Rational Distance Sets on Conic Sections

Megan D. Ly Shawn E. Tsosie Lyda P. Urresta Loyola Marymount UMASS Amherst Union College

July 2010

Abstract Leonhard Euler noted that there exists an inﬁnite set of rational points on the unit circle such that the pairwise distance of any two is also rational; the same statement is nearly always true for lines and other circles. In 2004, Garikai Campbell considered the question of a rational distance set consisting of four points on a parabola. We introduce new ideas to discuss a rational distance set of four points on a hyperbola. We will also discuss the issues with generalizing to a rational distance set of ﬁve points on an arbitrary conic section.

1 Introduction

A rational distance set is a set whose elements are points in R2 which have pairwise rational distance. Finding these sets is a difficult problem, and the search for rational distance sets with rational coordinates is even more difficult. The search for rational distance sets has a long history. Leonard Euler noted that there exists an infinite set of rational points on the unit circle such that the pairwise distance of any two is also rational. In 1945 Stanislaw Ulam posed a question about a rational distance set: is there an everywhere dense rational distance set in the plane [2]? Paul Erdösalso considered the problem and he conjectured that the only irreducible algebraic curves which have an infinite rational distance set are the line and the circle [4]. This was later proven to be true by Jozsef Solymosi and Frank de Zeeuw [4]. First, we provide the formal definition of a rational distance set: 2 Definition 1.1. A rational distance set S is a set of elements Pi = (xi, yi) ∈ R , 1 ≤ i ≤ n, such that for all Pi,Pj ∈ S, ||Pi − Pj|| is a rational number. We are primarily concerned with finding rational distance sets of rational points. p 2 2 The distance between two points on a graph is (xi − xj) + (yi − yj) . It will prove useful to rewrite this formula as s 2 yi − yj |xi − xj| 1 + (1) xi − xj Now, consider the following lemma:

1 Lemma 1.2. The rational solutions to the equation

α2 = 1 + β2 (2)

m2 + 1 m2 − 1 can be parameterized by α = , β = , for m ∈ , m 6= 0. 2m 2m Q For the proof, refer to Campbell [1, Lemma 2.1].

2 Main results

2.1 Line

Theorem 2.1. For any line L: ax + by + c = 0, for a, b, c ∈ Q, there exists a dense rational distance set with rational coordinates if and only if there exists m ∈ Q such that a2 + b2 = m2.

Proof. Suppose L has a dense rational distance set with rational coordinates. Then a c we can express the line L as y = − b x − b . Consider two points on L,(xi, yi) and (xj, yj). Using Lemma 1.2, we can write the distance between these two points as r a2 |x − x | 1 + . (3) i j b a2 2 2 2 2 2 The distance is rational only if 1 + b2 = n for n ∈ Q. That is, if a + b = b n . Hence, we may set m = bn. a c Suppose a2 +b2 = m2 for m ∈ Q. Then line L can be written as y = − x− . The bq b a 2 distance between two rational points on L,(xi, yi) and (xj, yj), is |xi − xj| 1 + b . 2 2 2 a2 m2 Note that from Lemma 1.2 we can write a + b = m as 1 + b2 = b2 . Hence, we see q a2 m 1 + b2 = b , which is rational.

2.2 Circle To show that there exists a rational distance set of rational points on a circle, we identify the cartesian plane R2 with the complex plane C.

Theorem 2.2. Let r ∈ Q with r 6= 0 and let C : kz − z0k = r be a circle in the complex plane with z0 = x+iy, x, y ∈ Q. Then, there exists a dense rational distance set on C consisting of rational points.

2 1 Sketch of Proof. Consider the vertical line L: Re z = 2r . Let f : C → C be the Möbiustransformation defined by (z − r)z − 1 f(z) = 0 . z Then f maps the line L into the circle C. The following figure shows this for a circle with radius r < 1 centered at z0.

