇒ (t, v) 7→ (w, z)

• Line B2C1

◦ A1 7→ C1 =⇒ (α, β) 7→ (e, ζ)

At this time, it is useful to substitute expressions of parameters into the unknowns. Since (α, β) = (γ, δ) is a degenerate situation, Solution 1 and Solution 2, together, apply to almost all parameter sequences of the form (α, β, γ, δ, e, ζ, η, θ) with the exceptions noted in constraints1 and2. Solution 1: For any parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints in1, we take:

t = t (t−η)(β−δ) v = θ + α−γ t(α−e)−η(γ−e) w = α−γ (t−η)(δ−ζ) z = v + α−γ

36 Solution 2: For any parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints2, we take:

(v−θ)(α−γ) t = η + β−δ v = v

(v−θ)(γ−e) w = t + β−δ v(β−ζ)−θ(δ−ζ) z = β−δ

Since at least one of t or v is free in each solution, in planes coordinatized over basefields of large enough order, any restrictions on t and v arising from the following computations will not impair one’s ability to construct a Pappus configuration for any parameter sequence not excluded by the constraints1 and2. Furthermore, those constraints are the only requirements on the parameter sequence to avoid zero in the denominator of any component of the slope or y-intercept of any of the lines A1B2, A2B1, A1C2, A2C1, B1C2 and B2C1 for either solution due to free unknown situations. There are no additional constraints on the parameter sequence to avoid zero in the second component of the slope of these type 2 lines due to Hall radical situations. We continue building the Pappus configuration. Note that the polynomial f is the defining polynomial of H.

• Point C3 (Solution 1)

((x1, x2), (y1, y2)) = (t−η)(tγ−αη) (t−η)(tδ−δη+(−α+γ)θ) f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , (µ(t−η)−(α−γ) f (µ))(tγ−αη) (µ(t−η)−(α−γ) f (µ))((t−η)δ−(α−γ)θ) f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2

37 • Point C3 (Solution 2)

((x1, x2), (y1, y2)) = (v−θ)(vγ−βη+δη−γθ) (v−θ)(vδ−βθ) (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , ( f (µ)(β−δ)+µ(θ−v))(γ(h−v)+(β−δ)η) − ( f (µ)(β−δ)+µ(θ−v))(vδ−βθ) (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v)

• Point B3 (Solution 1)

((x1, x2), (y1, y2)) = (t−η)(te−(α−γ+e) η) (t−η)(tζ−ζη+(−α+γ)h) f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , ((− f (µ)(α−γ)+µ(t−η))((t−η)e−(α−γ)η) (− f (µ)(α−γ)+µ(t−η))((t−η)ζ−(α−γ)h) f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2

• Point B3 (Solution 2)

((x1, x2), (y1, y2)) = (v−θ)(ve−βη+δη−eh) (v−θ)(vζ−(β−δ+ζ)θ) (h−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (− f (µ)(β−δ)+µ(v−θ))((v−θ)e−(β−δ)η) (−(µ2−µρ−ς)(β−δ)+µ(v−θ))(vζ−(β−δ+ζ)θ) (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v)

• Point A3 (Solution 1)

((x1, x2), (y1, y2)) = − (t−η)(t(α−γ−e)+eη) − (t−η)(t(β−δ−ζ)−βη+δη+ζη+αθ−γθ) f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , ( f (µ)(α−γ)−µ(t−η))(t(α−γ)−(t−η)e) , ( f (µ)(α−γ)−µ(t−η))((t−η)(β−δ−ζ)+(α−γ)θ) − (t−η)(t(β−δ−ζ)−βη+δη+ζη+αh−γh) f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2 , f (µ)(α−γ)2+(2µ−ρ)(α−γ)(η−t)+(η−t)2

38 • Point A3 (Solution 2)

((x1, x2), (y1, y2)) = − (v−θ)(v(α−γ−e)+βη−δη−αθ+γθ+eθ) − (v−θ)(v(β−δ−ζ)+ζθ) (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , ((v−θ)(α−γ−e)+(β−δ)g)( f (µ)(β−δ)−µ(v−θ)) (v(β−δ−ζ)+ζθ)( f (µ)(β−δ)−µ(v−θ)) (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v) , (θ−v)2+ f (µ)(β−δ)2+(β−δ)(2µ−ρ)(θ−v)

Notice that the expression in the denominator of each component of each point is quadratic in an unknown. Therefore, no constraints are added on the parameter sequences covered by these solutions due to the free unknown situation.

Finally, we ﬁnd the slopes of the lines A3B3 and A3C3 and verify that they are equal (here the lines are not vertical). This solution yields a type 1 Pappus line.

• The Pappus line: A3B3 = A3C3 (Solution 1)

µ(t−η)− f (µ)(α−γ) (y1, y2) = (x1, x2) t−η , 0 + (0, 0) for α 6= γ, t 6= η

• The Pappus line: A3B3 = A3C3 (Solution 2)

µ(v−θ)− f (µ)(β−δ) (y1, y2) = (x1, x2) v−h , 0 + (0, 0) for β 6= δ, v 6= θ

To compute the slope of the lines A3B3 and A3C3, we had to consider each of the 24 combinations of the following constraints. None lead to any additional constraints on the parameter sequence. They are provided here for the sake of completeness.

39 γ = α or γ 6= α, and

e = α or e 6= α, and

δ = β or δ 6= β, and

ζ = β or ζ 6= β

Of the sixteen possible sets of constraints, seven lead to degenerate Pappus conﬁgurations and nine work with Solution 1 or Solution 2. This is provided that

(η, θ) 6= (t, v) which must be true for the points A2 and B2 to be distinct. Notice that the parameter sequence in item 4 is the most general. The parameter sequences that are not covered by sequence in item 4 are covered by the remaining eight sequences or are among the seven degenerate sequences of parameters.

1. (α, β, γ, β, e, β) with γ 6= α, e 6= α, or

2. (α, β, γ, β, e, ζ) with γ 6= α, e 6= α, ζ 6= β, or

3. (α, β, γ, δ, e, β) with γ 6= α, e 6= α, δ 6= β, or

4. (α, β, γ, δ, e, ζ) with γ 6= α, e 6= α, δ 6= β, ζ 6= β, or

5. (α, β, γ, β, α, ζ) with γ 6= α, ζ 6= β, or

6. (α, β, γ, δ, α, ζ) with γ 6= α, δ 6= β, ζ 6= β, or

7. (α, β, α, δ, e, β) with e 6= α, δ 6= β, or

8. (α, β, α, δ, e, ζ) with e 6= α, δ 6= β, ζ 6= β, or

9. (α, β, α, δ, α, ζ) with δ 6= β, ζ 6= β

Thus, we have proven the 3 + 1 Conjecture asymptotically for any pair of intersecting lines from BF.

40 If our goal in this case was to prove the 3 + 0 Conjecture for every parameter sequence, then these solutions would be sufﬁcient. Indeed, for the 3 + 0 Conjecture, the parameters η and θ would be recast as unknowns and the constraints they impose

(γ, δ) 6= c(η, θ) and (e, ζ) 6= k(η, θ) for any c, k ∈ F would not be listed as constraints on the parameter sequence.

3.1.1.2 Parallel `1, `2 We construct the Pappus conﬁguration in the middle of Figure 1.3 to prove the 3 + 1 Conjecture asymptotically. This is done with one solution which is valid for all nonzero (κ, λ) ∈ H (arbitrary y-intercept of `1) and all possible parameter sequences (α, β, γ, δ, e, ζ, η, θ). The constraints on the parameter sequences for this solution are listed below.

1. θ 6∈ {δ, ζ, −β + δ + ζ}, and κθ 6∈ {κδ + λ(η − γ), κζ + λ(η − e), κ(−β + δ + ζ) + λ(α − γ − e + η)}

These constraints were determined after we made the substitutions for t, v, w, z based on the solution.

Proof. We begin our verification by using Proposition 2.1.15 to fix the lines `1 and `2 to y = (κ, λ) for some nonzero (κ, λ) ∈ H and y = 0, respectively. We will build a Pappus configuration using the following three distinct points on each line.

41 `1 : (y1, y2) = (κ, λ)

A1 : ((α, β), (κ, λ))

B1 : ((γ, δ), (κ, λ))

C1 : ((e, ζ), (κ, λ))

`2 : (y1, y2) = (0, 0)

A2 : ((η, θ), (0, 0))

B2 : ((t, v), (0, 0))

C2 : ((w, z), (0, 0))

At this time, it is convenient to substitute expressions of parameters into the unknowns. Solution: For the parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints in1, we take:

t = η + α − γ

v = θ + β − δ

w = t + γ − e

z = v + δ − ζ

We build parallel pairs of type 2 lines between the appropriate points on `1 and

`2 and compare their slopes to check for the existence of a Pappus configuration. For the six lines below, the constraints on the parameter sequences listed in1 are sufficient to ensure that the components of the denominators in the slope and y-intercept are nonzero and the slope is not from the basefield.

42 • Line A1B2 (parallel to Line A2B1)

2 2 ( ) = ( ) ς(γ−η)(δ−θ)−κ(λ+ρ(−δ+θ)) λ −ς(δ−θ) +λρ(−δ+θ) + y1, y2 x1, x2 λ(−γ+η)+κ(δ−θ) , λ(γ−η)+κ(−δ+θ) −κ2(β−δ+θ)+ς(γ−η)(β(γ−η)+α(−δ+θ))+κ(ρβ(γ−η)+λ(α−γ+η)+ρα(−δ+θ)) λ(−γ+η)+κ(δ−θ) , λ2(α−γ+η)+ς(δ−θ)(β(γ−η)+α(−δ+θ))−λ(ρβ(−γ+η)+ρα(δ−θ)+κ(β−δ+θ)) λ(−γ+η)+κ(δ−θ)

• Line A2B1 (parallel to Line A1B2)

2 2 ( ) = ( ) ς(γ−η)(δ−θ)−κ(λ+ρ(−δ+θ)) λ −ς(δ−θ) +λρ(−δ+θ) + y1, y2 x1, x2 λ(−γ+η)+κ(δ−θ) , λ(γ−η)+κ(−δ+θ) −κ2θ+ς(γ−η)(−δη+γθ)+κ(λη−ρδη+ργθ) −λ2η+ς(δ−θ)(δη−γθ)+λ(ρδη+κθ−ργθ) λ(−γ+η)+κ(δ−θ) , λ(γ−η)+κ(−δ+θ)

• Line A1C2 (parallel to Line A2C1)

2 2 ( ) = ( ) ς(e−η)(ζ−θ)−κ(λ+ρ(−ζ+θ)) λ −ς(ζ−θ) +λρ(−ζ+θ) + y1, y2 x1, x2 λ(−e+η)+κ(ζ−θ) , λ(e−η)+κ(−ζ+θ) −κ2(β−ζ+θ)+ς(e−η)(β(e−η)+α(−ζ+θ))+κ(ρβ(e−η)+λ(α−e+η)+ρα(−ζ+θ)) λ(−e+η)+κ(ζ−θ) , λ2(α−e+η)+ς(ζ−θ)(β(e−η)+α(−ζ+θ))−λ(ρβ(−e+η)+ρα(ζ−θ)+κ(β−ζ+θ)) λ(−e+η)+κ(ζ−θ)

• Line A2C1 (parallel to Line A1C2)

2 2 ( ) = ( ) ς(e−η)(ζ−θ)−κ(λ+ρ(−ζ+θ)) λ −ς(ζ−θ) +λρ(−ζ+θ) + y1, y2 x1, x2 λ(−e+η)+κ(ζ−θ) , λ(e−η)+κ(−ζ+θ) −κ2θ+ς(e−η)(−ζη+eθ)+κ(λη−ρζη+ρeθ) −λ2η+ς(ζ−θ)(ζη−eθ)+λ(ρζη+κθ−ρeθ) λ(−e+η)+κ(ζ−θ) , λ(e−η)+κ(−ζ+θ)

• Line B1C2 (parallel to Line B2C1)

2 2 ( ) = ( ) −ς(α−γ−e+η)(β−δ−ζ+θ)+κ(λ+ρ(β−δ−ζ+θ)) λ +λρ(β−δ−ζ+θ)−ς(β−δ−ζ+θ) + y1, y2 x1, x2 λ(−α+γ+e−η)+κ(β−δ−ζ+θ) , λ(−α+γ+e−η)+κ(β−δ−ζ+θ) κλ(α−e+η)−κ2(β−ζ+θ)+ς(α−γ−e+η)(−βγ+αδ−δe+γζ+δη−γθ)+κρ(βγ−αδ+δe−γζ−δη+γθ) λ(α−γ−e+η)+κ(−β+δ+ζ−θ) , −λ2(α−e+η)+κλ(β−ζ+θ)+λρ(−βγ+αδ−δe+γζ+δη−γθ)+ς(β−δ−ζ+θ)(βγ−αδ+δe−γζ−δη+γθ) λ(−α+γ+e−η)+κ(β−δ−ζ+θ)

43 • Line B2C1 (parallel to Line B1C2)

2 2 ( ) = ( ) −ς(α−γ−e+η)(β−δ−ζ+θ)+κ(λ+ρ(β−δ−ζ+θ)) λ +λρ(β−δ−ζ+θ)−ς(β−δ−ζ+θ) + y1, y2 x1, x2 λ(−α+γ+e−η)+κ(β−δ−ζ+θ) , λ(−α+γ+e−η)+κ(β−δ−ζ+θ) κλ(α−γ+η)−κ2(β−δ+θ)+ς(α−γ−e+η)(−βe+δe+αζ−γζ+ζη−eθ)+κρ(βe−δe−αζ+γζ−ζη+eθ) λ(α−γ−e+η)+κ(−β+δ+ζ−θ) , −λ2(α−γ+η)+κλ(β−δ+θ)+λρ(−βe+δe+αζ−γζ+ζη−eθ)+ς(β−δ−ζ+θ)(βe−δe−αζ+γζ−ζη+eθ) λ(−α+γ+e−η)+κ(β−δ−ζ+θ)

Hence, the Pappus conﬁguration exists, and we have proven the 3 + 1 Conjec- ture asymptotically for any pair of parallel lines from BF.

