On a Diophantine Equation That Generates All Integral Apollonian Gaskets

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On a Diophantine equation that generates all integral Apollonian Gaskets

Jerzy Kocik Department of Mathematics, Southern Illinois University, Carbondale, IL 62901 [email protected] On a Diophantine equation that generates all Apollonian Gaskets

AbstractJerzy Kocik Department of Mathematics, SIU, Carbondale,IL62901 A remarkably simple [email protected] quadratic equation is known to generate all Apollonian integral gaskets (disk packings). A new derivation of this formula isAbstract: presented A here remarkably based on inversivesimple Diophantine geometry. Alsoquadr occurrencesatic equation of is Pythagorean known to generate triplesall Apollonian in such gaskets integral is discussed.gaskets (disk packings). A new derivation of this formula is presented here based on inversive geometry. Also occurrences of Pythagorean triples Keywords:in such gasketsIntegral is discussed. Apollonian disk packings, inversive geometry, Pythagorean

triples.Keywords: Integral Apollonian disk packings, inversive geometry, Pythagorean triples.

AMS: 28A80, 52C26, 11Y50. 1. Introduction 1. Introduction Apollonian disk packing (or Apollonian gasket) is a pattern obtained by starting with Apollonianthree mutually disk tangent packing circles (or Apollonian of which one gasket contains) is a thepattern other obtained two, then by recursively starting with three mutuallyinscribing tangent new circles circles (disks) of whic in theh one curvilinear contains triangularthe other two, regions then (called recursively “ideal inscribing trian- new circlesgles”) formed(disks) betweenin the curvilinear the circles. triangular Figure 1regi showsons (called a few examples,“ideal triangles”) including formed (a) a between thespecial circles. case ofFigure the noncompact 1 shows a few “Apollonian examples, Strip”, including (b) the (a) Apollonian a special Windowcase of the which noncompact “Apollonian Strip”, (b) the Apollonian Window which is the only case that has symmetry is the only case that has symmetry D2, (c) the regular threefold gasket, which has symmetryD2, (c) theD regular, and (d) threefold a general gasket, gasket which that may has have symmetry no mirror D3, symmetry. and (d) a general gasket that may have no3 mirror symmetry.

(a) (b) (c) (d)

arXiv:2008.04440v1 [math.MG] 10 Aug 2020

Regular Apollonian A general Apollonian Apollonian strip Apollonian window gasket gasket

FigureFigure1.1: 1: Various Various Apollonian Apollonian disk disk packings packings

It turnsIt turns out outthat thatthat therethere exist exist an an inﬁnite infinite number number of ofintegral integralApollonian Apollonian packings, packings, where thewhere curvature the curvature of every of everycircle/disk circle is/disk an integer. is an integer.

RecallRecall that thatfour four pairwise pairwise tangent tangent circles circles are are said said to tobe be in in DescartesDescartes configuration conﬁguration. . Suppose Supposea,b,c,d area, curvaturesb, c, d are curvaturesof four such of circles. four such Then circles. Descartes’ Then Descartes’ circle formula circle states formula

2(a2 + b2 + c2 + d2) = (a + b + c + d)2 (1.1) 1 One may use this formula to obtain the curvature of the fourth circle d, given the first three. Being a quadratic equation, it yields two solutions, say d1 and d2, such that

d1 + d2 = 2(a + b + c) (1.2)

Disks d1 and d2 will be called Boyd-dual. As observed by Boyd [B], this “linearized” version of (1.1) provides a handy tool for determining the curvatures in a particular Apollonian gasket starting with the first four values. In particular, it follows that if the first four values are integers, so are all.

Remark: For consistency, the curvature of the circle that contains all of the remaining ones must be given a negative value for equations (1) and (2) to hold. It is customary to replace the term “curvature” by “bend” to account for this convention. Another way of looking at

1 Jerzy Kocik Diophantine equation of Apollonian gakets 2 states 2(a2 + b2 + c2 + d2) = (a + b + c + d)2. (1.1) One may use this formula to obtain the curvature of the fourth circle d, given the ﬁrst three. Being a quadratic equation, it yields two solutions, say d1 and d2, such that

d1 + d2 = 2(a + b + c). (1.2)

Disks d1 and d2 will be called Boyd-dual. As observed by Boyd [1], this “linearized” version of (1.1) provides a handy tool for determining the curvatures in a particular Apollonian gasket starting with the ﬁrst four values. In particular, it follows that if the ﬁrst four values are integers, so are all.

Remark 1.1. For consistency, the curvature of the circle that contains all of the remaining ones must be given a negative value for equations (1.1) and (1.2) to hold. It is customary to replace the term “curvature” by “bend” to account for this convention. Another way of looking at this is to think of disks rather than circles, where the greatest circle is the boundary of an exterior, unbounded, region. This way no two disks in an Apollonian gasket overlap.

In this note, Apollonian gaskets will be labeled by the bends of the five greatest circles, i.e., by the five least bends. Why five will become clear when we consider symmetries, Appendix A. Examples of integral Apollonian gaskets include these:

(0, 0, 1, 1, 1) − Apollonian belt (Figure 1.1a) (−1, 2, 2, 3, 3) − Apollonian window (Figure 1.1c)

(−2, 3, 6, 7, 7) − less regular gasket, but with D1 symmetry (−6, 11, 14, 15, 23) − quite irregular case

Note that the regular gasket (see Figure√ 1.1) cannot have√ integer bends, as its quintet − of curvatures is, up to scale,√ (1 3, 2, 2, 2, 10 + 2 3), hence its curvatures are populated by elements of Z[ 3]. Due to (1.2), the integrality of the ﬁrst four circles determines integrality of all disks in the packing. An integral Apollonian packing is irreducible if the bends have no common factor except 1. The problem is to classify and determine all irreducible integral Apollonian gaskets.