A dense rational distance set of rational points on L is given by

1 s2 − 1 S = z = 1 + i s ∈ , s 6= 0 . 2r 2s Q

Notice that Lemma 1.2 implies that each z ∈ S has rational norm kzk. The codomain f(S) = {f(z) z ∈ S} will then be a dense set of points on C with rational coordinates. Finally notice that f is a translation composed with the map z 7→ −1/z. Then, the rationality of the distance between the points on the codomain follows from the identity

1 1 kz − wk − = . z w kzk · kwk

2.3 Rational Distance Set of Three Points on a Parabola

Consider the parabola y = ax2 + bx + c where a, b, c ∈ Q. We provide a geometric intuition of the problem. An important note for this construction is that while the points are at rational distance, the points themselves do not have to be rational. We present a method for constructing a rational distance set of three points on a parabola. Note that we can ﬁnd a rational distance set using

3 (a) Two concentric circles with (b) Moving the construction rational radii are centered at along the parabola, we note some point on the parabola. that the length of segment P2P3 changes

Figure 1: While P1P2 and P1P3 are rational, we cannot guarantee that P2P3 is also rational. Since the length of P2P3 changes, the segment P2P3 must have a rational length at some point on the parabola. this method on any conic section. First, we choose some point on the parabola and construct two circles of rational radii around it, as shown in Figure 1. In Figure 1, we see that the line segments P1P2 and P1P3 are rational. All we require now is that P2P3 be rational. This segment is not necessarily rational. We can, however, make it rational. If we allow the construction to move along the parabola, we notice that the length of segment P2P3 changes. We see from this construction that the length of segment P2P3 is a continuous function on an interval. Therefore, we can find some point in the parabola where the segment P2P3 is rational. Now we turn to the more difficult problem of finding rational distance sets of rational points on the parabola. By equation 1, the distance dij between two rational points Pi and Pj on a parabola is q 2 dij = |xi − xj| 1 + (axi + axj + b) .

2 mij −1 1 m2−1 By Lemma 1.2, axi + axj + b = . We let g(m) = ( − b), which leads us to 2mij a 2m the following theorem:

2 Theorem 2.3. Given a parabola y = ax +bx+c, let S = {P1,P2,P3}. S is a rational distance set of rational points if and only if 1 ≤ i < j ≤ 3 we can ﬁnd a nonzero

4 rational value mij such that

g(m ) + g(m ) − g(m ) x = 12 13 23 1 2 g(m ) − g(m ) + g(m ) x = 12 13 23 2 2 −g(m ) + g(m ) + g(m ) x = 12 13 23 3 2

Proof. For each 1 ≤ i < j ≤ 3 where xi + xj = g(mij), we have three equations:

x1 + x2 = g(m12)

x1 + x3 = g(m13)

x2 + x3 = g(m23)

This is equivalent to the following row reduced augmented matrix:

 g(m12)+g(m13)−g(m23)  1 0 0 2  0 1 0 g(m12)−g(m13)+g(m23)   2  −g(m12)+g(m13)+g(m23) 0 0 1 2 which gives us our desired values.

2.4 Rational Distance Set of Four Points on a Parabola We can follow a similar argument to ﬁnd a rational distance set of four rational points on a parabola.

2 Theorem 2.4. Given a parabola y = ax + bx + c, let S = {P1,P2,P3,P4}. S is a rational distance set of rational points if and only if there are rational values mij, 1 ≤ i < j ≤ 4 such that 1 x = (g(m ) + g(m ) − g(m )) 1 2 12 13 23 1 x = (g(m ) − g(m ) + g(m )) 2 2 12 13 23 1 x = (−g(m ) + g(m ) + g(m )) 3 2 12 13 23 1 x = (−g(m ) − g(m ) + g(m ) + 2g(m )) 4 2 12 13 23 14

and g(m13) + g(m24) = g(m23) + g(m14) = g(m12) + g(m34)

5 Proof. For each 1 ≤ i < j ≤ 4 where xi + xj = g(mij), we have six equations:

x1 + x2 = g(m12)

x1 + x3 = g(m13)

x1 + x4 = g(m14)

x2 + x3 = g(m23)

x2 + x4 = g(m24)

x3 + x4 = g(m34)