If our goal in this case was to prove the 3 + 0 Conjecture for every parameter sequence, then this solution would be sufﬁcient. Indeed, for the 3 + 0 Conjecture, the parameter θ would be recast as an unknown and the constraints imposed on it

θ 6∈ {δ, ζ, −β + δ + ζ}, and

κθ 6∈ {κδ + λ(η − γ), κζ + λ(η − e), κ(−β + δ + ζ) + λ(α − γ − e + η)}. would not be listed as constraints on the parameter sequence. They would be considered as a free unknown situation and ignored.

3.1.2 Case 2: NBF/NBF For Case 2, we consider any pair of lines from NBF. This pair can be intersecting or parallel.

3.1.2.1 Intersecting `1, `2 We construct the Pappus configuration on the left in Figure 1.3 to prove the 3 + 1 Conjecture asymptotically. This is done by the presentation of the first of two solutions. The second solution is not necessary for an asymptotic proof, but it is included since it satisfies most of the parameter sequences that the first solution does not. Together they are valid for all (µ, ψ) ∈ H with ψ 6= 0 (arbitrary slope of type 2 line `1) and almost all possible parameter sequences (α, β, γ, δ, e, ζ, η, θ). The

44 constraints on the parameter sequences covered by Solution 1 and Solution 2 are listed below.

1. βδ 6= 0

2. δ = ζ = 0

These constraints were determined after we ﬁxed point A2 and made the substitutions for t, v, w, z based on Solution 1 and Solution 2, respectively.

Proof. We begin by using Proposition 2.1.16 to ﬁx the lines `1 and `2 to y = x(µ, ψ) for some (µ, ψ) ∈ H with ψ 6= 0 and y = x(0, 1), respectively. We will build a Pappus conﬁguration using the following three distinct points on each line.

`1 : (y1, y2) = (x1, x2)(µ, ψ)

A1 : ((α, β), (αµ − f (µ)β/ψ, β(ρ − µ) + αψ))

B1 : ((γ, δ), (γµ − f (µ)δ/ψ, δ(ρ − µ) + γψ))

C1 : ((e, ζ), (eµ − f (µ)ζ/ψ, ζ(ρ − µ) + eψ))

`2 : (y1, y2) = (x1, x2)(0, 1)

A2 : ((η, θ), (θς, η + θρ))

B2 : ((t, v), (vς, t + vρ))

C2 : ((w, z), (zς, w + zρ))

For both solutions in this case of intersecting lines from NBF, we make use of

Corollary 2.1.8 and without loss of generality, fix the point A2 : ((0, 1), (ς, ρ)). The two solutions are presented consecutively since the formulae for the points A3, B3, C3 are too large to present for the first solution, but not the second. At this time, it is useful to substitute expressions of parameters and values from the basefield into the unknowns for Solution 1.

45 Solution 1: For any parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints in1, we take:

t = 0

v = v

w = w

z = 0

Since point A2 is ﬁxed with x-coordinate (η, θ) = (0, 1) and the lines `1 and `2 meet at the origin, if v = 0, 1 or w = 0, then there is a degenerate situation. We ignore these possibilities.

First we build type 2 lines between the points on `1 and `2.

• Line A1B2

− 3 2+ (− + ) + ( − )− ( + )+ 2 ( + ) (y y ) = (x x ) µ β ςvα 1 ψ ψ µςβ β vψ µαψ ρβ αψ µ β ρβ 2αψ 1, 2 1, 2 µ2(v−β)β+µρβ(−v+β)−µα(v−2β)ψ+ς(v−β)(−β+vψ)−αψ(ρ(−v+β)+αψ) , ψ(−ς(v−β)2+(µβ−αψ)(ρ(v−β)+µβ−αψ)) µρ(v−β)β+µ2β(−v+β)+µα(v−2β)ψ−ς(v−β)(−β+vψ)+αψ(ρ(−v+β)+αψ) 4 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + µ vβ −2µ ρvβ +µ ρ vβ −2µ ςvβ +2µρςvβ +ς vβ +µ ςv βψ−µρςv βψ−ς v βψ ··· −µ2vβψ+µρvβψ+ςvβψ+µ2β2ψ−µρβ2ψ−ςβ2ψ 3 2 2 2 2 2 2 2 ··· −2µ vαβψ+3µ ρvαβψ−µρ vαβψ+2µςvαβψ−ρςvαβψ+µ ςvβ ψ−µρςvβ ψ−ς vβ ψ ··· −ςv2ψ2+µvαψ2−ρvαψ2+ςvβψ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 ··· −µςv αψ +µ vα ψ −µρvα ψ −ςvα ψ +ς v βψ −2µςvαβψ +ρςvαβψ +ςvα ψ −2µαβψ2+ραβψ2+α2ψ3 , −v(µ3β2−ςvα(−1+ψ)ψ+µςβ(−β+vψ)+µαψ(ρβ+αψ)−µ2β(ρβ+2αψ)) µ2(v−β)β+µρβ(−v+β)−µα(v−2β)ψ+ς(v−β)(−β+vψ)−αψ(ρ(−v+β)+αψ)

46 • Line A2B1

3 2− (− + ) + ( + )− 2 ( + )+ (− + ) (y y ) = (x x ) µ δ ςγ 1 ψ ψ µγψ ρδ γψ µ δ ρδ 2γψ µςδ δ ψ 1, 2 1, 2 µ2δ(δ−1)+µρδ(−δ+1)+µγ(−2δ+1)ψ+γψ(ρ( δ−1)+γψ)−ς(δ−1)(δ−ψ) , ψ(µ2δ2−ς(δ−1)2+µδ(ρ(−δ+1)−2γψ)+γψ(ρδ−ρ+γψ)) µ2δ(δ−1)+µρδ(−δ+1)+µγ(−2δ+1)ψ+γψ(ρ( δ−1)+γψ)−ς(δ−1)(δ−ψ) 4 2 3 2 2 2 2 2 2 2 2 2 3 + µ δ −2µ ρδ 1+µ ρ δ −2µ ςδ +2µρςδ +ς δ −2µ γδψ ··· µ2δ2ψ−µρδ2ψ−ςδ2ψ−µ2δψ+µρδψ 2 2 2 2 2 ··· +3µ ργδψ−µρ γδψ+2µςγδψ−ρςγδψ+µ ςδ ψ−µρςδ ψ ··· +ςδψ−2µγδψ2+ργδψ2 2 2 2 2 2 2 2 2 2 2 2 ··· −ς δ ψ+µ ςδψ−µρςδψ−ς δψ+µ γ ψ −µργ ψ −ςγ ψ ··· +µγψ2−ργψ2+ςδψ2 2 2 2 2 2 2 3 ··· −2µςγδψ +ρςγδψ −µςγψ +ς δψ +ςγ ψ −ςψ2+γ2ψ3 , µ3δ2−ςγ(−1+ψ)ψ+µγψ(ρδ+γψ)−µ2δ(ρδ+2γψ)+µςδ(−δ+ψ) µ2δ(δ−1)+µρδ(−δ+1)+µγ(−2δ+1)ψ+γψ(ρ( δ−1)+γψ)−ς(δ−1)(δ−ψ)

• Line A1C2

3 2+ (− + )+ 2 ( − − )− ( 2+ ( − )+ ( − )) (y y ) = (x x ) µ β ςwβ 1 ψ µ β w ρβ 2αψ µ ςβ ρβ w αψ αψ w αψ 1, 2 1, 2 µ2β2−ςβ2+µβ(−ρβ+(w−2α)ψ)+(w−α)ψ(w−ρβ−αψ) , ψ(w2+µ2β2−ςβ2+ραβψ+α2ψ2+w(2µβ−ρβ−2αψ)−µβ(ρβ+2αψ)) µ2β2−ςβ2+µβ(−ρβ+(w−2α)ψ)+(w−α)ψ(w−ρβ−αψ) 3 2 2 2 + w(−µ β −ςwβ(−1+ψ)+µ β(−w+ρβ+2αψ)+µ(ςβ +ρβ(w−αψ)+αψ(w−αψ))) µ2β2−ςβ2+µβ(−ρβ+(w−2α)ψ)+(w−α)ψ(w−ρβ−αψ) , −w(µ2β2(−1+ψ)+(−1+ψ)(−ςβ2+α ψ(−w+ρβ+αψ))+µβ(ψ(w+2α−2αψ)+ρ(β−βψ))) µ2β2−ςβ2+µβ(−ρβ+(w−2α)ψ)+(w−α)ψ(w−ρβ−αψ)

47 • Line A2C1

3 2− (− + ) + ( + )− 2 ( + )+ (− + ) (y y ) = (x x ) µ ζ ςe 1 ψ ψ µeψ ρζ eψ µ ζ ρζ 2eψ µςζ ζ ψ 1, 2 1, 2 µ2ζ(ζ−1)+µρζ(−ζ+1)+µe(−2ζ+1)ψ+eψ(ρ( ζ−1)+eψ)−ς(ζ−1)(ζ−ψ) , ψ(µ2ζ2−ς(ζ−1)2+µζ(ρ(−ζ+1)−2eψ)+eψ(ρζ−ρ+eψ)) µ2ζ(ζ−1)+µρζ(−ζ+1)+µe(−2ζ+1)ψ+eψ(ρ( ζ−1)+eψ)−ς(ζ−1)(ζ−ψ) 4 2 3 2 2 2 2 2 2 2 2 2 3 2 + µ ζ −2µ ρζ +µ ρ ζ −2µ ςζ +2µρςζ +ς ζ −2µ eζψ+3µ ρeζψ ··· µ2ζ2ψ−µρζ2ψ−ςζ2ψ−µ2ζψ 2 2 2 2 2 2 ··· −µρ eζψ+2µςeζψ−ρςeζψ+µ ςζ ψ−µρςζ ψ−ς ζ ψ ··· +µρζψ+ςζψ−2µeζψ2 2 2 2 2 2 2 2 2 2 ··· +µ ςζψ−µρςζψ−ς ζψ+µ e ψ −µρe ψ −ςe ψ ··· +ρeζψ2+µeψ2−ρeψ2 2 2 2 2 2 2 3 ··· −2µςeζψ +ρςeζψ −µςeψ +ς ζψ +ςe ψ +ςζψ2−ςψ2+e2ψ3 , µ3ζ2−ςe(−1+ψ)ψ+µeψ(ρζ+eψ)−µ2ζ(ρζ+2eψ)+µςζ(−ζ+ψ) µ2ζ(ζ−1)+µρζ(−ζ+1)+µe(−2ζ+1)ψ+eψ(ρ(ζ−1)+eψ)−ς(ζ−1)(ζ−ψ)

• Line B1C2

3 2+ (− + )+ 2 ( − − )− ( 2+ ( − )+ ( − )) (y y ) = (x x ) µ δ ςwδ 1 ψ µ δ w ρδ 2γψ µ ςδ ρδ w γψ γψ w γ ψ 1, 2 1, 2 µ2δ2−ςδ2+µδ(−ρδ+(w−2γ)ψ)+(w−γ)ψ(w−ρδ−γψ) , ψ(w2+µ2δ2−ςδ2+ργδψ+γ2ψ2+w(2µδ−ρδ−2γψ)−µδ(ρδ+2γψ)) µ2δ2−ςδ2+µδ(−ρδ+(w−2γ)ψ)+(w−γ)ψ(w−ρδ−γψ) 3 2 2 2 + w(−µ δ −ςwδ(−1+ψ)+µ δ(−w+ρδ+2γψ)+µ(ςδ +ρδ(w−γψ)+γψ(w−γ ψ))) µ2δ2−ςδ2+µδ(−ρδ+(w−2γ)ψ)+(w−γ)ψ(w−ρδ−γψ) , −w(µ2δ2(−1+ψ)+(−1+ψ)(−ςδ2+ γψ(−w+ρδ+γψ))+µδ(ψ(w+2γ−2γψ)+ρ(δ−δψ))) µ2δ2−ςδ2+µδ(−ρδ+(w−2γ)ψ)+(w−γ)ψ(w−ρδ−γψ)

• Line B2C1

− 3 2+ (− + ) + ( − )− ( + )+ 2 ( + ) (y y ) = (x x ) µ ζ ςve 1 ψ ψ µςζ ζ vψ µeψ ρζ eψ µ ζ ρζ 2eψ 1, 2 1, 2 µ2(v−ζ)ζ+µρζ(−v+ζ)−µe(v−2ζ)ψ+ς(v−ζ)(−ζ+vψ)−eψ(ρ(−v+ζ)+eψ) , ψ(−ς(v−ζ)2+(µζ−eψ)(ρ(v− ζ)+µζ−eψ)) µρ(v−ζ)ζ+µ2ζ(−v+ζ)+µe(v−2ζ)ψ−ς(v−ζ)(−ζ+vψ)+eψ(ρ(−v+ζ)+eψ) 4 2 3 2 2 2 2 2 2 2 2 2 2 2 2 + µ vζ −2µ ρvζ +µ ρ vζ −2µ ςvζ +2µρςvζ +ς vζ +µ ςv ζψ−µρςv ζψ ··· −µ2vζψ+µρvζψ+ςvζψ+µ2ζ2ψ 2 2 3 2 2 2 2 2 ··· −ς v ζψ−2µ veζψ+3µ ρveζψ−µρ veζψ+2µςveζψ−ρςveζψ+µ ςvζ ψ−µρςvζ ψ ··· −µρζ2ψ−ςζ2ψ−ςv2ψ2+µveψ2−ρveψ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 ··· −ς vζ ψ−µςv eψ +µ ve ψ −µρve ψ −ςve ψ +ς v ζψ −2µςveζψ +ρςveζψ +ςve ψ +ςvζψ2−2µeζψ2+ρeζψ2+e2ψ3 , −v(µ3ζ2−ςve(−1+ψ)ψ+µςζ(−ζ+vψ)+µeψ(ρζ+eψ)−µ2ζ(ρζ+2eψ)) µ2(v−ζ)ζ+µρζ(−v+ζ)−µe(v−2ζ)ψ+ς(v−ζ)(−ζ+vψ)−eψ(ρ(−v+ζ)+eψ)

48 The components of the slope and y-intercept of the lines A1B2, A1C2, B1C2 and

B2C1 have nonzero denominators by virtue of the free unknown situation. Whereas, the components of the slope and y-intercept of the lines A2B1 and A2C1 have nonzero denominator and all six type 2 lines have slope not in the baseﬁeld due to the Hall radical situation.