2. Integral disk packing – the formula

All integral Apollonian disk packings may be determined using a simple Diophantine quadratic equation with constraints. The derivation of this formula is a much simpler alternative to that of “super-Apollonian packing” [2]–[4] and is based on inversive geometry. this is to think of disks rather than circles, where the greatest circle is the boundary of an exterior, unbounded, region. This way no two disks in an Apollonian gasket overlap.

In this note, Apollonian gaskets will be labeled by the bends of the five greatest circles, i.e., by the five least bends. Why five will become clear when we consider symmetries, Appendix A.

Examples of integral Apollonian gaskets include these:

(0, 0, 1, 1, 1) – Apollonian belt (Figure 1 a) (–1, 2, 2, 3, 3) – Apollonian window (Figure 1 c) (–2, 3, 6, 7, 7) – less regular gasket, but with D2 symmetry (–6, 11, 14, 15, 23) – quite irregular case

Note that the regular gasket (see Figure 1) cannot have integer bends, as its quintet of curvatures is, up to scale, (1–3, 2, 2, 2, 10+23), hence its curvatures are populated by elements of Z[3]. Due to (1.2), the integrality of the first four circles determines integrality of all disks in the packing. An integral Apollonian packing is irreducible if the bends have no common factor except 1.

The problem is to classify and determine all irreducible integral Apollonian gaskets.

2. Integer disk packing – the formula

Theorem 2.1: There is a one to one correspondence between the irreducible integral Apollonian gaskets and the irreducible quadruples of non-negative integers B,k,n,  N that are solutions to quadratic equation 2 2 (2.1) B +  = kn with constrains (i) 0    B/3, (ii) 2  k  n.

Every solution to (2.1) corresponds to an Apollonian gasket with the following quintet of the major bends (curvatures): Jerzy Kocik Diophantine equation of Apollonian gakets 3 (B0, B1, B2, B3, B4) = (–B, B + k, B + n, B + k + n – 2, B + k + n + 2)

B0 = –B main B = circle 4 B+k+n+2

B2 = B+n B = B+k 1

B3 =

B+k+n–2

Figure 2.1: An Apollonian gasket and its four greatest circles (smallest curvatures)

Theorem 2.1. There is a one to one correspondence between the irreducible integral Apollonian gaskets and the irreducible quadruples of non-negative integers B, k, n, µ ∈ N that are solutions to quadratic equation

B2 + µ2 = kn (2.1) with constraints √ (i) 0 ≤ µ ≤ B/ 3, (ii) 2µ ≤ k ≤ n. Every solution to (2.1) corresponds to an Apollonian gasket with the following quintet of the major bends (curvatures):

(B0, B1 , B2, B3, B4) = (−B, B + k, B + n, B + k + n − 2µ, B + k + n + 2µ)

Corollary 2.2. The Apollonian gasket is integral iﬀ gcd(B, k, n) = 1.

Figure 2.1 locates the curvatures of the theorem in the Apollonian gasket. Note that the triple of integers (B, µ, k) is also a good candidate for a label that uniquely identiﬁes an Apollonian gasket (since n is determined by: n = (B2 + µ2)/k). Equation (2.1) leads to an algorithm producing all integral Apollonian gaskets, ordered by the curvatures, presented in Figure 2.2: Appendix A shows the ﬁrst 183 entries for the main bend varying from 0 through 32.

Example 2.1. The table below shows that there are only 3 Apollonian gaskets for B = 6.

Corollary 2.2: The Apollonian gasket is integral iff gcd(B,k,n) = 1

Figure 2 locates the curvatures of the theorem in the Apollonian gasket. Note that the triple of integers (B, , k) is also a good candidate for a label that uniquely identifies an Apollonian gasket (since n is determined by: n = (B2 + 2 )/k).

Jerzy KocikEquation (2.1) leads Diophantine to an algorithm equation of producing Apollonian all gakets integral Apollonian gaskets,4 ordered by the curvatures, presented in Figure 3:

for B = 1 to whatever you wish for  = 0 to int(B/3) evaluate H = B2 + 2 for k = 2 to int(H) if n = H/k is an integer and gcd(B,k,n) = 1 then the quintet of main curvatures is (–B, B+k, B+n, B+k+n–2, B+k+n+2) next k next  next B

Figure 3: An algorithm that produces all integral disk packings. Figure 2.2: An algorithm that produces all integral disk packings.

Appendix A shows the first 183 entries for the main bend varying from 0 through 32. B µ k n formula major curvatures of Apollonian gasket

2 2 6 0 Example: 1 36 6The +0 table = 1 below36 shows that there are (–6, only 7, 342, Apollonian 43, 43) gaskets for B = 6. 2 2 6 0 2 18 6 +0 = 218 (multiples of 2) 2 2 6 0 3 12 6 +0 = 312 (multiples of 3) B  k2 n 2 formula major bends of Apollonian gasket 6 0 4 9 6 +0 = 45  (–6, 10, 15, 19, 19) 2 2 2 2 6 0 6 6 6 0 16 +0 36 = 4 65 +0 =(multiples 136 of 6)  (–6, 7, 42, 43, 43) 2 2 (multiples of 2) 6 0 2 2 182 6 +0 = 218 2 2 (k is too small) 6 1 1 6 37 0 36 +1 12 = 1 637+0 = 312 (multiples of 3) 2 2 6 2 2 6 20 0 46 2 +2 9 2 = 2640+0 = (multiples 45 of 2)  (–6, 10, 15, 19, 19) 2 2 2 2 (multiples of 6) 6 2 4 6 10 0 6 +2 6 = 4 610+0 =(multiples 45 of 2)  2 2 2 2 6 2 5 6 8 1 16 +2 37 = 5 68 +1 = 137 (no good  k. 1(–6, is too 11, small) 14, 15, 23)