This is equivalent to the following row reduced augmented matrix:

 g(m12)+g(m13)−g(m23)  1 0 0 0 2  0 1 0 0 g(m12)−g(m13)+g(m23)   2   −g(m12)+g(m13)+g(m23)   0 0 1 0 2   −g(m12)−g(m13)+g(m23)+2g(m14)   0 0 0 1   2   0 0 0 0 g(m13) + g(m24) − g(m23) − g(m14)  0 0 0 0 g(m12) + g(m34) − g(m23) − g(m14) which gives us our desired equations. We know that any three points lie on a common circle. However, in the case of four points, this does not always occur. Deﬁnition 2.5. A set of points is concyclic if they lie on a common circle. Similarly, a set of points is nonconcyclic if one cannot construct a common circle through them. We will now examine the cases of concyclic and nonconcyclic points on a parabola.

2.5 Concyclic Points on a Parabola

2 Proposition 2.6. Let Pi = (xi, axi + bxi + c) for 1 ≤ i ≤ 4 be four rational points. 2b Then they are concyclic if and only if x1 + x2 + x3 + x4 = − a . 2 2 2 Proof. Let Cα,β,ρ be the circle (x − α) + (y − β) = ρ , where α, β, ρ ∈ R. This circle intersects the parabola y = ax2 + bx + c at the points whose x-coordinates are the roots of (x − α)2 + (ax2 + bx + c − β)2 − ρ2 = a2x4 + 2abx3 + (2a(c − β) + b2 + 1)x2 + 2b(c − β)x + (c − β)2 + α2 − ρ2 2 = a (x − x1)(x − x2)(x − x3)(x − x4) 3 2 Since the coeﬃcient of x is 2ab, then −a (x1 + x2 + x3 + x4) = 2ab. Thus, x1 + x2 + 2b x3 + x4 = − a . 2b Conversely, if we have xi, 1 ≤ i ≤ 4 such that x1 + x2 + x3 + x4 = − a , we can solve: 2 a (x − x1)(x − x2)(x − x3)(x − x4) =

6 a2x4 + 2abx3 + (2a(c − β) + b2 + 1)x2 + 2b(c − β)x + (c − β)2 + α2 − ρ2 for some α, β, and ρ.

Theorem 2.7. Suppose that P1,P2,P3, and P4 are rational and concyclic points on the parabola y = ax2 + bx + c. Then, these points are at rational distance if and only if there are nonzero rational values m12, m13, and m23 such that the equations 1 of Theorem 2.4 hold. If we have this condition, then x4 = − 2 (g(m12) + g(m13) + 4b g(m23) + a ). 2b Proof. By Proposition 2.6, x4 = −(x1 + x2 + x3 + a ). Therefore, we can directly solve for x4 using the values given for the x1, x2, x3. Inputting our values, we get 1 4b x4 = − 2 (g(m12) + g(m13) + g(m23) + a ).

2.6 Non-Concyclic Points on a Parabola We now explore the more diﬃcult case of ﬁnding a rational distance set of four nonconcyclic rational points on the parabola. We therefore need to examine the last condition of Theorem 2.4:

g(m13) + g(m24) = g(m23) + g(m14) = g(m12) + g(m34), in order to ﬁnd our six mij values. Consider the surface g(mij) + g(mkl) = t where t is an arbitrary rational number. We can make the following substitutions:

X2 + X X + 1 T − 2b m = , m = , t = , ij Y − TX kl Y − TX a to obtain the elliptic curve E : Y 2 = X3 + (T 2 + 2)X2 + X. Thus if we ﬁnd rational points on the elliptic curve we can ﬁnd rational mij values that satisfy the last condition.