Next, we ﬁnd the intersection points C3 = A1B2 ∧ A2B1, B3 = A1C2 ∧ A2C1, and A3 = B1C2 ∧ B2C1. They are too long to list here. There are no new constraints on the parameter sequences required to avoid zero in the denominators due to the free unknown situation. The Mathematica subroutines which take the lines listed above as inputs are provided in Section 5.2 and can be used to ﬁnd the points.

Finally, we compute the slope of the lines A3B3 and A3C3 and check if they are equal to determine if the Pappus line exists. This computation yields a slope not from the baseﬁeld for a type 2 Pappus line. The formula is also too long to include here. Due to the free unknown situation, there are no new constraints on the parameter sequence generated by this computation. The formula to calculate the slope (mAB1, mAB2) of A3B3 is provided below, and the associated code is provided in Section 5.2. The formula for the slope of A3C3 is analogous.

Let (xA1, xA2) ∈ H be the x-coordinate of point A3. The other coordinates will be denoted in a similar way.

ς(xA1−xB1)(xA2−xB2)+ρ(xA2−xB2)(yA1−yB1)−(yA1−yB1)(yA2−yB2) (mAB1, mAB2) = , (3.1) (xA2−xB2)(yA1−yB1)−(xA1−xB1)(yA2−yB2) 2 2 ς(xA2−xB2) +ρ(xA2−xB2)(yA2−yB2)−(yA2−yB2) (xA2−xB2)(yA1−yB1)−(xA1−xB1)(yA2−yB2)

As stated before the proof, Solution 1 is sufficient to prove the 3 + 1 Conjecture asymptotically. However, we will present Solution 2 in order to build a Pappus configuration using a large portion of the parameter sequences that were not included in the first solution.

49 Solution 2: For any parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints in2, we take:

t = 0

v = v

w = 0

z = z

Since point A2 is ﬁxed with x-coordinate (η, θ) = (0, 1) and the lines `1 and `2 meet at the origin, if v = 0, 1 or z = 0, 1, then there is a degenerate situation. We ignore these possibilities.

First we build type 2 lines between the points on `1 and `2.

• Line A1B2

− 3 2+ 2 2+ 2− − + 2 − + 2− 2 2 (y y ) = (x x ) µ β µ ρβ µςβ ςvαψ µςvβψ 2µ αβψ µραβψ ςvαψ µα ψ 1, 2 1, 2 µ2vβ−µρvβ−ςvβ−µ2β2+µρβ2+ςβ2+ςv2ψ−µvαψ+ρvαψ−ςvβψ+2µαβψ−ραβψ−α2ψ2 , ψ(−ςv2+µρvβ+2ςvβ+µ2β2−µρβ2−ςβ2−ρvαψ−2µαβψ+ραβψ+α2ψ2) −µ2vβ+µρvβ+ςvβ+µ2β2−µρβ2−ςβ2−ςv2ψ+µvαψ−ρvαψ+ςvβψ−2µαβψ+ραβψ+α2ψ2 4 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + µ vβ −2µ ρvβ +µ ρ vβ −2µ ςvβ +2µρςvβ +ς vβ +µ ςv βψ−µρςv βψ−ς v βψ ··· −µ2vβψ+µρvβψ+ςvβψ+µ2β2ψ−µρβ2ψ 3 2 2 2 2 2 2 2 ··· −2µ vαβψ+3µ ρvαβψ−µρ vαβψ+2µςvαβψ−ρςvαβψ+µ ςvβ ψ−µρςvβ ψ−ς vβ ψ ··· −ςβ2ψ−ςv2ψ2+µvαψ2−ρvαψ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 ··· −µςv αψ +µ vα ψ −µρvα ψ −ςvα ψ +ς v βψ −2µςvαβψ +ρςvαβψ +ςvα ψ +ςvβψ2−2µαβψ2+ραβψ2+α2ψ3 , −µ3vβ2+µ2ρvβ2+µςvβ2−ςv2αψ−µςv2βψ+2µ2vαβψ−µρvαβψ+ςv2αψ2−µvα2ψ2 µ2vβ−µρvβ−ςvβ−µ2β2+µρβ2+ςβ2+ςv2ψ−µvαψ+ρvαψ−ςvβψ+2µαβψ−ραβψ−α2ψ2

• Line A2B1

2 2 2 ( ) = ( ) −ςγ−µγ ψ+ςγψ −ς−ργψ+γ ψ y1, y2 x1, x2 −µγ+ργ+ς−γ2ψ , µγ−ργ−ς+γ2ψ 2 2 2 2 2 2 + µ γ −µργ −ςγ −µςγ+ςγ ψ −ςγ−µγ ψ+ςγψ µγ−ργ−ς+γ2ψ , −µγ+ργ+ς−γ2ψ

50 • Line A1C2

− 3 2+ 2 2+ 2− − + 2 − + 2− 2 2 (y y ) = (x x ) µ β µ ρβ µςβ ςzαψ µςzβψ 2µ αβψ µραβψ ςzαψ µα ψ 1, 2 1, 2 µ2zβ−µρzβ−ςzβ−µ2β2+µρβ2+ςβ2+ςz2ψ−µzαψ+ρzαψ−ςzβψ+2µαβψ−ραβψ−α2ψ2 , ψ(−ςz2+µρzβ+2ςzβ+µ2β2−µρβ2−ςβ2−ρzαψ−2µαβψ+ραβψ+α2ψ2) −µ2zβ+µρzβ+ςzβ+µ2β2−µρβ2−ςβ2−ςz2ψ+µzαψ−ρzαψ+ςzβψ−2µαβψ+ραβψ+α2ψ2 4 2 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 + µ zβ −2µ ρzβ +µ ρ zβ −2µ ςzβ +2µρςzβ +ς zβ +µ ςz βψ−µρςz βψ−ς z βψ ··· −µ2zβψ+µρzβψ+ςzβψ+µ2β2ψ−µρβ2ψ 3 2 2 2 2 2 2 2 ··· −2µ zαβψ+3µ ρzαβψ−µρ zαβψ+2µςzαβψ−ρςzαβψ+µ ςzβ ψ−µρςzβ ψ−ς zβ ψ ··· −ςβ2ψ−ςz2ψ2+µzαψ2−ρzαψ2+ςzβψ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 ··· −µςz αψ +µ zα ψ −µρzα ψ −ςzα ψ +ς z βψ −2µςzαβψ +ρςzαβψ +ςzα ψ −2µαβψ2+ραβψ2+α2ψ3 , −µ3zβ2+µ2ρzβ2+µςzβ2−ςz2αψ−µςz2βψ+2µ2zαβψ−µρzαβψ+ςz2αψ2−µzα2ψ2 µ2zβ−µρzβ−ςzβ−µ2β2+µρβ2+ςβ2+ςz2ψ−µzαψ+ρzαψ−ςzβψ+2µαβψ−ραβψ−α2ψ2

• Line A2C1

( ) = ( ) e(µeψ+ς(1−ψ)) ς+eψ(ρ−eψ) y1, y2 x1, x2 µe−ρe−ς+e2ψ , −µe+ρe+ς−e2ψ 2 + e(µ e−µ(ρe+ς)+ςe(−1+ψ)) e(µeψ+ς(1−ψ)) µe−ρe−ς+e2ψ , µe−ρe−ς+e2ψ

• Line B1C2

2 ( ) = ( ) γ(µγψ+ς(z−zψ)) ςz +γψ(ρz−γψ) y1, y2 x1, x2 −ςz2+γ(µz−ρz+γψ) , ςz2−γ(µz−ρz+γψ) 2 + zγ(µςz−µ γ+µργ+ςγ−ςγψ) zγ(µγψ+ς(z−zψ)) ςz2−γ(µz−ρz+γψ) , −ςz2+γ(µz−ρz+γψ)

• Line B2C1

2 ( ) = ( ) e(µeψ+ς(v−vψ)) ςv +eψ(ρv−eψ) y1, y2 x1, x2 −ςv2+e(µv−ρv+eψ) , ςv2−e(µv−ρv+eψ) 2 + ve(µςv−µ e+µρe+ςe−ςeψ) ve(µeψ+ς(v−vψ)) ςv2−e(µv−ρv+eψ) , −ςv2+e(µv−ρv+eψ)

Similar to the previous solution, the lines A1B2, A1C2, B1C2 and B2C1 have nonzero denominators by virtue of the free unknown situation. Whereas, the lines

A2B1 and A2C1 have nonzero denominator and all six type 2 lines have slope not in

51 the baseﬁeld due the Hall radical situation.

Next, we build the points C3 = A1B2 ∧ A2B1, B3 = A1C2 ∧ A2C1, and A3 =

B1C2 ∧ B2C1.

• Point C3

( − )(− 2 2+ ( − )+ ( + )− ( − + )) ((x x ) (y y )) = γ v 1 µ β ςβ β vψ µβ ρβ 2αψ αψ ρβ vγψ αψ 1, 2 , 1, 2 µ2β2−ςβ2+(vγ−α)ψ(−ρβ+vγψ−αψ)−µβ(ρβ−2vγψ+2αψ) , 2 2 2 2 2 2 µ vβ −µρvβ −ςvβ +µv βγψ−ρv βγψ−2µvαβψ+ρvαβψ ··· µ2β2−µρβ2−ςβ2+2µvβγψ−ρvβγψ−2µαβψ 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· +µvβγψ+v βγ ψ −v αγψ +v γ ψ −vβγ ψ +vα ψ −vαγψ +ραβψ+v2γ2ψ2−2vαγψ2+α2ψ2 , 3 2 2 2 2 2 2 2 2 2 3 2 −µ vβ γ+µ ρvβ γ+µςvβ γ+µ ςvβ −µρςvβ −ς vβ +µ β γ ··· µ2β2−µρβ2 2 2 2 2 2 2 2 2 2 2 2 ··· −µ ρβ γ−µςβ γ−µ v βγ ψ+µρv βγ ψ+ςv βγ ψ+µςv βγψ ··· −ςβ2+2µvβγψ −ρςv2βγψ+2µ2vαβγψ−µρvαβγψ+µ2vβγ2ψ−µρvβγ2ψ−ςvβγ2ψ ··· −ρvβγψ−2µαβψ ··· 2 2 2 2 ··· −2µςvαβψ+ρςvαβψ+µςvβγψ−2µ αβγψ+µραβγψ+µv αγ ψ ··· +ραβψ+v2γ2ψ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ςv αγψ −µvα γψ +ςv γ ψ −µvαγ ψ +ςvα ψ −ςvαγψ +µα γψ −2vαγψ2+α2ψ2 , 3 2 2 2 2 2 2 2 2 2 3 2 2 2 −µ vβ γ+µ ρvβ γ+µςvβ γ+µ ςvβ −µρςvβ −ς vβ +µ β γ−µ ρβ γ ··· µ2β2−µρβ2−ςβ2 −µςβ2γ−µ2v2βγ2ψ+µρv2βγ2ψ+ςv2βγ2ψ+µςv2βγψ−ρςv2βγψ+2µ2vαβγψ ··· +2µvβγψ ··· −µρvαβγψ+µ2vβγ2ψ−µρvβγ2ψ−ςvβγ2ψ−2µςvαβψ+ρςvαβψ+µςvβγψ ··· −ρvβγψ−2µαβψ+ραβψ ··· 2 2 2 2 2 2 2 2 2 2 2 ··· −2µ αβγψ+µραβγψ+µv αγ ψ −ςv αγψ −µvα γψ +ςv γ ψ ··· +v2γ2ψ2 2 2 2 2 2 2 2 ··· −µvαγ ψ +ςvα ψ −ςvαγψ +µα γψ −2vαγψ2+α2ψ2