2 2 2 2 6 3 5 6 9 2 26 +3 20 = 1 645+2 =(k 2to40o small )(multiples of 2) 2 2 6 2 4 10 6 +2 = 410 (multiples of 2) 2 2 6 2 5 8 6 +2 = 58 –B B+k (–6, 11,B+n 14, 15, 23)B+k +n ± 2µ

2 2 6 3 5 9 6 +3 = 145 (no good k )

Proof Theorem 2.1. (We denote a circle and its curvature by–B the sameB+k symbol).B+n Con- B+k+n 2 sider an Apollonian gasket of disks inscribed inside a circle of curvature B (bend equal to −B).Draw an axis throughk starts the with center at least of this2 and circle cannot and exceed the center n of the next largest circle B1 (the horizontal axis A in Figure 2.3). Inverting the gasket through a circle K of radius 2/ variesB will from produce 0 through an Apollonian 6/3 = 3.46… belt, shown in the ﬁgure on the right side of the gasket. Denote its width by 2ρ. Lines L and L are the images of B and B , 0 1 1 respectively. 0 The next largest circle, B2, will show up in the strip as a circle B2 . It will intersect the axis A. Various Apollonian gaskets will result by varying the strip’s width (that is 3 0 the parameter ρ) and the height h of the center of the circle B2 above the axis. Clearly, the range of h is the interval [0, ρ), as going beyond would result in repetitions of the Apollonian arrangements. Recall the basic formulas of inversive geometry: inversion through a circle of radius K centered at O carries a point at distance d from O to a point at distance d0 Proof Theorem 2.1: (We denote a circle and its curvature by the same symbol). Consider an Apollonian gasket of disks inscribed inside a circle of curvature B (bend equal to –B). Draw an axis through the center of this circle and the center of the next largest circle B1 (the Jerzy Kocikhorizontal axis Diophantine A in Figure equation 4). Inverting of Apollonian the gasket gakets through a circle K of radius 2/5B will produce an Apollonian belt, shown in the figure on the right side of the gasket. Denote its

width by 2. Lines L0 and L1 are the images of B and B1, respectively.

L0 = B’ L1 = B’1 K

B’2 B1 B2 axis A h

B3 2/B B’3

2

The next largest circle, B2, will show up in the strip as a circle B’2. It will intersect the axis A. Various ApollonianFigure 2.3: gaskets The will method result of by inversion. varying the strip’s width (that is the parameter

) and the height h of the center of the circle B2’ above the axis. Clearly, the range of h is the interval [0, ), as going beyond would result in repetitions of the Apollonian according to arrangements. 0 2 Recall the basic formulas ofdd inversive= K geometry: inversion through a circle of(2.2) radius K centered at O carries a point at distance d from O to a point at distance d’ according to 0 and a disk of radius r with center distance d from O is mapped to a disk of radius r 0 d d’ = K2 (2.2) with its center located at distance d according to and a disk of radius r with center distance d from O is mapped to a disk of radius r’ with its rR2 dR2 center located(i) atr0 distance= d’ according, ( iito) d0 = . (2.3) 2 − 2 rR 2 2 − 2 dR 2 d (i) r’r = , (ii) d d’ =r (2.3) d 2  r 2 d 2  r 2 We shall analyze the Apollonian gasket as the inversive image of the Apollonian strip, disk by disk: We shall analyze the Apollonian gasket as the inversive image of the Apollonian strip, disk by disk: Step 0. Recall that B ∈ N represents the curvature of the encompassing circle of the Step 0. Recall that B  N represents the curvature of the encompassing circle of the gasket. gasket. Step 1. To calculate the curvature B1 of the greatest circle inside (the image of L1), apply Step 1. To calculate(2.3 i) to the point curvature of intersectionB1 of of the L1 greatestand axis A circle to get: inside (the image of L1),

2 apply (2.3)i) to the point of intersection of L1 Band1 = axis B +A Bto get:

2 k B1 = B + B ρ . Since B is an integer, so is the last|{z} term; denote it by k = B2 . Clearly, k can be any k nonnegative integer (to make B1  B).

Since B is an integer, so is the last term; denote it by k = B2ρ. Clearly, k can be any nonnegative integer (to make B1 ≥ B). 4 0 Step 2. For B2, the image of B2, use (2.3)ii to get 1 1 d2 − ρ2 B2 B = = · = · (d2 − ρ2) (simpliﬁcation) 2 r0 ρ (2/B)2 4ρ B2 = · (2/B + ρ)2 + h2 − ρ2 (Pythagorean thm) 4ρ 4 + h2B2 = B + 4ρ | {z } n

Jerzy Kocik Diophantine equation of Apollonian gakets 6

As before, we conclude that the last term must be an integer; denote it by n. Clearly, n ≥ k (to make B2 ≥ B1).

Step 3. Similarly, we get a formula for the next largest circle B3 located below B2, the 0 0 image of B3. Simply use the above formula with h = h − 2ρ instead of h to get

1 d2 − ρ2 B2 B = · = · (2/B + ρ)2 + (h2 − ρ2)2 − ρ2 3 ρ (2/B)2 4ρ 4 + h2B = B + ρB2 + − hB2 4ρ |{z} | {z } |{z} k n m Quite pleasantly, the ﬁrst three terms coincide with terms from previous steps. Since we have already established that they must be integers, so is the last one; denote it by m. Thus we have three integers deﬁned by the geometry of the construction:

4 + h2B2 n = , k = ρB2, m = hB2. (2.4) 4ρ Integers k, n and m are not independent; take the deﬁnition for n and eliminate h and r from it to get 4nk = 4B2 + m2, from which it follows immediately that m must be even, say m = 2µ. Reduce the common factor of 4 to get the “master equation” (2.1). As to the constraints, the order of the curvatures gives three inequalities:

B1 ≥ B ⇒ k ≥ 0, B2 ≥ B1 ⇒ n ≥ 0, B3 ≥ B2 ⇒ k ≥ 2µ. (2.5)

The additional upper bound for µ comes from the fact that k takes its greatest value at p k = B2 + µ2 . Thus the last inequality of (2.5), k ≥ 2µ, implies: q B2 + µ2 ≥ 2µ and therefore (after squaring) B2 > 3µ2. This ends the proof.