2.7 Examples In order to illustrate this process, we include the following example: 2 13 Given the parabola y = x , let T = 6 in order to get the following elliptic curve 2 3 241 2 13 E : Y = X + 36 X + X. We chose T = 6 because it is a small rational value that gives us an elliptic curve of positive rank. We get the rational points Q1 = (6: 19: 18),Q2 = (3: 13: 36),Q3 = (30: 169: 750) on E. Now our mij have the following values: m12 = 3/10, m13 = 1/2, m23 = 4/3, m14 = 4, m24 = 6, m34 = 15/2. Which yields the following rational distance set: 307 94249 19 361 127 16129 757 573049 − , , − , , , , , . 240 57600 80 6400 240 57600 240 57600 We illustrate this in the following ﬁgure: We now show an example of a rational distance set on the parabola y = x2 + 2x. 13 2 If we let T = 6 then we get the same elliptic curve as in Example 1: E : Y =

7 Figure 2: Rational distance set of four points on a parabola y = x2.

3 241 2 13 X + 36 X + X. Again, we chose T = 6 because it is a small rational value that gives us an elliptic curve of positive rank. We get the rational points Q1 = (6: 19: 18),Q2 = (3: 13: 36),Q3 = (30: 169: 750) on E. Now our mij have the following values: m12 = 3/10, m13 = 1/2, m23 = 4/3, m14 = 4, m24 = 6, m34 = 15/2. This gives us diﬀerent values for g(mij): 211 11 41 g(m ) = − , g(m ) = − , g(m ) = − , 12 60 13 4 14 24 1 11 101 g(m ) = − , g(m ) = , g(m ) = . 23 8 24 12 34 60 These values give us the following rational distance set:

737 189409 107 39911 77 42889 109 29321 − , , − , − , , , , . 240 57600 240 57600 240 57600 80 6400

We can see this in the following ﬁgure:

2.8 Rational Distance Set of Five Points on a Parabola Next, we provide the necessary conditions for a rational distance set of ﬁve rational points on a parabola.

2 Theorem 2.8. Given a parabola y = ax + bx + c, let S = {P1,P2,P3,P4,P5}. S is a rational distance set of rational points if and only if there are rational values mij,

8 Figure 3: Rational distance set of four points on a parabola y = x2 + 2x.

1 ≤ i < j ≤ 5 such that

g(m ) + g(m ) − g(m ) x = 12 13 23 1 2 g(m ) − g(m ) + g(m ) x = 12 13 23 2 2 −g(m ) + g(m ) + g(m ) x = 12 13 23 3 2 −g(m ) − g(m ) + g(m ) + 2g(m ) x = 12 13 23 14 4 2 −g(m ) − g(m ) + g(m ) + 2g(m ) x = 12 13 23 15 5 2

as well as,

g(m13) + g(m24) = g(m14) + g(m23)

g(m13) + g(m25) = g(m15) + g(m23)

g(m12) + g(m34) = g(m14) + g(m23)

g(m12) + g(m35) = g(m15) + g(m23)

g(m12) + g(m13) + g(m45) = g(m14) + g(m15) + g(m23)

9 Proof. For each 1 ≤ i < j ≤ 5 where xi + xj = g(mij), we have ten equations:

x1 + x2 = g(m12)

x1 + x3 = g(m13)

x1 + x4 = g(m14)

x1 + x5 = g(m15)

x2 + x3 = g(m23)

x2 + x4 = g(m24)

x2 + x5 = g(m25)

x3 + x4 = g(m34)

x3 + x5 = g(m35)

x4 + x5 = g(m34)

This is equivalent to the following row reduced augmented matrix:

 g(m12)+g(m13)−g(m23)  1 0 0 0 0 2  0 1 0 0 0 g(m12)−g(m13)+g(m23)   2   0 0 1 0 0 −g(m12)+g(m13)+g(m23)   2   −g(m12)−g(m13)+g(m23)+2g(m14)   0 0 0 1 0 2   −g(m12)−g(m13)+g(m23)+2g(m15)   0 0 0 0 1   2   0 0 0 0 0 g(m13) − g(m14) − g(m23) + g(m24)     0 0 0 0 0 g(m13) − g(m15) − g(m23) + g(m25)     0 0 0 0 0 g(m12) − g(m14) − g(m23) + g(m34)     0 0 0 0 0 g(m12) − g(m15) − g(m23) + g(m35)  0 0 0 0 0 g(m12) + g(m13) − g(m14) − g(m15) − g(m23) + g(m45) which gives us our desired equations.