52 • Point B3

( − )(− 2 2+ ( − )+ ( + )− ( − + )) ((x x ) (y y )) = e z 1 µ β ςβ β zψ µβ ρβ 2αψ αψ ρβ zeψ αψ 1, 2 , 1, 2 µ2β2−ςβ2+(ze−α)ψ(−ρβ+zeψ−αψ)−µβ(ρβ−2zeψ+2αψ) , 2 2 2 2 2 2 µ zβ −µρzβ −ςzβ +µz βeψ−ρz βeψ−2µzαβψ+ρzαβψ+µzβeψ ··· µ2β2−µρβ2−ςβ2+2µzβeψ−ρzβeψ +z2βe2ψ2−z2αeψ2+z2e2ψ2−zβe2ψ2+zα2ψ2−zαeψ2+ρzαβψ ··· −2µαβψ+ραβψ ··· 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· +µzβeψ+z βe ψ −z αeψ +z e ψ −zβe ψ +zα ψ −zαeψ +z2e2ψ2−2zαeψ2+α2ψ2 , 3 2 2 2 2 2 2 2 2 2 3 2 −µ zβ e+µ ρzβ e+µςzβ e+µ ςzβ −µρςzβ −ς zβ +µ β e ··· µ2β2−µρβ2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −µ ρβ e−µςβ e−µ z βe ψ+µρz βe ψ+ςz βe ψ+µςz βeψ−ρςz βeψ ··· −ςβ2+2µzβeψ +2µ2zαβeψ−µρzαβeψ+µ2zβe2ψ−µρzβe2ψ−ςzβe2ψ+µραβeψ ··· −ρzβeψ−2µαβψ ··· 2 2 2 2 2 2 ··· −2µςzαβψ+ρςzαβψ+µςzβeψ−2µ αβeψ+µz αe ψ −ςz αeψ ··· +ραβψ+z2e2ψ2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −µzα eψ +ςz e ψ −µzαe ψ +ςzα ψ −ςzαeψ +µα eψ −2zαeψ2+α2ψ2 , 2 2 2 2 2 2 2 2 2 2 2 µ ρzβ −µρ zβ −ρςzβ +µρz βeψ−ρ z βeψ−ςz βeψ−µ zβ eψ ··· µ2β2−µρβ2 2 2 2 ··· +µρzβ eψ+ςzβ eψ−2µρzαβψ+ρ zαβψ+µρzβeψ+ςzβeψ ··· −ςβ2+2µzβeψ +µ2β2eψ−µρβ2eψ−ςβ2eψ−µz2βe2ψ2+ρz2βe2ψ2−ρz2αeψ2 ··· −ρzβeψ−2µαβψ ··· 2 2 2 2 2 2 2 2 2 2 2 ··· +2µzαβeψ −ρzαβeψ +ρz e ψ +µzβe ψ −ρzβe ψ +ρzα ψ ··· +ραβψ+z2e2ψ2 2 2 2 2 2 3 2 3 2 3 2 3 ··· −ρzαeψ −2µαβeψ +ραβeψ +z αe ψ −zα eψ −zαe ψ +α eψ −2zαeψ2+α2ψ2

• Point A3

(v−z)γe vz(γ−e) ςvz(γ−e)+µ(v−z)γe ρvz(γ−e)+(v−z)γeψ ((x1, x2), (y1, y2)) = vγ−ze , vγ−ze , vγ−ze , vγ−ze

Notice that there are no constraints on the parameter sequences generated to avoid zero denominators in points A3, B3, and C3. This is due to the free unknown situation.

Finally, we compute the slope of the lines A3B3 and A3C3 and check if they are equal to determine if the Pappus line exists. This computation yields a slope not from

53 the baseﬁeld for a type 2 Pappus line. The formula is too long to include here. Due to the free unknown situation, there are no new constraints on the parameter sequence generated by this computation. To compute the slope of A3B3 one may use Formula 3.1 at the end of the previous construction. Alternatively, the Mathematica code in

Section 5.2 can be used to compute an equation of either A3B3 or A3C3 and verify that it contains the third point. Note that the constraints listed as2 above ensure that neither the denominator of any component of slope (or y-intercept), nor the second component of the slope of the type 2 Pappus line is zero due to this computation. Together, these two solutions prove the 3 + 1 Conjecture asymptotically for any pair of intersecting lines from NBF.

3.1.2.2 Parallel `1, `2 We construct the Pappus configuration on the left in Figure 1.3 for the first two of three solutions and the Pappus configuration in the middle of Figure 1.3 for third solution. Solution 1 is enough to prove the 3 + 1 Conjecture asymptotically. The second and third solutions are valid for almost all of the parameter sequences that the first solution does not cover. Together the solutions are valid for all (κ, λ) ∈ H

(arbitrary y-intercept of `1) and almost all parameter sequences (α, β, γ, δ, e, ζ, η, θ). The constraints on the parameter sequences covered by Solution 1, Solution 2, and Solution 3 are listed below.

δ−ρβδ+δe+ρβζ+ρδζ−eζ−ρζ2 β−ρβδ+βe+ρβζ+ρδζ−eζ−ρζ2 1. (α, γ) 6= ( δ−ζ , β−ζ ), α 6∈ {γ, 1 + e}, γ 6= 1 + e, and at least one of β, δ, ζ nonzero

2. β = δ = ζ = 0 and θ 6= 0

54 3. α = γ such that

2 2 2 2 6= κλγ−κ δ+λςγδ−κςδ −κλe+κγe+κρδe−2λςδe−κe +κ ζ+λςγζ−κρeζ β ς(λγ−κδ−λe+κζ) 2 + κςζ −κγη−κρδη+λςδη+κeη+κρζη−λςζη−λςγθ+κςδθ+λςeθ−κςζθ ς(λγ−κδ−λe+κζ) , and −κλ−λςζ+κη+λςθ β 6= ζ or e 6= κ , and κ2β+κλe−κρβe+λςβe+κe2−κ2ζ+κςβζ+κρeζ−κςζ2+κρβη−λςβη γ 6= κλ+κe+λςζ−κη−λςθ −κeη−κρζη+λςζη−κςβθ−λςeθ+κςζθ + κλ+κe+λςζ−κη−λςθ , and κδ−κζ+λη γ 6= λ or e 6= η, and (−κρ+λς)γ κ2+κρη−λςη−κςθ δ 6= − κς − κς , and −κ+ςθ e 6= η or ζ 6= ς

These constraints were determined after we made the substitutions for t, v, w, z based on Solution 1, Solution 2, and Solution 3, respectively. For Solution 1, we adjust which point on line `2 is parameterized and ﬁx it. The constraints reﬂect these adjustments.

Proof. We begin by using Proposition 2.1.16 to ﬁx lines `1 and `2 to y = x(0, 1) + (κ, λ) for some (κ, λ) ∈ H and y = x(0, 1), respectively. We will build a Pappus conﬁguration using the following three distinct points on each line.

55 `1 : (y1, y2) = (x1, x2)(0, 1) + (κ, λ)

A1 : ((α, β), (ςβ + κ, α + ρβ + λ))

B1 : ((γ, δ), (ςδ + κ, γ + ρδ + λ))

C1 : ((e, ζ), (ςζ + κ, e + ρζ + λ))

`2 : (y1, y2) = (x1, x2)(0, 1)

A2 : ((η, θ), (ςθ, η + ρθ))

B2 : ((t, v), (ςv, t + ρv))

C2 : ((w, z), (ςz, w + ρz))

The three solutions are presented consecutively, and only Solution 1 is presented with details.

Solution 1: When we labelled the points on lines `1 and `2, we did it in the usual way. Only for this solution, we will change which point on `2 has x- coordinates that are parameters and which has unknowns. We do for the sake of ﬁxing a parameterized point while maintaining the notation involving Greek and Latin letters. By mapping (η, θ) to (g, h) and (w, z) to (π, φ), we obtain:

`2 : (y1, y2) = (x1, x2)(0, 1)

A2 : ((g, h), (ςh, g + ρh))

B2 : ((t, v), (ςv, t + ρv))

C2 : ((π, φ), (ςφ, π + ρφ))

Now, we make use of Corollary 2.1.8 and without loss of generality, ﬁx the

56 point C2 : ((1, 0), (0, 1)). For the eight element parameter sequence (α, β, γ, δ, e, ζ, π, φ) satisfying all of the constraints in1, we take:

g = 0

h = −β + ζ

t = 0

v = −δ + ζ

Since line `1 contains the origin, a degenerate case occurs if β = ζ or δ = ζ.

Now, we build type 2 lines between the points on `1 and `2.

• Line A1B2

( ) = ( ) − −κ(λ+α)−λς(β+δ−ζ) y1, y2 x1, x2 λα+α2+ρα(β+δ−ζ)−(κ+ς(β+δ−ζ))(β+δ−ζ) , λ2+2λα+α2+λρ(β+δ−ζ)+ρα(β+δ−ζ)−ς(β+δ−ζ)2 λα+α2+ρα(β+δ−ζ)−(κ+ς(β+δ−ζ))(β+δ−ζ) 2 + − (κ −κρα+λςα+κς(β+δ−ζ))(δ−ζ) λα+α2+ρα(β+δ−ζ)−(κ+ς(β+δ−ζ))(β+δ−ζ) , − (κ(λ+α)+λς(β+δ−ζ))(δ−ζ) λα+α2+ρα(β+δ−ζ)−(κ+ς(β+δ−ζ))(β+δ−ζ)

• Line A2B1

( ) = ( ) − κ(λ+γ)+λς(β+δ−ζ) y1, y2 x1, x2 −γ(λ+γ+ρ(β+δ−ζ))+κ(β+δ−ζ)+ς(β+δ−ζ)2 , λ2+2λγ+γ2+λρ(β+δ−ζ)+ργ(β+δ−ζ)−ς(β+δ−ζ)2 γ(λ+γ+ρ(β+δ−ζ))−κ(β+δ−ζ)−ς(β+δ−ζ)2 2 + (κ −κργ+λςγκς(β+δ−ζ))(β−ζ) −γ(λ+γ+ρ(β+δ−ζ))+κ(β+δ−ζ)+ς(β+δ−ζ)2 , (κ(λ+γ)+λς(β+δ−ζ))(β−ζ) −γ(λ+γ+ρ(β+δ−ζ))+κ(β+δ−ζ)+ς(β+δ−ζ)2

57 • Line A1C2

(− + + )+ (y y ) = (x x ) − κ 1 λ α λςβ 1, 2 1, 2 −1+λ−λα−α2+κβ+ρβ+ςβ2+α(2−ρβ) , 1+λ2+α2−ρβ−ςβ2+α(−2+ρβ)+λ(−2+2α+ρβ) 1+λ(−1+α)+α2−κβ−ρβ−ςβ2+α(−2+ρβ) + κ(−1+λ+α)+λςβ −1+λ−λα−α2+κβ+ρβ+ςβ2+α(2−ρβ) , λ2+κβ+λ(−1+α+ρβ) −1+λ−λα−α2+κβ+ρβ+ςβ2+α(2−ρβ)

• Line A2C1

2 2 2 ( ) = ( ) λςβ+κ(λ+e) λ +λρβ−ςβ +2λe+ρβe+e y1, y2 x1, x2 −κβ−ςβ2+e(λ+ρβ+e) , −κβ−ςβ2+e(λ+ρβ+e) 2 + (κ +κςβ−κρe+λςe)(β−ζ) (λςβ+κ(λ+e))(β−ζ) κβ+ςβ2−e(λ+ρβ+e) , κβ+ςβ2−e(λ+ρβ+e)

• Line B1C2

(− + + )+ (y y ) = (x x ) − κ 1 λ γ λςδ 1, 2 1, 2 −1+λ−λγ−γ2+κδ+ρδ+ςδ2+γ(2−ρδ) , 1+λ2+γ2−ρδ−ςδ2+γ(−2+ρδ)+λ(−2+2γ+ρδ) 1+λ(−1+γ)+γ2−κδ−ρδ−ςδ2+γ(−2+ρδ) + κ(−1+λ+γ)+λςδ −1+λ−λγ−γ2+κδ+ρδ+ςδ2+γ(2−ρδ) , λ2+κδ+λ(−1+γ+ρδ) −1+λ−λγ−γ2+κδ+ρδ+ςδ2+γ(2−ρδ)

• Line B2C1

2 2 2 ( ) = ( ) λςδ+κ(λ+e) λ +λρδ−ςδ +2λe+ρδe+e y1, y2 x1, x2 −κδ−ςδ2+e(λ+ρδ+e) , −κδ−ςδ2+e(λ+ρδ+e) 2 + (κ +κςδ−κρe+λςe)(δ−ζ) (λςδ+κ(λ+e))(δ−ζ) κδ+ςδ2−e(λ+ρδ+e) , κδ+ςδ2−e(λ+ρδ+e)

These type 2 lines have slope not in the baseﬁeld and nonzero denominators due to the Hall radical situation.