3. Symbols of the circles in an Apollonian gasket

The symbol of a circle [5, 6] is a formal fraction x˙, y˙ b where b = 1/r denotes the bend (signed curvature) of the circle and the position of the center is x˙ y˙ (x, y) = , . b b 3. Symbols of the circles in an Apollonian gasket

The symbol of a circle [jk] is a formal fraction

x, y

b 3. Symbols of the circles in an Apollonian gasket where b = 1/r denotes the bend (signed curvature) of the circle and the position of the center is The symbol of a circle [jk] is a formal fraction    x y  x, y (x, y) =  ,  .  b b  b

By reduced coordinateswhere we b mean= 1/r denotesthe pair the x ,bendy . (signedIn the casecurvature) of the of Apollonian the circle andWindow the position of the center (packingJerzy with Kocik the majoris curvatures Diophantine (–1, equation 2, 2, of3, Apollonian3)), the reduced gakets coordinates and the7 bend of each circle are integers, see Figure 5.  x y  (x, y) =  ,  .  b b 

By reduced coordinates we mean the pair x,y . In the case of the Apollonian Window -5,12 5,12 (packing with14 the major curvatures (–1, 142, 2, 3, 3)), the reduced coordinates and the bend of each circle are integers, see Figure 5. -21,20 21,20 30 -3, 4 0, 2 3, 4 30 6 3 6 -8,6 8,6 5,12 11 -8,12 -5,12 8,12 11 23 14 23 14 -15, 8 15 8 18 18 -21,20 21,20 30 -3, 4 0, 2 3, 4 30 0,46 3 6 -8,6 15 8,6 11 -8,12 8,12 11 23 23 -1,0 -15, 8 1,0 15 8 18 18 2 2

0,4 Figure 5: Apollonian Window with labels15 Figure 3.1: Apollonian Window with labels. -1,0 1,0 Boyd’s linearization (1.2) for bends in Descartes2 configurations holds for the reduced2 coordinates,By reduced since coordinates they satisfywe a mean quadratic the pair equation (x ˙, y˙). In thedue case to the of theextended Apollonian Descartes Win- Circle dow (packing with the major curvatures (−1, 2, 2, 3, 3)), the reduced coordinates and Theorem [LGM]. Thus: Figure 5: Apollonian Window with labels the bend of each circle are integers, see Figure 3.1. Boyd’sx linearization x   2(x (1.2) x for x bends) and in Descartes y  y conﬁgurations 2(y  y  holdsy ) for the (3.1) 4 Boyd’s4 1linearization2 3 (1.2) for4 bends4 in Des1 cartes2 configurations3 holds for the reduced reduced coordinates, since they satisfy a quadratic equation due to the extended Descartes Circle Theoremcoordinates, [2, 3, since 4]. Thus: they satisfy a quadratic equation due to the extended Descartes Circle Remark: The geometricTheorem interpretation [LGM]. Thus: of the integrality of the labels. Every circle in the Apollonian Window is an0 integer multiple of its radius0 above the horizontal axis and from x˙4 + x˙ = 2(x ˙1 + x˙2 + x˙3) andy ˙4 + y˙ = 2(˙y1 + y˙2 + y˙3). (3.1) the vertical axis. This is4 a generalization of Pappus’4 Arbelos Theorem [Wil] for a single x4  x4  2(x1  x2  x3 ) and y 4  y 4  2(y1  y 2  y3 ) (3.1) chain Remarkin an arbelos-like 3.1. The geometric figure, illustrated interpretation in Figure of the integrality6. of the labels. Every circle in the ApollonianRemark: Window is anThe integer geometric multiple interpretation of its radius aboveof the the integr horizontalality of axis the labels. Every circle in the and from the verticalApollonian axis. This Window is a generalization is an integer of Pappus’multiple Arbelos of its radius Theorem above [11] the horizontal axis and from for a single chainthe in an vertical arbelos-like axis. ﬁgure, This illustratedis a generalizati in Figureon 3.2.of Pappus’ Arbelos Theorem [Wil] for a single chain in an arbelos-like figure,Figure illustrated 6: Classical in Figure Pappus’ 6. Arbelos Theorem: the base is divided randomly, yet every circle in the chain is an even multiple of its radius above the axis. Figure 6: Classical Pappus’ Arbelos Theorem: the base is divided randomly, yet every circle in the chain is an even multiple of its radius above the axis. 6

Figure 3.2: Classical Pappus’ Arbelos Theorem: the base is divided randomly, yet every circle in the chain is an even multiple of its radius above the axis. 6 The question is whether the same may be expected for other integer Apollonian disk packings, that is: Are the three numbers in the label all integers? The question is whether the same may be expected for other integer Apollonian disk packings, that is: Are the three numbers in the label all integers?

Proposition 3.1: In the case of the coordinate system with the center located at the center of circle B (see Figure 4), the labels for the Apollonian gasket generated from (B, n, k, ) are as presented in Figure 7. Jerzy Kocik Diophantine equation of Apollonian gakets 8

Main circle B2−(k+µ)2 2(k+µ) 0, 0 Bk , k

B + k + n + 2µ  B

2 2 B −µ , 2µ Bk k k B + n  , 0 B B  k

2− − 2 − B (k µ) , − 2(k µ) Bk k B + k + n − 2µ

Figure 7: Labels of circles in the Apollonian packing [B, , n, k] Figure 3.3: Labels of circles in the Apollonian packing [B, µ, n, k]

Proof: Direct calculations. Proposition 3.2. In the case of the coordinate system with the center located at the One centercan check of circle that the B (see circles Figure of curvatures 2.3), the labels B + k for+ n the   Apollonian are Boyd-dual gasket with generated respect to from the triple( Bof, nmutually, k, µ) are tangent as presented circles inof Figurebends ( 3.3.–B, B+n, B+k).