2.9 Rational Distance Sets of Three Points on a Hyperbola At this point, we would like to examine a rational distance set of three rational points on a hyperbola. Theorem 2.9. Let a, b, c, d ∈ Q such that a(ad−bc) 6= 0. Then there exists a rational distance set of three rational points, S = {P1,P2,P3} on the hyperbola axy + bx + cy + d = 0. Proof. The proof proceeds in several parts. First, we rewrite the distance formula as m2 −1 in Equation 1. So, yi−yj can be parameterized as yi−yj = ij . xi−xj xi−xj 2mij bx+d Then, we solve for y in our hyperbola equation to get the expression: y = − ax+c . Substituting y into our modiﬁed distance formula yields 2(ad − bc) y − y m2 − 1 = 2 i j = ij . (axi + c)(axj + c) xi − xj mij

10 m2−1 2 X Deﬁne D = 2(ad − bc) and let m = Dn . Making the substitutions m = D and Y (D) 2 3 2 n = DX gives us the elliptic curve E : Y = X − D X. Note that we can also m2−1 2 (D) obtain the expression m = Dn from E with the substitutions X = Dm and Y = D2mn. (D) Letting Qi = (Xi : Yi : 1) be rational points on E , we deﬁne the following rational points:

Y1 Y2 Y3 n23 = , n13 = , n12 = DX1 DX2 DX3 in order to allow the following relation to hold:

2 D 2 mij − 1 = Dnij = . (axi + c)(axj + c) mij We can now deﬁne an explicit formula for a rational distance set of three rational points on a hyperbola: n23 c b D n12n13 P1 = − , − − an12n13 a a 2a n23 n13 c b D n12n23 P2 = − , − − an12n23 a a 2a n13 n12 c b D n13n23 P3 = − , − − an13n23 a a 2a n12

2.10 Example In order to illustrate this process, we include the following example: Given the hyperbola 2xy + x + y + 2 = 0, we get the following elliptic curve E (6) : Y 2 = X3 − 36X. From this curve, we get the rational points Q1 = (−3: 9: 1) ,Q2 = (50: − 35: 8) ,Q3 = (−58719: − 321057: 50653). Now our nij have the following values: n23 = −1/2, n13 = −7/60, n12 = 1551/1709. This yields the following rational distance set:

40203 27877 8654 37876 87103 36977 , − , − , − , , − . 21714 34040 23265 5957 11914 62040

We show this in the following ﬁgure:

2.11 Rational Distance Set of Four Points on a Hyperbola The system of equations we derive to find a rational distance set of four rational points on a hyperbola will be overdetermined, as in the four point parabola case. In order to handle the extra constraints, we must consider a new result. Before lauching into the proposition, however, we should first state some relevant defitions:

11 Figure 4: The hyperbola 2xy + x + y + 2 = 0 with a rational distance set of rational points.

Deﬁnition 2.10. Let X,Y,Z be groups and f : X → Z and g : Y → Z be two homomorphisms. Then the ﬁbered of f and g is the subgroup X × Y consisting of all pairs (X,Y ), x ∈ X, y ∈ Y such that f(X) = g(Y )[3].