58 Next, we ﬁnd the intersection points C3 = A1B2 ∧ A2B1, B3 = A1C2 ∧ A2C1, and A3 = B1C2 ∧ B2C1.

• Point C3

((x1, x2), (y1, y2)) = (β−δ)(κ+ς(β+δ−ζ)) −ρβ2+γδ+ρδ2+λ(−β+δ)+ρβζ−γζ−ρδζ+α(−β+ζ) − α−γ , α−γ , κρ(−β+δ)+ς(γ(δ−ζ)−ρ(β−δ)(β+δ−ζ)+α (−β+ζ)) α−γ , λρ(−β+δ)−ρ2(β−δ)(β+δ−ζ)−ς(β−δ)(β+δ−ζ)+ρ(−αβ+γδ+αζ− γζ) α−γ

• Point B3

((x1, x2), (y1, y2)) = e+κ(β−ζ)+ςβ(β−ζ) λβ+αβ+ρβ2+ζ−λζ−αζ−ρβζ 1−α+e , 1−α+e , κ+κρ(β−ζ)+ς(α(β−ζ)+ρβ(β−ζ)+ζ) λ+ςβ2+e+λρ(β−ζ)+ρ2β(β−ζ)−ςβζ+ρ(αβ+ζ−αζ) 1−α+e , 1−α+e−α+e

• Point A3

((x1, x2), (y1, y2)) = e+κ(δ−ζ)+ςδ(δ−ζ) λδ+γδ+ρδ2+ζ−λζ−γζ−ρδζ 1−γ+e , 1−γ+e , κ+κρ(δ−ζ)+ς(γ(δ−ζ)+ρδ(δ−ζ)+ζ) λ+ςδ2+e+λρ(δ−ζ)+ρ2δ(δ−ζ)−ςδζ+ρ(γδ+ζ−γζ) 1−γ+e , 1−γ+e

Due to the free unknown situation, there are no additional constraints on the parameter sequences required to ensure that the denominators of the components of points A3, B3, and C3 are nonzero.

Finally, we create the lines A3B3 and A3C3 and compare the slopes to determine if the Pappus line exists. This solution yields a type 2 Pappus line with a formula that is too long to include here. To compute the slope of A3B3 one may

59 use Formula 3.1 from the previous case. Alternatively, the Mathematica code in

Section 5.2 can be used to compute an equation of either A3B3 or A3C3 and verify that it contains the third point. Note that the constraints in1 ensure that neither the denominators of any component of slope (or y-intercept), nor the second component of the slope of the type 2 Pappus line is zero due to this computation. Though Solution 1 is sufﬁcient to prove the 3 + 1 Conjecture asymptotically, two additional solutions are provided to improve our result. We do not to provide details for these solutions.

We return to the standard notation for the points A2 and C2. Solution 2: For any parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints in2, we take:

t = t

v = θ

w = w

z = θ

This solution works with the construction of a type 2 Pappus line.

Solution 3: For any parameter sequence (α, β, γ, δ, e, ζ, η, θ) satisfying all of the constraints in3, we take:

t = η

v = β − δ + θ

w = γ − e + η

z = β − ζ + θ

60 This solution works with the construction of a non-generic Pappus conﬁgu- ration consisting of three parallel pairs of type 2 lines: A1B2 k A2B1, A1C2 k A2C1,

B1C2 k B2C1. Hence, the Pappus conﬁguration exists, and we have proven the 3 + 1 Conjec- ture asymptotically for any pair of parallel lines from NBF.

3.2 Proof of Theorem 3.0.2

For case 3a, we consider any pair of lines with `1 from NBF and `2 from BF. This pair of lines must intersect.

3.2.1 Case 3a: NBF/BF We construct the Pappus conﬁguration on the left in Figure 1.3 to prove the 3 + 0 Conjecture. We will verify two solutions which are valid for all parameter sequences (α, β, γ, δ, e, ζ) with no constraints.

Proof. We begin by using Proposition 2.1.17 to ﬁx the lines `1 and `2 to y = x(0, 1) and x = 0, respectively. We will build a Pappus conﬁguration using the following three distinct points on each line.

`1 : (y1, y2) = (x1, x2)(0, 1)

A1 : ((α, β), (βς, α + βρ))

B1 : ((γ, δ), (δς, γ + δρ))

C1 : ((e, ζ), (ζς, e + ζρ))

`2 : (x1, x2) = (0, 0)

A2 : ((0, 0), (g, h))

B2 : ((0, 0), (t, v))

C2 : ((0, 0), (w, z))

61 We build type 2 lines between the appropriate points on `1 and `2 to check for the existence of a Pappus conﬁguration. The substitution of the solutions for g, h, t, v, w, z is not implemented here so that we do not need to compute each line separately.

• Line A1B2

2 2 2 ( ) = ( ) t(v−α)−ςvβ v +α +ραβ−ςβ −v(2α+ρβ) + ( ) y1, y2 x1, x2 −vα+α2+ραβ+β(t−ςβ) , −vα+α2+ραβ+β(t−ςβ) t, v

First, we build type 2 lines between the appropriate points `1 and `2 to check for the existence of a Pappus conﬁguration. The formulae for these lines is presented before expressing the unknowns in terms of elements from the baseﬁeld and parameters.

• Line A2B1

◦ A1 7→ B1 =⇒ (α, β) 7→ (γ, δ)

◦ B2 7→ A2 =⇒ (t, v) 7→ (g, h)

We do not need to recalculate to obtain the equations of the remaining lines

AiBj, AiCk, BjCk, i, j, k ∈ {1, 2}. We just permute the points from the line A1B2 as was done in Section 3.1.1.1.

• Line A1C2

◦ B2 7→ C2 =⇒ (t, v) 7→ (w, z)

• Line A2C1

◦ A1 7→ C1 =⇒ (α, β) 7→ (e, ζ)

◦ B2 7→ A2 =⇒ (t, v) 7→ (g, h)

• Line B1C2

62 ◦ A1 7→ B1 =⇒ (α, β) 7→ (γ, δ)

◦ B2 7→ C2 =⇒ (t, v) 7→ (w, z)

• Line B2C1

◦ A1 7→ C1 =⇒ (α, β) 7→ (e, ζ)

At this time, it is useful to substitute expressions of parameters into the unknowns. Since `1 and `2 meet at the origin, (α, β) = (0, 0) is a degenerate case for which a Pappus conﬁguration trivially exists. Therefore, we assume that (α, β) 6= (0, 0). Solution 1: For any parameter sequence (α, β, γ, δ, e, ζ), we take:

g = g gβ h = α t = t tβ v = α w = w wβ z = α

Solution 2: For any parameter sequence (α, β, γ, δ, e, ζ), we take:

g = g gα h = β t = t

tα v = β w = w

wα z = β

63 After taking these substitutions into account, we determine that the lines

A1B2, A2B1, A1C2, A2C1, B1C2, and B2C1 have slope not in the baseﬁeld and that the components of their slope and y-intercept are nonzero due to both the free unknown situation and the Hall radical situation. So, their computation does not introduce any constraints on the parameter sequence. We continue building the Pappus conﬁguration.

• Point C3 (Solution 1)

((x1, x2), (y1, y2)) = 2 2 2 2 2 2 2 2 2 2 2 2 2 t αγ +ρt αγδ−ςt αδ −tα γg−t βγg−ρtαβγg+ςtβ γg−tαγ g+t αδg−ρtαγδg+ςtαδ g+α γg ··· t2γ2+ρt2γδ−ςt2δ2−2tαγg−ρtβγg−ρtαδg 2 2 2 2 2 ··· +tβγg +ραβγg −ςβ γg −tαδg +2ςtβδg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 3 2 2 2 t αβγ +ρt αβγδ−ςt αβδ −t β γg−tαβγ g−tα δg+t αβδg−ρtα βδg+ςtαβ δg−ρtαβγδg ··· t2αγ2+ρt2αγδ−ςt2αδ2−2tα2γg−ρtαβγg 2 2 2 3 2 2 2 2 2 2 ··· +ςtαβδ g+tβ γg +α δg −tαβδg +ρα βδg −ςαβ δg −ρtα2δg+2ςtαβδg+α3g2+ρα2βg2−ςαβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ςt βγ +ρςt βγδ−ς t βδ −t αγg−ρt βγg+t γ g−ςtβγ g−ςtα δg+ςt βδg−ρςtαβδg+ς tβ δg+ρt γδg ··· t2γ2+ρt2γδ−ςt2δ2−2tαγg−ρtβγg 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ρςtβγδg−ςt δ gς tβδ g+tα g +ρtαβg −ςtβ g −tαγg −ρtαδg +ςα δg +ςtβδg +ρςαβδg −ς β δg −ρtαδg+2ςtβδg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 t α γ +ρt αβγ +ρt α γδ+ρ t αβγδ−ςt α δ −ρςt αβδ −tα γg−t αβγg−ρtα βγg−ρt β γg+ςtαβ γg ··· t2αγ2+ρt2αγδ−ςt2αδ2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 ··· −tα γ g+t βγ g−ρtαβγ g−ρtα δg−ρ tα βδg+ςt β δg+ρςtαβ δg−ρtα γδg+ρt βγδg−ρ tαβγδg ··· −2tα2γg−ρtαβγg−ρtα2δg 2 2 2 2 2 2 2 2 2 3 2 3 2 2 2 2 ··· +ςtα δ g−ςt βδ g+ρςtαβδ gtα βg +ρtαβ g −ςtβ g +α γg −tαβγg +ρα βγg ··· +2ςtαβδg+α3g2 2 2 3 2 2 2 2 2 2 2 2 2 ··· −ςαβ γg +ρα δg −ρtαβδg +ρ α βδg +ςtβ δg −ρςαβ δg +ρα2βg2−ςαβ2g2

64 • Point C3 (Solution 2)

((x1, x2), (y1, y2)) = 2 2 2 2 2 2 2 2 2 t αγ +ρt αγδ−ςt αδ −tα γg−t βγg−ρtαβγg+ςtβ γg−tαγ g ··· t2γ2+ρt2γδ−ςt2δ2−2tαγg−ρtβγg 2 2 2 2 2 2 2 2 2 ··· +t αδg−ρtαγδg+ςtαδ g+α γg +tβγg +ραβγg −ςβ γg −tαδg −ρtαδg+2ςtβδg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 −t β γ −ρt β γδ+ςt β δ +t αβγg+tβ γ g−t α δg+tα βδg+ρtαβ δg−ςtβ δg ··· −t2βγ2−ρt2βγδ+ςt2βδ2+2tαβγg+ρtβ2γg 2 2 2 2 2 2 2 2 2 2 3 2 ··· +ρtβ γδg−ςtβ δ g−tαβγg +tα δg −α βδ g −ραβ δg +ςβ δg +ρtαβδg−2ςtβ2δg−α2βg2−ραβ2g2+ςβ3g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ςt βγ +ρςt βγδ−ς t βδ −t αγg−ρt βγg+t γ g−ςtβγ g−ςtα δg+ςt βδg−ρςtαβδg+ς tβ δg+ρt γδg ··· t2γ2+ρt2γδ−ςt2δ2−2tαγg 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ρςtβγδg−ςt δ g+ς tβδ g+tα g +ρtαβg −ςtβ g −tαγg −ρtαδg +ςα δg +ςtβδg +ρςαβδg −ς β δg −ρtβγg−ρtαδg+2ςtβδg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 −t αβγ −ρt β γ −ρt αβγδ−ρ t β γδ+ςt αβδ +ρςt β δ +t α γg+ρt αβγg+tα βγg ··· −t2βγ2−ρt2βγδ+ςt2βδ2 2 3 2 2 2 2 2 2 2 2 2 3 ··· +ρtαβ γg−ςtβ γg−t αγ g+tαβγ g+ρtβ γ g−ςt αβδg+ρtα βδg+ρ tαβ δg−ρςtβ δg ··· +2tαβγg+ρtβ2γg+ρtαβδg 2 2 2 2 2 2 2 2 3 2 2 2 2 2 ··· −ρt αγδg+ρtαβγδg+ρ tβ γδg+ςt αδ g−ςtαβδ g−ρςtβ δ g−tα g −ρtα βg +ςtαβ g ··· −2ςtβ2δg−α2βg2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 3 2 ··· +tα γg −α βγg −ραβ γg +ςβ γg +ρtα δg −ςtαβδg −ρα βδg −ρ αβ δg +ρςβ δg −ραβ2g2+ςβ3g2

65 • Point B3 (Solution 1)

((x1, x2), (y1, y2)) = 2 2 2 2 2 2 2 2 2 w αe +ρw αeζ−ςw αζ −wα eg−w βeg−ρwαβeg+ςwβ eg−wαe g ··· w2e2+ρw2eζ−ςw2ζ2−2wαeg−ρwβeg 2 2 2 2 2 2 2 2 2 ··· +w αζg−ρwαeζg+ςwαζ g+α eg +wβeg +ραβeg −ςβ eg −wαζg −ρwαζg+2ςwβζg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 3 2 2 w αβe +ρw αβeζ−ςw αβζ −w β eg−wαβe g−wα ζg+w αβζg−ρwα βζg ··· w2αe2+ρw2αeζ−ςw2αζ2−2wα2eg 2 2 2 2 2 2 2 2 3 2 2 ··· +w αβe +ρw αβeζ−ςw αβζ −w β eg−wαβe g−wα ζg+w αβζg−ρwα βζg −ρwαβeg−ρwα2ζg+2ςwαβζg+α3g2+ρα2βg2−ςαβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ςw βe +ρςw βeζ−ς w βζ −w αeg−ρw βeg+w e g−ςwβe g−ςwα ζg+ςw βζg−ρςwαβζg+ς wβ ζg+ρw eζg ··· w2e2+ρw2eζ−ςw2ζ2−2wαeg−ρwβeg 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· +ςw βe +ρςw βeζ−ς w βζ −w αeg−ρw βeg+w e g−ςwβe g−ςwα ζg+ςw βζg−ρςwαβζg+ς wβ ζg+ρw eζg −ρwαζg+2ςwβζg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 2 2 w α e +ρw αβe +ρw α eζ+ρ w αβeζ−ςw α ζ −ρςw αβζ −wα eg−w αβeg−ρwα βeg ··· w2αe2+ρw2αeζ−ςw2αζ2 2 2 2 2 2 2 2 2 3 2 2 2 2 2 ··· −ρw β eg+ςwαβ eg−wα e g+w βe g−ρwαβe g−ρwα ζg−ρ wα βζg+ςw β ζg+ρςwαβ ζg ··· −2wα2eg−ρwαβeg−ρwα2ζg 2 2 2 2 2 2 2 2 2 2 2 2 3 2 ··· −ρwα eζg+ρw βeζg−ρ wαβeζg+ςwα ζ g−ςw βζ g+ρςwαβζ g+wα βg +ρwαβ g −ςwβ g ··· +2ςwαβζg+α3g2 3 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 ··· +α eg −wαβeg +ρα βeg −ςαβ eg +ρα ζg −ρwαβζg +ρ α βζg +ςwβ ζg −ρςαβ ζg +ρα2βg2−ςαβ2g2