The above proposition implies that the reduced coordinates for circles are fractional unless Proof. Direct calculations. B|k (see the symbol for B1 in the figure). Since we have also 2  k (constraints) and the need forOne k|2 can (see check the symbols that the circlesfor the of reduced curvatures y-coordinate),B+k+n±µ weare conclude Boyd-dual that with we respect would needto  the= 0 tripleor k = of 2 mutually. tangent circles of bends (−B, B + n, B + k).

This happensThe above in two proposition cases: Apollonian implies that Wi thendow reduced (–1,2,2,3,3), coordinates and for Apollonian circles are frac-Strip | ≤ (0,0,1,1,1),tional unlessthat is Bfork (see[B,,k,n the] symbolequal [1,0,1,1] for B1 inor the[0,0,0,1], ﬁgure). respectively. Since we have However, also 2µ if kk| (which(constraints) happens “often”, and the whenever need for k |2=0),µ (see all the reduced symbols y-coordinates for the reduced are yinteger.-coordinate), This weis a specialconclude (integral) that case we would of Pappus’ need µ observation,= 0 or k = 2 µsince. =0 corresponds pattern shown in Figure 6,This see Appendix). happens in two cases: Apollonian Window (−1, 2, 2, 3, 3), and Apollonian Strip (0, 0, 1, 1, 1), that is for [B, µ, k, n] equal [1, 0, 1, 1] or [0, 0, 0, 1], respectively. However, if k|µ (which happens “often”, whenever µ = 0), all reduced y-coordinates are integer. This is a special (integral) case of Pappus’ observation, since µ = 0 corresponds to the pattern shown in Figure 3.2, see Appendix).

7 Jerzy Kocik Diophantine equation of Apollonian gakets 9

4. Pythagorean triples in Apollonian gaskets 4. Pythagorean triples in Apollonian gaskets Given two tangent circles C1 and C2, we construct a triangle the hypotenuse of which joins Given4. Pythagorean two tangent triples circles in ApollonianC1 and C2 ,gaskets we construct a triangle whose hypotenuse joins thethe centers centers and and the otherother sides sides of whichof which are horizontal are horiz orontal vertical or with vertical respect with to some respect to some fixed axes.ﬁxed Given axes. two tangent circles C1 and C2, we construct a triangle the hypotenuse of which joins the centers and the other sides of which are horizontal or vertical with respect to some fixed axes.

We shall associate with this triangle a triple of numbers, namely WeWe shall shall associate associate withwith this this triangle triangle a triple a triple of numbers, of numbers, namely namely (, , ) = ( b1 x2  b2 x1 , b1 y 2  b2 y1 , b1  b2 ) (4.1) (∆, Γ, H) = (b1 x˙2 − b2 x˙1, b1y˙2 − b2y˙1, b1 + b2). (4.1) (, , ) = ( b1 x2  b2 x1 , b1 y 2  b2 y1 , b1  b2 ) (4.1) The actual size of the triangle’s sides is the above triple scaled down by the factor of b1b2. TheHence actual the sizesymbol of thefor such triangle’s triangles: sides is the above triple scaled down by the factor of Theb 1actualb 2. Hence size the ofsymbol the triangle’sfor such triangles: sides is the above triple scaled down by the factor of b1b2. , ,  Hence the symbol for such triangles:∆, Γ, H (4.2) b1b2 . (4.2) b1b2 Clearly, 2 + 2 = 2. If the reduced coordinates are integers, then the triples (4.1) are 2 2 2 , ,  Clearly,Pythagorean∆ + triples! Γ = HThe. Apollonian If the reduced Window coordinates thus contains are infinitely integers, many then Pythagorean the triples (4.2) (4.1)triples are [jk]. Pythagorean A few are displayed triples! Thein Figure Apollonian 8. Windowb1b2 thus contains inﬁnitely many

Pythagorean2 triples2 [5, 6].2 A few are displayed in Figure 4.1. Clearly,  +  =  . If the reduced coordinates are integers, then the triples (4.1) are Pythagorean triples! The Apollonian Window thus contains infinitely many Pythagorean 12, 16, 20 triples [jk]. A few are displayed in Figure 8. 5,12 84 15, 8, 17 14 66 -3, 4 0, 2 3, 4 6 3 6

-8,6 8,6 11 -8,12 11 12, 16, 20 23 84 5,12 14 15, 8, 17 3, 4, 5 66 -3, 46 0, 2 6 3 5, 12, 13 3, 4 -1,0 1,0 22 6 2 2 -8,6 8,6 0, 0 11 -8,12 11 23 –1 (Exterior circle)

Figure 4.1: Pythagorean triples in the Apollonian Window Figure 8: Pythagorean3, 4, triples 5 in the Apollonian Window 6 Now, for the arbitrary integer packing. Consider the four major circles B , B , B , 0 1 25, 12, 13 B3 in an Apollonian gasket (Figure 3.2). Pairwise, they determine six right triangles. Each of the thick segments-1,0 in Figure 4.2 represents the hypotenuse of1,0 one of them. 22 This sextet will be called2 the principal frame for the gasket. 2

We shall prove later that if the corresponding0 triples, 0 ∆, Γ, H are integers, so are all triples in the gasket, But ﬁrst: –1 (Exterior circle) 8

Figure 8: Pythagorean triples in the Apollonian Window

8 Now, for the arbitrary integer packing. Consider the four major circles B0, B1, B2, B3 in an Apollonian gasket (Figure 6). Pairwise, they determine six right triangles. Each of the thick segments inJerzy Figure Kocik 9 represents the hypotenuse Diophantine of equation one of them. of Apollonian This sextet gakets will be called the 10 principal frame for the gasket.