Deﬁnition 2.11. Given two curves E : f(x, y) = 0 and D : g(u, v) = 0 a map φ: E → D is called a rational map if there exist polynomials r(x, y), s(x, y), and r(x,y) s(x,y) t(x, y) such that u = t(x,y) and v = t(x,y) . Moreover, we say that E and D are birationally equivalent if there exist two rational maps φ: E → D and ψ : D → E such that the following two conditions hold:

a. ψ ◦ φ = idE, that is the composition of ψ and φ is the identity on E.

b. φ ◦ ψ = idD, that is the composition of φ and ψ is the identity on D. With these deﬁnitions in hand, we introduce the following proposition:

Proposition 2.12. Fix a rational number T, and consider the curve

2 3 2 4 Yi = Xi − D Xi XiXj ET = (Xi,Yi,Xj,Yj) ∈ A 2 3 2 and 2D = T (4) Yj = Xj − D Xj YiYj

−1 (D) (D) a. The curve ET ' π ({T }) is a ﬁber over the product E ×π E associated to the morphism π : E(D) × E(D) → P1, which sends

XiXj ((Xi : Yi : 1), (Xj : Yj : 1)) 7→ 2D . YiYj

b. Additionally, ET is birationally equivalent to the curve

0 4 2 2 4 2 2 2 2 2 2 ET : y = x(x + D )(x + D T )(x − D T ) (x + D T ) .

12 Proof. Part (a) is clear from the definition of a fiber product. For part (b), we consider a point (x, y) on the curve, and define 1 y2 X = − i T 2 (x + D2)(x + D2T 2)(x − D2T 2) 1 y Y = − i T 3 x + D2 2Dy2 − (x − D2T 2)(x + D2T 2)(x2 + 2D2x + D4T 4) X = D j (x − D2T 2)2(x + D2T 2)2 2Dy2 − (x − D2T 2)(x + D2T 2)(x2 + 2D2x + D4T 4) Y = 2D2 j (x − D2T 2)3(x + D2T 2)3

From the substitutions, we see that this point is on ET . Conversely, we consider a point (Xi,Yi,Xj,Yj) on ET , and deﬁne

3 3 2 3 Xi Xj (Xj − D) x = −4D 2 4 Yi Yj 3 3 3 3 3 2 3 4 3 5 2 4 5 Xi Xj (4D Xi Xj − 8D Xi Xj + 4DXi Xj − Yi Yj ) y = 8D 2 7 Yi Yj

0 Again, from the substitutions we see that this point is on the curve ET . We are now ready to examine a method for ﬁnding a rational distance set of four rational points on a hyperbola.

Theorem 2.13. Let a, b, c, d ∈ Q such that a(ad − bc) 6= 0. There exists a rational distance set of four rational points on the hyperbola axy + bx + cy + d = 0 if and only if we can ﬁnd three rational points on the curve

0 4 2 2 4 2 2 2 2 2 2 2 ET : y = x(x + D )(x + D T ) (x − D T ) (x + D T ) . Proof. This proof is similar to that of a rational distance set of three rational points on a hyperbla. First, we express the distance formula as in Equation 1. So, yi−yj can xi−xj m2 −1 be parameterized as yi−yj = ij . xi−xj 2mij bx+d Then, we solve for y in our hyperbola equation to get y = − ax+c . Substituting y into our modiﬁed distance formula yields

2(ad − bc) y − y m2 − 1 = 2 i j = ij . (axi + c)(axj + c) xi − xj mij

m2−1 2 X Deﬁne D = 2(ad − bc) and let m = Dn . Making the substutions m = D and Y (D) 2 3 2 n = DX gives us the elliptic curve E : Y = X − D X. Note that we can also m2−1 2 (D) obtain the expression m = Dn from E with the substitutions X = Dm and Y = D2mn.

13 (D) Letting Qi = (Xi : Yi : 1) be rational points on E , we deﬁne the rational values

Y1 Y2 Y3 Y4 Y5 Y6 n13 = , n24 = , n23 = , n14 = , n12 = , n34 = , DX1 DX2 DX3 DX4 DX5 DX6 such that the following relation holds:

2 D 2 mij − 1 = Dnij = . (axi + c)(axj + c) mij Therefore, the four rational points in the rational distance set must be in the form: n23 c b D n12n13 P1 = − , − − an12n13 a a 2a n23 n13 c b D n12n23 P2 = − , − − an12n23 a a 2a n13 n12 c b D n13n23 P3 = − , − − an13n23 a a 2a n12 n23 c b D n24n34 P4 = − , − − an24n34 a a 2a n23