66 • Point B3 (Solution 2)

((x1, x2), (y1, y2)) = 2 2 2 2 2 2 2 2 2 w αe +ρw αeζ−ςw αζ −wα eg−w βeg−ρwαβeg+ςwβ eg−wαe g ··· w2e2+ρw2eζ−ςw2ζ2−2wαeg−ρwβeg 2 2 2 2 2 2 2 2 2 ··· +w αζg−ρwαeζg+ςwαζ g+α eg +wβeg +ραβeg −ςβ eg −wαζg −ρwαζg+2ςwβζg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 3 −w β e −ρw β eζ+ςw β ζ +w αβeg+wβ e g−w α ζg+wα βζg+ρwαβ ζg−ςwβ ζg ··· −w2βe2−ρw2βeζ+ςw2βζ2+2wαβeg 2 2 2 2 2 2 2 2 2 2 3 2 ··· +ρwβ eζg−ςwβ ζ g−wαβeg +wα ζg −α βζg −ραβ ζg +ςβ ζg +ρwβ2eg+ρwαβζg−2ςwβ2ζg−α2βg2−ραβ2g2+ςβ3g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ςw βe +ρςw βeζ−ς w βζ −w αeg−ρw βeg+w e g−ςwβe g−ςwα ζg+ςw βζg−ρςwαβζg+ς wβ ζg+ρw eζg ··· w2e2+ρw2eζ−ςw2ζ2−2wαeg−ρwβeg 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ρςwβeζg−ςw ζ g+ς wβζ g+wα g +ρwαβg −ςwβ g −wαeg −ρwαζg +ςα ζg +ςwβζg +ρςαβζg −ς β ζg −ρwαζg+2ςwβζg+α2g2+ραβg2−ςβ2g2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 −w αβe −ρw β e −ρw αβeζ−ρ w β eζ+ςw αβζ +ρςw β ζ +w α eg+ρw αβeg+wα βeg ··· −w2βe2−ρw2βeζ+ςw2βζ2 2 3 2 2 2 2 2 2 2 2 2 3 ··· +ρwαβ eg−ςwβ eg−w αe g+wαβe g+ρwβ e g−ςw αβζg+ρwα βζg+ρ wαβ ζg−ρςwβ ζg ··· +2wαβeg+ρwβ2eg 2 2 2 2 2 2 2 2 3 2 2 2 2 2 ··· −ρw αeζg+ρwαβeζg+ρ wβ eζg+ςw αζ g−ςwαβζ g−ρςwβ ζ g−wα g −ρwα βg +ςwαβ g ··· +ρwαβζg−2ςwβ2ζg−α2βg2 2 2 2 2 2 2 3 2 2 2 2 2 2 2 2 2 3 2 ··· +wα eg −α βe g −ραβ eg +ςβ eg +ρwα ζg −ςwαβζg −ρα βζg −ρ αβ ζg +ρςβ ζg −ραβ2g2+ςβ3g2

67 • Point A3 (Solution 1)

((x1, x2), (y1, y2)) = 2 2 2 2 2 2 2 2 2 2 2 2 2 t γ e−twγ e+t wδe−tw δe+ρt γδe−ρtwγδe−ςt δ e+ςtwδ e−twγe +w γe −t wγζ ··· t2γ2+ρt2γδ−ςt2δ2−2twγe−ρtwδe 2 2 2 2 2 ··· +tw γζ−ρtwγeζ+ρw γeζ+ςtwγζ −ςw γζ +w2e2−ρtwγζ+2ςtwδζ+ρw2eζ−ςw2ζ2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 t wβδe−tw βδe−twαδe +w αδe −t wβγζ+tw βγζ+t αγ ζ−twαγ ζ+ρt αγδζ−ρtwαγδζ−ςt αδ ζ ··· t2αγ2+ρt2αγδ−ςt2αδ2−2twαγe−ρtwαδe 2 2 2 2 2 ··· +ςtwαδ ζ−ρtwαδeζ+ρw αδeζ+ςtwαδζ −ςw αδζ +w2αe2−ρtwαγζ+2ςtwαδζ+ρw2αeζ−ςw2αζ2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 t wγ +ρt wγδ−ςt wδ −t wγe−tw γe−ρtw δe+tw e −ςtwδe +ςw δe −ρt wγζ+ςt γ ζ−ςtwγ ζ+ςt wδζ ··· t2γ2+ρt2γδ−ςt2δ2−2twγe−ρtwδe 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· +ςtw δζ+ρςt γδζ−ρςtwγδζ−ς t δ ζ+ς twδ ζ+ρtw eζ−ρςtwδeζ+ρςw δeζ−ςtw ζ +ς twδζ −ς w δζ +w2e2−ρtwγζ+2ςtwδζ+ρw2eζ−ςw2ζ2 , 2 2 2 2 2 2 2 2 2 2 2 2 t wβγ +ρt wβγδ−ςt wβδ −t wβγe−tw βγe+t αγ e−twαγ e−ρtw βδe+ρt αγδe−ρtwαγδe ··· t2αγ2+ρt2αγδ−ςt2αδ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ςt αδ e+ςtwαδ e+tw βe −twαγe +w αγe −ρtwαδe +ρw αδe −ρt wβγζ+ρt αγ ζ ··· −2twαγe−ρtwαδe+w2αe2 −ρtwαγ2ζ+ςt2wβδζ+ςtw2βδζ+ρ2t2αγδζ−ρ2twαγδζ−ρςt2αδ2ζ+ρςtwαδ2ζ+ρtw2βeζ ··· −ρtwαγζ+2ςtwαδζ ··· 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ρtwαγeζ+ρw αγeζ−ρ twαδeζ+ρ w αδeζ−ςtw βζ +ςtwαγζ −ςw αγζ +ρςtwαδζ −ρςw αδζ +ρw2αeζ−ςw2αζ2

68 • Point A3 (Solution 2)

((x1, x2), (y1, y2)) = 2 2 2 2 2 2 2 2 2 t γ e−twγ e+t wδe−tw δe+ρt γδe−ρtwγδe−ςt δ e+ςtwδ e ··· t2γ2+ρt2γδ−ςt2δ2−2twγe−ρtwδe+w2e2 2 2 2 2 2 2 2 2 2 ··· −twγe +w γe −t wγζ+tw γζ−ρtwγeζ+ρw γeζ+ςtwγζ −ςw γζ −ρtwγζ+2ςtwδζ+ρw2eζ−ςw2ζ2 , 2 2 2 2 2 2 2 2 2 2 t wαδe−tw αδe−twβδe +w βδe −t wαγζ+tw αγζ+t βγ ζ−twβγ ζ ··· t2βγ2+ρt2βγδ−ςt2βδ2−2twβγe−ρtwβδe 2 2 2 2 2 2 2 2 ··· +ρt βγδζ−ρtwβγδζ−ςt βδ ζ+ςtwβδ ζ−ρtwβδeζ+ρw βδeζ+ςtwβδζ −ςw βδζ +w2βe2−ρtwβγζ+2ςtwβδζ+ρw2βeζ−ςw2βζ2 , 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 t wγ +ρt wγδ−ςt wδ −t wγe−tw γe−ρtw δe+tw e −ςtwδe +ςw δe −ρt wγζ+ςt γ ζ−ςtwγ ζ+ςt wδζ ··· t2γ2+ρt2γδ−ςt2δ2−2twγe−ρtwδe 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· +ςtw δζ+ρςt γδζ−ρςtwγδζ−ς t δ ζ+ς twδ ζ+ρtw eζ−ρςtwδeζ+ρςw δeζ−ςtw ζ +ς twδζ −ς w δζ +w2e2−ρtwγζ+2ςtwδζ+ρw2eζ−ςw2ζ2 , 2 2 2 2 2 2 2 2 2 2 2 2 t wαγ +ρt wαγδ−ςt wαδ −t wαγe−tw αγe+t βγ e−twβγ e−ρtw αδe+ρt βγδe−ρtwβγδe ··· t2βγ2+ρt2βγδ−ςt2βδ2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ςt βδ e+ςtwβδ e+tw αe −twβγe +w βγe −ρtwβδe +ρw βδe −ρt wαγζ+ρt βγ ζ ··· −2twβγe−ρtwβδe+w2βe2 −ρtwβγ2ζ+ςt2wαδζ+ςtw2αδζ+ρ2t2βγδζ−ρ2twβγδζ−ρςt2βδ2ζ+ρςtwβδ2ζ+ρtw2αeζ ··· −ρtwβγζ+2ςtwβδζ ··· 2 2 2 2 2 2 2 2 2 2 2 2 ··· −ρtwβγeζ+ρw βγeζ−ρ twβδeζ+ρ w βδeζ−ςtw αζ +ςtwβγζ −ςw βγζ +ρςtwβδζ −ρςw βδζ +ρw2βeζ−ςw2βζ2

Observing the expression in each component of each denominator, we see that there are no new constraints on the parameter sequence due to the free unknown situation.

5.2 can be used to compute an equation of either A3B3 or A3C3 and verify that it contains the third point. Note that there are no new constraints on the parameter sequence due to the free unknown situation. Since there are no constraints involving only parameters for the pair of lines from the NBF and BF orbits, respectively, we have proven the 3 + 0 Conjecture for any pair of lines with `1 from NBF and `2 from BF.

69 3.3 Proof of Theorem 3.0.3

For case 3b, we consider any pair of lines with `1 from BF and `2 from NBF. This pair of lines must intersect.

3.3.1 Case3b: BF/NBF We construct the Pappus configuration in the middle of Figure 1.3 to prove the 2.5 + 1 Conjecture asymptotically. We will verify two solutions. The first solution is sufficient to prove the conjecture asymptotically, and the second solution is provided to improve our result. Solution 1 is valid for almost all parameter sequences (α, β, γ, δ, e, η, θ). The constraints on the parameter sequences covered by Solution 1 and Solution 2 are listed below.

1. α 6= γ such that

δη−η2−ρηθ+ςθ2 γ 6= θ , and 2 2 2 2 6= αδη−δeη−αη +γη −αeθ+γeθ−ραηθ+ργηθ+ςαθ −ςγθ β (γ−e)η , and e 6= ςθ or η 6= 0

βζ 2. β 6∈ {0, δ} and δ 6∈ {0, β+ζ } and ζ 6= 0

Proof. We begin by using Proposition 2.1.17 to ﬁx the lines `1 and `2 to x = 0 and y = x(0, 1), respectively. We will build a Pappus conﬁguration using the following three distinct points on each line.

70 `1 : (x1, x2) = (0, 0)

A1 : ((0, 0), (α, β))

B1 : ((0, 0), (γ, δ))

C1 : ((0, 0), (e, ζ))

`2 : (y1, y2) = (x1, x2)(0, 1)

A2 : ((η, θ), (θς, η + θρ))

B2 : ((t, v), (vς, t + vρ))

C2 : ((w, z), (zς, w + zρ))

At this time it is useful to substitute expressions of parmeters into the unknowns. Since (α, β) = (γ, δ) is a degenerate situation, Solution 1 and Solution 2 complement each other.

Solution 1: When labelling the points on lines `1 and `2 we did it in the usual way. Only for this solution, we will exchange ζ for j to denote the 0.5 point which is now an unknown in the point C1.