2 2 B −µ , 2µ k Bk k  B , 0 B + n B  k C b a A c

B2−(k−µ)2 2(k−µ) Bk , − k B + n + k − 2µ

Figure 9: Labels of circles in the Apollonian packing [B, , n, k] Figure 4.2: Labels of circles in the Apollonian packing [B, µ, n, k]

We shall prove later that if the corresponding triples , ,  are integers, so are all triples in the gasket, ButProposition first: 4.1. The labels of the six right triangles in an Apollonian gasket whose hypotenuses join the pairs of the ﬁrst four greatest circles are, in notation as in Figure Proposition3.3 4.1: and The 4.2, labels as follows: of the six right triangles in an Apollonian gasket whose hypotenuses join the pairs of the first four greatest circles are, in notation as in Figure 7 and 9, as follows: k, 0, k a = −B(B + k) a = k, 0, k –B(B+k)n − 2B2/k, −2µB/k, n b = 2 −B(B + n) b = n–2B /k, –2B/k, n –B(B+n)k + n − 2µ − 2B2/k, 2B − 2µB/k, k + n − 2µ c = 2 −B(B + k + n − 2µ) k+n–2–2B /k, 2B–2B/k, k+n–2 (4.3) (4.3) c = −2b − k + 2µ + 4Bµ/k, −2(B − n + µ + 2B2/k), 2B + 2n + k − 2µ A = –B(B+k+n–2) (B + n)(B + k + n − 2µ) -2B–k+2+4B/k, -2(B–n++2B2/k), 2B+2n+k–2 A = 2 −2B + n(B+n)(B+k+n–2− 2µ − 2B /k,) 2(B + k − µ − 2Bµ/k), 2B + n + 2k − 2µ B = −B(B + k + n − 2µ) -2B+n –2 –2B2/k, 2(B+k––2B/k), 2B+n+2k–2 B = 2B + k − n + 2B2/k, 2µ + 2µB/k, 2B + k + n C = –B(B+k+n-2) (B + k)(B + n) 2 C = 2B+k–n+2B /k, 2+2B/k, 2B+k+n Proof. Direct calculation from the data of Proposition 3.2. (B+k)(B+n)

Proof: Direct calculationThe master from equation the data is of used Proposition to bring 3.1. the triples to a form where the only fractional terms contain B/k as a factor. Note that, in general, each of the entries for the reduced coordinates is a linear combination with integer coeﬃcients of B, k, n and µ plus the only, possibly, non-integer term, a multiple of 2Bµ/k or 2B2/k. The fact that it can be done proves the following theorem: 9

Theorem 4.2. If k|2B2 (or, equivalently k|2Bµ) then all triples (4.1) in the integer Apollonian gasket (B, µ, k) are integers (form Pythagorean triples). Jerzy Kocik Diophantine equation of Apollonian gakets 11

Proof. The following implication is direct k|2B2 ⇒ k|(2µk − 2µ2) ⇒ k|2µ2. Combining the premise with the result: k2|2B22µ2 ⇒ k2|4(Bµ)2 ⇒ k|2µB. Every entry of (4.3) may be expressed as a linear integral combination of B, k, n, µ and an extra term of either 2Bµ/k or 2Bµ/k. This proves the claim. Integer Pythagorean triples happen frequently. For instance, each of the conditions: µ = 0, or k = 1, or k = 2, is suﬃcient.

5. Linear recurrence for Pythagorean triples in an Apollonian gasket

Consider four circles in Descartes conﬁguration, C1,..., C4 (see Figure 5.1). The four centers determine six segments that we view as hypotenuses of triangles. They give, after rescaling, six Pythagorean triangles with sides denoted as follows:

horizontal side: ∆i j = bib j(xi − x j) = x˙ib j − x˙ jbi vertical side: Γi j = bib j(yi − y j) = y˙ib j − y˙ jbi (5.1) hypotenuse: Bi j = bib j(1/bi + 1/b j) = bi + b j with i, j = 1,... 4. Note that ∆i j = −∆ ji and Γi j = −Γ ji but Bi j = B ji. They form the frame of the Descartes configuration. 0 Now, complement the picture with a circle C4, Boyd-dual to C1 (see Figure 5.1). 0 This leads to a new frame, the frame of the Descartes configuration C4, C2, C3, C4. Quite interestingly, the elements of the new frame are linear combinations of the elements of the initial frame. Theorem 5.1. Following the notation of (5.1) and Figure 3.3, the transition of frames is given by 0 ∆41 = −∆41 + 2∆21 − 2∆13 0 0 0 Γ41 = −Γ41 + 2Γ21 − 2Γ13 0 B41 = −B41 + 2B12 + 2B13 Proof. Using the linear relations (1.2) and (3.1), we can express the position and the curvature of the new, fifth, circle in terms of the initial four: 0 x˙4 = 2x ˙1 + 2x ˙2 + 2x ˙3 − x˙4 0 y˙4 = 2˙y1 + 2˙y2 + 2˙y3 − y˙4 0 b4 = 2b1 + 2b2 + 2b3 − b4 0 We can calculate the Pythagorean vectors for each pair C4Ci, i = 1, 2, 3. By some magic, due to the following regroupings and adding zeros, these can be expressed in 0 terms of the initial four Pythagorean vectors only, as shown here for the pair C4C1: 0 0 0 ∆41 = x˙4b1 − x˙1b4 = (2x ˙1 + 2x ˙2 + 2x ˙3 − x˙4)b1 − x˙1(2b1 + 2b2 + 2b3 − b4)

= x˙1b4 − x˙4b1 + 2(x ˙2b1 − x˙1b21) + 2(x ˙3b1 − x˙1b3)

= −Delta41 + 2∆21 − 2∆13 .