Additionally, we have the constraint n13n24 = n23n14 = n12n34. In order to satisfy this constraint, we want to ﬁnd six rational points on E(D) satisfying the relation: Y Y Y Y Y Y 1 2 = 3 4 = 5 6 . DX1 DX2 DX3 DX4 DX5 DX6

To solve any pair of such equations, we look for rational points on ET , deﬁned in Equation 4. Recall Propostition 2.12, which states that ET is birationally equivalent to the curve

0 4 2 2 4 2 2 2 2 2 2 ET : y = x(x + D )(x + D T )(x − D T ) (x + D T ) .

Thus in order to find a six rational points on E (D), it suffices to find a set of three 0 rational points on the curve ET .

0 One way to ﬁnd rational points on the curve ET is to consider the superelliptic curve 4 2 2 4 2 4 2 4 4 ET : w = 2D z(z + T )(z + 2z + T )(z + 2T z + T ). Deﬁnition 2.14. A superelliptic curve is a curve of the form wn = f(z), where f(z) is a polynomial with no repeated roots and the exponent n is relatively prime to the degree of f(z).

Let (zp : wp : 1) and (zq : wq : 1) be two rational points on the superelliptic curve. Upon deﬁning the rational numbers

14 2 4 2 4 Yij 1 zp − T Yik 1 zp − T nij = = nik = = DXij T wp DXik T wp Ykl 2 wp Yjl 2 wp nkl = = 2 4 njl = = 2 4 DXkl D zp − T DXjl D zp − T we have the relation

Yij Ykl 2 Yik Yjl nijnkl = = = = niknjl. DXij DXkl DT DXik DXjl

Finally, we want to show that ﬁnding three rational points on ET corresponds to ﬁnding at least three rational points on ET .

Proof. Using the morphism φ : ET → ET deﬁned by

(z : w : 1) 7→ (2D2(T 4 + z2)z : D3(T 4 − z2)w : 4z2) along with the substitutions above, we have that (Xij,Yij,Xkl,Ykl), where

1 w2 X = ij 2T 2 z(z2 + 2z + T 4) D w(z2 − T 4) Y = ij 2T 3 z(z2 + 2z + T 4) 2w2 − D(z4 + 4z3 + 6T 4z2 + 4T 4z + T 8) X = kl (z2 − T 4)2 2w2 − D(z4 + 4z3 + 6T 4z2 + 4T 4z + T 8) Y = 2w kl (z2 − T 4)3

is a rational point on ET . The rest follows.

0 This result asserts that one way to find three rational points on ET is to find three rational points on ET . There is a simple way of finding all rational points R = (z : w : 1) on ET . A theorem by Gerd Faltings states that if the genus of a curve is greater than one then ET (Q) is a finite set for any fixed rational T where T 6= −1, 0, 1. Our curve has a genus of nine. Hence one can use software such as SAGE to find all of them.

3 Acknowledgments

This work was conducted during the 2010 Mathematical Sciences Research Institute Undergraduate Program (MSRI-UP), a program supported by the National Science Foundation (grant No. DMS-0754872) and the National Security Agency (grant No. H98230-10-1-0233). We also thank Dr. Edray Goins, Dr. Luis Lomel´ı,Dr. Duane Cooper, Katie Ansaldi, Ebony Harvey, and the MSRI staﬀ.

15 References

[1] Campbell, Garikai. “‘Points on y = x2 at Rational Distance”’. Mathematics of Computation.

[2] Erd¨os,Paul. “‘Ulam, the Man and the Mathematician’”. Journal of Graph Theory

[3] Lang, Serge. “Algebra”. Graduate Texts in Mathematics.

[4] Solymosi, Jozsef and Frank De Zeeuw. “‘On a Question of Erd¨osand Ulam’”. Discrete Computational Geometry