C1 = ((0, 0), (e, j))

For the seven element parameter sequence {α, β, γ, δ, e, η, θ} satisfying all of the constraints in1, we take:

71 −βγ+αδ+βe−δe j = α−γ

2 = βγ −αγδ+αγη+ρβγη−ςβδη−ςβγθ+ςαδθ t γ2+ργδ−ςδ2

2 = βγδ−αδ −βγη+αδη+αγθ+ραδθ−ςβδθ v γ2+ργδ−ςδ2

2 − 2 − 3 + 2 − 2+ 2 2+ 2 2− 2+ 2 − 2 2 − 2 + + 3 w = α βγe αβγ e α δe α γ δe αβγe βγ e α δe αγδe ςαβ γη ςβ γ η ςα βδη ςαβγδη α eη ··· ςβ2γ2−2ςαβγδ+ςα2δ2+ραβγe−2ςβ2γe−ρβγ2e−ρα2δe 2 2 2 2 2 2 2 2 ··· +ρα βeη−ςαβ eη−2α γeη−2ραβγeη+ςβ γeη+αγ eη+ρβγ eη+ςαβδeη−ςβγδeη−ςα βγθ+ςαβγ θ ··· +2ςαβδe+ραγδe+2ςβγδe−2ςαδ2e−α2e2−ραβe2+ςβ2e2+2αγe2 3 2 2 2 ··· +ςα δθ−ςα γδθ+ςαβγeθ−ςβγ eθ−ςα δeθ+ςαγδeθ +ρβγe2−γ2e2+ραδe2−2ςβδe2−ργδe2+ςδ2e2

2 2− 2 + 3 2− 2 − 2 2 + 2 + − 2 2 + 2 2− 2− 2+ 2 2+ 2 z = αβ γ 2α βγδ α δ αβ γe β γ e α βδe 3αβγδe 2α δ e β γe αβδe βγδe αδ e α βγη ··· ςβ2γ2−2ςαβγδ+ςα2δ2+ραβγe−2ςβ2γe−ρβγ2e−ρα2δe+2ςαβδe 2 3 2 2 2 2 2 2 2 2 3 ··· −αβγ η−α δη+α γδη−αβγeη+βγ eη+α δeη−αγδeη+ρα βγθ−ςαβ γθ−ραβγ θ+ςβ γ θ−ρα δθ ··· +ραγδe+2ςβγδe−2ςαδ2e−α2e2−ραβe2 2 2 3 2 2 2 2 2 2 ··· +ςα βδθ+ρα γδθ−ςαβγδθ−α eθ−ρα βeθ+ςαβ eθ+2α γeθ+ραβγeθ−ςβ γeθ−αγ eθ+ρα δeθ ··· +ςβ2e2+2αγe2+ρβγe2−γ2e2+ραδe2 ··· −ςαβδeθ−ραγδeθ+ςβγδeθ −2ςβδe2−ργδe2+ςδ2e2

We build parallel pairs of type 2 lines between the appropriate points on `1 and `2 and compare their slopes to check for the existence of a Pappus configuration. We must build six lines. For the four lines with formulae short enough to present here, the constraints on the parameter sequences listed in1 are sufficient to ensure that the components of the denominators in the slope and y-intercept are nonzero and the slope is not from the basefield. For the remaining two lines, there are no new constraints on the parameter sequences due to the Hall radical situation.

72 • Line A1B2 (parallel to Line A2B1)

2 2 2 ( ) = ( ) γ(δ−η)−ςδθ δ +η +ρηθ−ςθ −δ(2η+ρθ) + ( ) y1, y2 x1, x2 −δη+η2+ρηθ+θ(γ−ςθ) , −δη+η2+ρηθ+θ(γ−ςθ) α, β

• Line A2B1 (parallel to Line A1B2)

2 2 2 ( ) = ( ) γ(δ−η)−ςδθ δ +η +ρηθ−ςθ −δ(2η+ρθ) + ( ) y1, y2 x1, x2 −δη+η2+ρηθ+θ(γ−ςθ) , −δη+η2+ρηθ+θ(γ−ςθ) γ, δ

• Line A1C2 (parallel to Line A2C1)

2 2 ( ) = ( ) −βγe+αδe+βe −δe −αeη+γeη+ςβγθ−ςαδθ−ςβeθ+ςδeθ y1, y2 x1, x2 βγη−αδη−βeη+δeη+αη2−γη2+αeθ−γeθ+ραηθ−ργηθ−ςαθ2+ςγθ2 , 2 2 2 2 2 2 2 2 2 2 2 β γ −2αβγδ+α δ −2β γe+2αβδe+2βγδe−2αδ e+β e −2βδe +δ e ··· αβγη−βγ2η−α2δη+αγδη−αβeη+βγe η+αδeη 2 2 2 2 2 ··· +2αβγη−2βγ η−2α δη+2αγδη−2αβeη+2βγeη+2αδeη−2γδeη+α η −2αγη ··· −γδeη+α2η2−2αγη2+γ2η2+α2eθ−2αγeθ+γ2eθ+ρα2ηθ 2 2 2 2 2 ··· +γ η +ραβγθ−ρβγ θ−ρα δθ+ραγδθ−ραβeθ+ρβγeθ+ραδeθ−ργδeθ+ρα ηθ ··· −2ραγηθ+ργ2ηθ−ςα2θ2 2 2 2 2 2 2 ··· −2ραγηθ+ργ ηθ−ςα θ +2ςαγθ −ςγ θ + ( ) +2ςαγθ2−ςγ2θ2 α, β

• Line A2C1 (parallel to Line A1C2)

2 2 ( ) = ( ) −βγe+αδe+βe −δe −αeη+γeη+ςβγθ−ςαδθ−ςβeθ+ςδeθ y1, y2 x1, x2 βγη−αδη−βeη+δeη+αη2−γη2+αeθ−γeθ+ραηθ−ργηθ−ςαθ2+ςγθ2 , 2 2 2 2 2 2 2 2 2 2 2 β γ −2αβγδ+α δ −2β γe+2αβδe+2βγδe−2αδ e+β e −2βδe +δ e ··· αβγη−βγ2η−α2δη+αγδη−αβeη+βγeη+αδeη 2 2 2 2 2 ··· +2αβγη−2βγ η−2α δη+2αγδη−2αβeη+2βγeη+2αδeη−2γδeη+α η −2αγη ··· −γδeη+α2η2−2αγη2+γ2η2+α2eθ−2αγeθ+γ2eθ+ρα2ηθ 2 2 2 2 2 ··· +γ η +ραβγθ−ρβγ θ−ρα δθ+ραγδθ−ραβeθ+ρβγeθ+ραδeθ−ργδeθ+ρα ηθ ··· −2ραγηθ+ργ2ηθ−ςα2θ2 2 2 2 2 2 2 ··· −2ραγηθ+ργ ηθ−ςα θ +2ςαγθ −ςγ θ + −βγ+αδ+βe−δe +2ςαγθ2−ςγ2θ2 e, α−γ

The equations of lines B1C2 and B2C1 are too long to list here, but may be created using the Mathematica code in Section 5.2. Though this proves the 2.5 + 1 conjecture for this pair of lines and almost all

73 parameter sequences of the form (α, β, γ, δ, e, η, θ), we present another solution to strengthen the result. For Solution 2, we must revert back to the original deﬁnitions of the points

A1, B1, C1 and A2, B2, C2 at the beginning of this proof. This solution can be charac- terized as applying to a "2.5 + 0" Conjecture. Solution 2: We exchange Greek letters for Latin letters to denote the elements which are unknowns in the points C1 and A2.

C1 = ((0, 0), (e, ζ)) and A2 = ((g, h), (hς, g + hρ))

For the parameter sequence (α, β, γ, δ, ζ) satisfying all of the constraints in2, we take:

βγ−αδ+αζ−γζ e = β−δ

4 2 − 3 2+ 2 2 3+ 3 + 4 − 3 2 − 2 2 2 − 3 2 − 2 2 − 3 2 g = − β γ δ 2αβ γδ α β δ 2αβ γδζ ρβ γδζ β γ δζ 2α β δ ζ ραβ δ ζ αβ γδ ζ ρβ γδ ζ ··· −β4γ2+2αβ3γδ−α2β2 δ2−2αβ3γζ−ρβ4γζ+2β3γ2ζ+2α2β2δζ+ραβ3δζ+2ρβ3γδζ−2β2γ2δζ−2α2βδ2ζ 2 2 2 2 3 2 3 3 2 2 2 3 2 4 2 2 2 3 2 2 2 2 ··· +β γ δ ζ+2α βδ ζ+ραβ δ ζ−αβγ δ ζ+α β δζ +ραβ δζ −ςβ δζ −αβ γ δζ −ρβ γδζ −2α βδ ζ ··· −2ραβ2δ2ζ+2αβγδ2ζ−ρβ2γδ2ζ+ραβδ3ζ−α2β2 ζ2−ραβ3ζ2+ςβ4ζ2+2αβ2γζ2+ρβ3γζ2−β2γ2ζ2 2 2 2 3 2 2 2 2 2 2 2 2 3 2 3 2 2 3 2 3 2 3 2 ··· −2ραβ δ ζ +3ςβ δ ζ +2αβγδ ζ +2ρβ γδ ζ +α δ ζ +ραβδ ζ −3ςβ δ ζ −αγδ ζ −ρβγδ ζ ··· +2α2βδζ2+3ραβ2δζ2−4ςβ3δζ2−4αβγδζ2−3ρβ2γδζ2+2βγ2δζ2−α2δ2 ζ2−3ραβδ2ζ2+6ςβ2δ2ζ2+2αγδ2ζ2 4 2 ··· +ςβδ ζ +3ρβγδ2ζ2−γ2δ2 ζ2+ραδ3ζ2−4ςβδ3ζ2−ργδ3ζ2+ςδ4ζ2

3 2 − 2 3 h = − β γδ ζ αβ δ ζ ··· β4γ2−2αβ3γδ+α2β2 δ2+2αβ3γζ+ρβ4γζ−2β3γ2ζ−2α2β2δζ−ραβ3δζ−2ρβ3γδζ+2β2γ2δζ+2α2βδ2ζ 2 3 4 ··· −β γδ ζ+αβ δ ζ ··· +2ραβ2δ2ζ−2αβγδ2ζ+ρβ2γδ2ζ−ραβδ3ζ+α2β2 ζ2+ραβ3ζ2−ςβ4ζ2−2αβ2γζ2−ρβ3γζ2+β2γ2ζ2 3 2 2 2 2 2 2 2 ··· −β γδζ +α β δ ζ +2β γδ ζ ··· −2α2βδζ2−3ραβ2δζ2+4ςβ3δζ2+4αβγδζ2+3ρβ2γδζ2−2βγ2δζ2+α2δ2 ζ2+3ραβδ2ζ2−6ςβ2δ2ζ2 3 2 3 2 4 2 ··· −2αβδ ζ −βγ δ ζ +αδ ζ −2αγδ2ζ2−3ρβγδ2ζ2+γ2δ2 ζ2−ραδ3ζ2+4ςβδ3ζ2+ργδ3ζ2−ςδ4ζ2

−β2γ+αβδ−αβ ζ−ρβ2ζ+αδζ+ρβδζ t = −βγ+αδ

74 β2ζ−βδζ v = − βγ−αδ

− 3 2+ 3 2 2+ 2 2 3− 2 3+ 3 − 2 2 2 − 3 2 + 4 2 + 3 2 + 2 3 + 2 3 w = − αβ γδ β γ δ α β δ αβ γδ αβ γδζ 2α β δ ζ ραβ δ ζ ςβ δ ζ ρβ γδ ζ 2α βδ ζ ραβ δ ζ ··· −α2β2δ2−ραβ3δ2+ςβ4δ2+2αβ2γδ2+ρβ3γδ2−β2γ2δ2+ραβ2δ3−2ςβ3δ3−ρβ2γδ3+ςβ2δ4+2α2β2δζ+2ραβ3δζ 3 3 3 2 3 2 4 2 2 2 3 2 4 2 2 2 2 2 2 2 3 2 2 ··· −2ςβ δ ζ−αβγ δ ζ−ρβ γδ ζ+ςβ δ ζ+α β δ ζ +ραβ δζ −ςβ δζ −2α βδ ζ −2ραβ δ ζ +2ςβ δ ζ ··· −2ςβ4δζ−2αβ2γδζ−ρβ3γδζ−2α2βδ2ζ−3ραβ2δ2ζ+4ςβ3δ2ζ+2αβγδ2ζ+ρβ2γδ2ζ+ραβδ3ζ−2ςβ2δ3ζ 2 3 2 3 2 2 3 2 ··· +α δ ζ +ραβδ ζ −ςβ δ ζ −α2β2ζ2−ραβ3ζ2+ςβ4ζ2+2α2βδζ2+2ραβ2δζ2−2ςβ3δζ2−α2δ2ζ2−ραβδ2ζ2+ςβ2δ2ζ2

4 2− 3 3− 3 3+ 2 4 z = β γδ αβ δ β γδ αβ δ ··· −α2β2δ2−ραβ3δ2+ςβ4δ2+2αβ2γδ2+ρβ3γδ2−β2γ2δ2+ραβ2δ3−2ςβ3δ3−ρβ2γδ3+ςβ2δ4+2α2β2δζ+2ραβ3δζ 4 3 2 3 2 ··· −β γδζ+αβ δ ζ+2β γδ ζ ··· −2ςβ4δζ−2αβ2γδζ−ρβ3γδζ−2α2βδ2ζ−3ραβ2δ2ζ+4ςβ3δ2ζ+2αβγδ2ζ+ρβ2γδ2ζ+ραβδ3ζ−2ςβ2δ3ζ 2 3 2 3 4 ··· −2αβ δ ζ−β γ δ ζ+αβδ ζ −α2β2ζ2−ραβ3ζ2+ςβ4ζ2+2α2βδζ2+2ραβ2δζ2−2ςβ3δζ2−α2δ2ζ2−ραβδ2ζ2+ςβ2δ2ζ2

This solution works with the construction of a type 1 Pappus line. Hence, the Pappus conﬁguration exists, and we have proven the 2.5 + 1 Conjecture asymptotically for any pair of lines with the ﬁrst from BF and the second from NBF.