We can calculate the Pythagorean vectors for each pair C’4 Ci, i=1,2,3. By some magic, due Jerzy Kocikto the following regroupings Diophantine and equationadding zeros, of Apollonianthese can be expressed gakets in terms of the initial 12 four Pythagorean vectors only, as shown here for the pair C’4C1:

Δ=''41xb 4 1 − xb 1 ' 4 =++−−(2x 2xxxbxbbbb 2 0 ) (2 ++− 2 2 ) The Γ’s follow the same argument.1234111234 Only B s are slightly diﬀerent: =−+xb14 x 41 b2( x 21 b − xb 1 2 1 ) + 2(xb 31 − xb 13 ) =−Δ+Δ−Δ0 41022 21 13 B41 = b4 + b1 The Γ’s follow the same argument. Only B’s are slightly different: = −b4 + 3b1 + 2b2 + 2b3 Bbb''41=+ 4= 4 1b + 2b + 2b − (b + b ) =−bbbb +321 + + 2 3 1 4 412 3 =++−+42bb12314= −B41 2( bbb+ 2B12 + )2B13 . =−BBB41 +22 12 + 13

Thus the transformationThus the transformation is linear is linear in thein the entries entries ofof ∆Δ,, ΓΓ, ,and, and, Β, Bwith, with integer integer coefficients, coeﬃ ascients, as stated. stated.

2 2 1 1 4 4

3 3

Figure 5.1: (a) A frame (b) and an adjacent frame. Circle 40 is shaded. Figure 10: (a) A frame (b) and an adjacent frame. Circle 4' is shaded

Actually,Actually, we havewe have proved proved a little little more: more:

Corollary 5.2: If the (scaled) triangles formed by the initial Descartes configuration are Corollaryinteger, 5.2. soIf are the all (scaled) of the triangles triangles of the Apollonian formed by window. the initial Descartes conﬁguration are integer, so are all of the triangles of the Apollonian window.

Matrix description: Now we may rephrase our ﬁndings on transitions between frames in terms of matrices. The data for the Pythagorean triangles in each frame may be expressed as three vectors (columns), e.g., for the initial frame: ∆ =

T [∆41, ∆42, ∆43, ∆12, ∆23, ∆31] and similarly for Γ and B. Similarly for the second frame we have ∆ 0, Γ0 and B0.

Here are the three matrices of the transition from a frame to the corresponding subsequent frame:

 1 0 0 0 0 0   0 0 0 0 0 −1       0 0 0 0 0 1   0 1 0 0 0 0 11  0 0 0 0 −1 0   0 0 0 1 0 0  A =   B =    0 0 0 1 −2 −2   0 −2 −1 2 0 0       2 0 1 0 2 0   0 0 0 −2 1 −2      −2 −1 0 0 0 2 1 2 0 0 0 2    0 0 0 0 1 0   −   0 0 0 1 0 0   0 0 1 0 0 0  C =    0 1 2 2 0 0     −1 0 −2 0 2 0    0 0 0 −2 2 1 Jerzy Kocik Diophantine equation of Apollonian gakets 13

(vertical and horizontal lines in the matrices are drawn for easier inspection). Now acting with them on the initial vectors ∆ and Γ one may reconstruct all the interior triangles. Vectors ∆ and Γ may be combined into a single matrix, T = [∆|Γ].     ∆41 Γ41 B41     ∆42 Γ42 B42     ∆43 Γ43 0 B43 T =   B =   ∆12 Γ12 B12     ∆23 Γ23 B23     ∆31 Γ31 B31

The vector B is transformed to B0 by matrices like A, B, C, where all of the entries are to be replaced by their absolute values. We conclude with two examples of Apollonian packings in which every triangle for adjacent circles determines a Pythagorean triple, see Figures 5.1 and 6.1. Only some of the triangles are displayed.

6. Concluding remarks

Integral Apollonian disk packings have a topic of much interest for a while [2]-[4], [8]–[9] and their occurrence has been analyzed. Yet it is not clear whether the integral Apollonian disk packings admit coordinate systems in which the reduced coordinates are integral, except the known cases of Apollonian Strip and Apollonian Window. We have however a frequent occurrence of gaskets that admit integral Pythagorean triples, constructed in Sections 4 and 5. Yet another intriguing remark concerns the master equation

B2 + µ2 − nk = 0.

If we see it as B2 + µ2 = hypotenuse2 – one may identify the corresponding (scaled) triangle in Figure 2.3 as the one with legs being the diameter 2/B of the encompassing circle and altitude h of the center of circle B2 – the inversive image of circle B2. Each solution to the master equation may also be viewed as an integral isotropic vector [B, µ, n, k]T in the Minkowski space R3,1 a la Pedoe map from circles to vectors, see [6]. Such vectors represent points in this Minkowski space – degenerated circles of radius 0. Jerzy Kocik Diophantine equation of Apollonian gakets 14

0, 4 Main circle 0, 0 10 3 , 2  2 2 13 , 4 7 2 19 3,4,5 Main circle 0, 4 14 5,12,13 1 0, 0 10 2, 0 42  2 , 0 3 , 2  2 2 6 3,4,5 3 13 , 4 7 2 14 19 8,6,10 3 21 3,4,5 2 ,  2 14 5,12,13 1 2, 0 42  2 , 0 7 15,8,17 6 3,4,570 3 14

8,6,10 3 21 2 ,  2 Figure 10: Arrangement (b ,b ,b ,b ) = (–2,3,6,7), or (B, , k, n) = (2, 4, 4, 5). Left side shows Figure 6.1: Arrangement (b0, b1,0b21, b273)3 = (−2, 3, 6, 7) or (B, µ, k, n) = (2, 4, 4, 5). Left side symbols of the principal and other circles, right side shows Pythagorean triples. 15,8,17 shows symbols of the principal and other circles, right side shows70 Pythagorean

triples.