75 Chapter 4

SUPPLEMENT

4.1 Numerical Search for Pappus Configurations In this section, the word plane refers affine plane or projective plane. We began this project trying to find a statement as close to the Pappus Theorem as possible in some projective planes. Using the terminology from the previous chapters, the Pappus Theorem would be called the 3 + 3 Theorem. It’s clear that it holds only in classical planes. We began with a 3 + 2 Conjecture and checked it on all planes of order less than 25 and on some of order 25 using the data available on Moorhouse’s database of projective planes [21]. The 3 + 1 Conjecture and the 3 + 2 Conjecture were tested in Magma. Some planes were constructed using built-in commands in Magma, some were constructed by me algebraically, and many were downloaded from [21]. I checked by computer that the 3 + 2 Conjecture did not work in the following projective planes. Let us consider a counterexample of the 3 + 2 Conjecture in a projective plane π. Suppose it is formed by points A1, B1, C1 on line `1 and points A2, B2 on line `2. This means that picking a point C2 on `2 and not on `1 different from A2 and B2 will not lead to a Pappus configuration in π. Therefore, the points A3 = B1C2 ∧ B2C1, B3 =

A1C2 ∧ A2C1, and C3 = A1B2 ∧ A2B1 are not collinear. Let ` be a line in π which does not contain Ai, Bi, Ci for i ∈ {1, 2, 3}. Clearly, if the number of points in π is large enough, such a line ` exists. Therefore, the 3 + 2 Conjecture will not hold in the afﬁne plane π \ `, and the same points A1, B1, C1 and A2, B2 will give a counterexample. A selection of projective planes with their orders for which the 3 + 2 Conjecture failed:

76 • Hall: 9, 16, 25, 49

• Hughes: 9, 25, 49

• Dickson Near-ﬁeld: 49

As of now, we have not found a nonclassical plane where the 3 + 2 Conjecture holds. We turned our attention to the 3 + 1 Conjecture. From among the many projective planes we tested and for which the 3 + 1 Conjecture passed, a selection of planes with their orders is provided below.

• Hall: 16, 25

• Hughes: 25

• Czerwinski & Oakden: 25: a1, a6, b3, b6

• Rao: 25: a5, a7

4.2 Alternative Methods for Constructing Hall planes For the construction technique of spreads, we follow [24, pp. 121-123]. To construct a Hall plane on q4 points, we begin with a 2-dimensional vector space

V over F = Fq. A vector space V4 = {(x, y) : x, y ∈ V} can be represented as the direct sum of two copies of the vector space V so that each point of V4 may be written as an ordered pair of elements of V.A spread in V4 is a class of 2-dimensional vector subspaces of V4 called the components of the spread such that each nonzero vector belongs to exactly one component. Let the components of the spread and their images under translations be called "lines" and let the vectors be called "points". A spread can be constructed in such a way that {(x, y) : x = 0}, {(x, y) : y = 0}, and {(x, y) : y = x} are components of the spread. For each other component, there is some non-singular linear transformation M so that the component consists of the set of vectors {(x, y) : y = xM}. We shall use M as a name for the component as well as the linear transformation. This system of points and lines is a translation plane, see

77 Section 2.1.3. The construction of a Hall plane requires a particular definition for the 2 × 2 nonsingular matrix over F particular to each component. The algebraic system {V, +, ·}, where + is the standard vector addition and · is quasifield multiplication informs us about the matrix M. We begin by choosing a nonzero vector from V which we denote as 1. For every m ∈ V, the component {(x, y) : y = xM} has 1M = m and we define x · m = xM. Details about quasifield multiplication specific to Hall planes can be found in Section 2.1.1. For the construction technique of derivation, we follow [16, pp. 202-218]. The technique of derivation may be thought of as a special case of the construction technique of spreads. A partial spread of V4 is a set of 2 dimensional subspaces

W1, W2, ... , Wt such that Wi ∩ Wj = {0} for i 6= j. Thus, a spread is a partial spread such that each vector of V4 is in one of the subspaces Wi. The subspaces Wi are called the components of the partial spread. The points of a plane A (not necessarily a Hall plane) are the elements of V4, and the lines are the translates of the components of a spread. Each component of a spread may be thought of as deﬁning a parallel class of the plane A. If we construct a new translation plane A0 from A by removing the lines of some parallel class in A and replacing them with "new lines", then the spread corresponding to A0 will have certain components in common with the spread of A. This motivates the following deﬁnition; two partial spreads W1, W2, ... , Wt and U1, U2, ... , Ut are called replacements for each other if W1 ∪ W2 ∪ · · · ∪ Wt =

U1 ∪ U2 ∪ · · · ∪ Ut and Wi 6= Uj for any i or j. Either one of the partial spreads is said to be replaceable. Details about how to construct a Hall plane coordinatized by a Hall system with a particular defining polynomial by deriving a classical plane can be found in [16, pp. 214-215]. In 1960, Ostrom [26, pp. 457-458] suggested that one may consider the sets of points and lines of non-Desarguesian planes as configurations in classical planes of the same order. He described congruent partitions. It can also be read about in [16, pp. 160-164]. To construct a Hall plane, AH, we refer to our plane from the spread construction. Take G as the additive group of V4 and define the group of translations

78 ∼ T : V4 → V4, see Section 2.1.3. Clearly, the group T = G. Furthermore, G is also isomorphic to the translation group of a classical plane. A congruence of T is a set of subgroups of T each with order q2 such that every element of T other than the identity, belongs to exactly one of the subgroups in the congruence, and the sum of any two of these subgroups generates T. The points of AH are the elements of

T. The lines of AH are the subgroups of the congruence and their cosets in T. If the subgroups of the congruence of T are isomorphic to the components of the spread described in the ﬁrst paragraph of this section, then we have a Hall plane. André proved that every translation plane is equivalent to a congruence of a group T [26, p. 458].

4.3 Possible Future Work • Continue to try to prove a non-asymptotical version of the 3 + 1 Conjecture for Hall planes, and perhaps ﬁnd an easier solution to this problem.

• Check the 2 + 2 Conjecture for Hall planes.

• Extend this methodology to prove the strongest conditions for the existence of Pappus configurations in other infinite families of finite planes, such as those coordinatized by semifields.

• Extend this methodology to find the existence of other configurations in infinite families of finite planes.

• Look for an axiomatic argument that the Pappus configuration exists in every finite plane of sufficiently large order, similar to Ostrom’s argument for the Desargues configuration.

• And finally, study what has been accomplished and try to extend results for a problem posed by Erd˝osabout whether every finite configuration can be found in some finite plane, rephrased in a paper by Moorhouse and Williford [22, pp. 1-2].

79 Chapter 5

CODE SAMPLES

5.1 Magma Some code to count the number of Pappus conﬁgurations in plane P, on lines L1, L2 for an ordered triple of points A1, B1, C1 on line L1 and another ordered triple of points A2, B2, C2 on line L2.

//------Beginning of Pappus Configuration Test------// //create lines A1B2, A1C2, B1A2, B1C2, C1A2, C1B2 TF,line := IsCollinear(P,{A1,B2}); i := Index(line); A1B2 := L.i;

TF,line := IsCollinear(P,{A1,C2}); i := Index(line); A1C2 := L.i;

TF,line := IsCollinear(P,{B1,A2}); i := Index(line); B1A2 := L.i;

TF,line := IsCollinear(P,{B1,C2}); i := Index(line); B1C2 := L.i;

TF,line := IsCollinear(P,{C1,A2}); i := Index(line); C1A2 := L.i;

TF,line := IsCollinear(P,{C1,B2}); i := Index(line);

80 C1B2 := L.i;

//------find points A3, B3, C3------// TF,point := IsConcurrent(P,{B1C2,C1B2}); i := Index(point); A3 := V.i;

TF,point := IsConcurrent(P,{A1C2,C1A2}); i := Index(point); B3 := V.i;

TF,point := IsConcurrent(P,{A1B2,B1A2}); i := Index(point); C3 := V.i;

//------Test for existence of Pappus configuration------// TFPC := IsCollinear(P,{A3,B3,C3});

//---output details: which lns, pts, existence?---// print Index(L1)," : ", A1,B1,C1,"; "; print Index(L2),": ",A2, B2, C2, ";", TFPC;

5.2 Mathematica Some code to multiply in a Hall system, verify incidence, ﬁnd a line on two points, ﬁnd a point on two lines.

Hall system multiplication rules F[x_, r_, s_]:=x∧2 − rx − s

M1[{a_, b_}, {c_, d_}, {r_, s_}]:={ac, bc} (* Note : in this case d = 0 *) M2[{a_, b_}, {c_, d_}, {r_, s_}]:={ac − bd∧(−1)F[c, r, s], ad − bc + br}

verify incidence pointOnLineVert[{x1_, x2_, y1_, y2_}, {m1_, m2_, b1_, b2_}, {r_, s_}]:={x1 − b1, x2 − b2} pointOnLine1[{x1_, x2_, y1_, y2_}, {m1_, m2_, b1_, b2_}, {r_, s_}]:=M1[{x1, x2}, {m1, m2}, {r, s}] + {b1, b2} − {y1, y2} pointOnLine2[{x1_, x2_, y1_, y2_}, {m1_, m2_, b1_, b2_}, {r_, s_}]:=M2[{x1, x2}, {m1, m2}, {r, s}] + {b1, b2} − {y1, y2}

81 ﬁnd a type 1 line or a type 2 line on two points and ﬁnd the associated constraints t1OneLine[{x1_, x2_, y1_, y2_}, {x3_, x4_, y3_, y4_}, {x5_, x6_, y5_, y6_}, {r_, s_}]:= n o y1−y3 −x3y1+x1y3 + x2(−y1+y3) x1−x3 , 0, x1−x3 , y2 + x1−x3 ; t1OnebLineTests[{x1_, x2_, y1_, y2_}, {x3_, x4_, y3_, y4_}, {x5_, x6_, y5_, y6_}, {r_, s_}]:= 6= 6= y1−y3 == y2−y4 x1 6=6 x3&&x2 6=6 x4&& x1−x3 == x2−x4 ; t2TwoLine[{x1_, x2_, y1_, y2_}, {x3_, x4_, y3_, y4_}, {x5_, x6_, y5_, y6_}, {r_, s_}]:= {(s(x1 − x3)(x2 − x4) + r(x2 − x4)(y1 − y3) − (y1 − y3)(y2 − y4))/((x2 − x4)(y1 − y3) − (x1 − x3)(y2 − y4)), ((x2 − x4)∧2F[(y2 − y4)/(x2 − x4), r, s])/((x1 − x3)(y2 − y4) − (x2 − x4)(y1 − y3)), −s(x1 − x3)(−x2x3 + x1x4) + x4y12 − x3y1y2 + r(x2x3 − x1x4)(y1 − y3) − x2y1y3 − x4y1y3 + x1y2y3+ . x2y32 + x3y1y4 − x1y3y4 (x4(y1 − y3) + x2(−y1 + y3) + (x1 − x3)(y2 − y4)), s(x2 − x4)(x2x3 − x1x4) + x4y1y2 − x3y22 − x4y2y3 + r(x2x3 − x1x4)(y2 − y4) − x2y1y4 + x1y2y4+ . o x3y2y4 + x2y3y4 − x1y42 (x4(y1 − y3) + x2(−y1 + y3) + (x1 − x3)(y2 − y4)) ; t2TwoLineTests[{x1_, x2_, y1_, y2_}, {x3_, x4_, y3_, y4_}, {x5_, x6_, y5_, y6_}, {r_, s_}]:= (x2 − x4) 66=6 0&&(x2 − x4)(y1 − y3) 66=6 (x1 − x3)(y2 − y4);

ﬁnd a point on two type 1 lines or on two type 2 lines with constraints pointT1T1[{m1_, m2_, k1_, k2_}, {m3_, m4_, k3_, k4_}, {r_, s_}]:= n n o o { } { } − k1−k3 − k2−k4 − −k3m1+k1m3 − −k4m1+k2m3 { == == } {m1, m2, k1, k2}, {m3, m4, k3, k4}, − m1−m3 , − m1−m3 , − m1−m3 , − m1−m3 , {m2 == 0&&m4 == 0} pointT2T2[{m1_, m2_, k1_, k2_}, {m3_, m4_, k3_, k4_}, {r_, s_}]:= {{m1, m2, k1, k2}, {m3, m4, k3, k4}, n . −(k1 − k3)m2(m1 − m3)m4 + k4 m2 m32 − m3r − s + m4 −m12 + m1r + s + k2 m4 m12 − m1r − s + m2 −m32 + m3r + s m42 m12 − m1r − s + m22 m32 − m3r − s + m2m4(m3r + m1(−2m3 + r) + 2s) , ( ( − )+ (− + )−( − )( − )) m2 k2 m1 m3 k4 m1 m3 k1 k3 m2 m4 m4 , m42(m12−m1r−s)+m22(m32−m3r−s)+m2m4(m3r+m1(−2m3+r)+2s) k1m22m32 − k1m1m2m3m4 − k3m1m2m3m4 + k3m12m42 − k1m22m3r + k3m1m2m4r + k1m2m3m4r − k3m1m42r− k1m22s + k1m2m4s + k3m2m4s − k3m42s + k4 −m12m3m4 + m1 m3m4r + m2 m32 − m3r − s + m3m4s + . k2 m12m3m4 − m3m4s + m1 −m3m4r + m2 −m32 + m3r + s m42 m12 − m1r − s + m22 m32 − m3r − s + m2m4(m3r + m1(−2m3 + r) + 2s) , k4m2 m2 m32 − m3r − s + m4(−m1m3 + m3r + s) + . m4 (k1 − k3)m2(m2m3 − m1m4 − m2r + m4r) + k2 −m1m2m3 + m12m4 + m1m2r − m1m4r + m2s − m4s o o m42 m12 − m1r − s + m22 m32 − m3r − s + m2m4(m3r + m1(−2m3 + r) + 2s) , {m2 66=6 0&&m4 66=6 0}

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