Figure 10: Arrangement (b ,b ,b ,b ) = (–2,3,6,7), or (B, , k, n) = (2, 4, 4, 5). Left side shows 0 1 2 3 symbols of the principal and other circles, right side shows Pythagorean triples. Main circle 0, 0  8 , 5 0, 3 3  3 12 20

16,12,20 4 ,1 Main circle 3 96

0, 0 8  8 , 5 0, 3  3 3 12,5,13 20 12 6, 0 40  2 , 0 3 21 4 16,12,204,3,5 5 ,1 96 3 12,5,13 24 20,21,29 4 8 40 , 1 168 3 12,5,13 8 6, 0 40  2 , 0 3 21 4,3,516,12,20 8 0,  3  3 ,5 5 96 12,5,13 24 20 20,21,29 12 4 40 , 1 168 3 8

16,12,20  8 ,  5 0,  3 Figure 11: Arrangement3 (b ,b , b , b ) = (–3, 5, 8, 8), or (B, , k, n) = (3, 4, 5, 5) Both96 symbols 0 1 2 3 (left) and Pythagorean20 triples12 (right). The triangles are reported with absolute values of sides.

igure − F 6.2:Figure Arrangement 11: Arrangement (b0, (bb10,,bb1,2 ,bb2, 3b)3)= = (–3,( 3 ,5,5 8,, 8 8),, 8) or or (B, (B ,, µ,k, kn), n=) (3,= 4,(3 5,, 45), 5 Both, 5). symbols Both13 sym- (left) bols and (left)Pythagorean and Pythagorean triples (right). triples The triangl (right).es areThe reported triangles with absolute are reported values of with sides. absolute values of sides.

13 Jerzy Kocik Diophantine equation of Apollonian gakets 15

Appendix: List of integral packings A. List of integral packings HereAppendix: we provide List a list of of integral the first packings 200 intege r Apollonian packings. Fist, however, a few wordsHere on we symmetry provide aof listthese of arrangements. the ﬁrst 200 integer Apollonian packings. First, however, a fewHere words we provide on symmetry a list of the of first these 200 arrangements. integer Apollonian packings. Fist, however, a few skew words on symmetry of these arrangements.

c skew

b b b b c b b

b b b b e c d c e d c c (a, a, b, b, b) (a, b, b, c, c) (a, b, c, d, e)

Apollonian belt Apollonian Window General Apollonian gasket (a, a, b, b, b) (a, b, b, c, c) (a, b, c, d, e) Rich symmetry Symmetry: D Trivial symmetry: C Apollonian belt Apollonian Window2 General Apollonian gasket1 Rich symmetry Symmetry: D2 Trivial symmetry: C1

odd even even* odd even even* c b b c b b d b b b d d d b d c d c c c c c d d (a, b, c, d, d) (a, b, c, c, d) (a, b, b, c, d) (a, b, c, d, d) (a, b, c, c, d) (a, b, b, c, d) Odd symmetric Even symmetric Even symmetric inverted Symmetry:Odd symmetric D EvenSymmetry: symmetric D Even symmetricsymmetry inverted D Symmetry: D1 Symmetry: D 1 symmetry D 1  = 0 1 k =  1 k = n 1  = 0 k =  k = n

Figure A1: Types of symmetries and arrangements. The second circle, b, is thicker. FigureFigure A1: A1: Types Types of of symmetries symmetries andand arrangements.arrangements. TheThe second second circle, circle, b ,b ,is is thicker thicker

Only two integral Apollonian packings have rich symmetry: that of Apollonian OnlyStrip,Only two two and integral integral of Apollonian Apollonian Apollonian Window packings packings (D havehave2). Otherwiserich symmetry: we have that that of twoof Apollonian Apollonian cases, that Strip, Strip, of and symme- and of of Apollonian Window (D ). Otherwise we have two cases, that of symmetry C and D , which tryApollonianC1 and DWindow1, which (D2 2 will). Otherwise be labeled we ashave follows: two cases, that of symmetry C1 and1 D1, 1which willwill be be labeled labeled as as follows: follows:

Skew means no nontrivial symmetry (thus symmetry of type C1). Skew Skew means means no no nontrivial nontrivial symmetry symmetry (thus(thus symmetrysymmetry ofof typetype C C1).1).

For For Forcases cases cases with with withsingle single single axis axis symmetry axissymmetry symmetry ((DD11)) wewe (D shall1) we distinguishdistinguish shall distinguish two two cases: cases: two cases:

Odd – mirror symmetry D1, with exactly three circles cut by the axis of symmetry Odd Odd – mirror– mirror symmetry symmetry D1, withD exactly1, with three exactly circles three cut circles by the cut axis by of the symmetry axis of sym- Even – mirror symmetry with infinitely many circles on the axis Even metry – mirror symmetry with infinitely many circles on the axis Even* Even* – –even even symmetry symmetry that that is is obtained obtained “accidentally”“accidentally” byby algorithm algorithm in in Figure Figure 3, 3, where the two greatest congruent circles are not images of the congruent disks Even where– mirrorthe two symmetry greatest congruent with inﬁnitely circles manyare not circles images on of thethe congruent axis disks in in the the Apollonian Apollonian strip strip

Even* – even symmetry that is obtained “accidentally” by algorithm in Fig- ConsultConsult Figure Figure A1 A1 for for visualization. visualization. ure 2.2, where the two greatest congruent circles are not images of the congruent

disks in the Apollonian strip

Consult Figure A1 for visualization. 14 14 Jerzy Kocik Diophantine equation of Apollonian gakets 16

Yet another attribute of an Apollonian packing is a shift, which we deﬁne as the degreeYet another the diskattribute on theof an horizontal Apollonian axis packing in the is Apollonian a shift, which strip we isdefine raised as abovethe degree this the axis indisk terms on the of horizontal the fraction axis of in its the radius. Apollonian strip is raised above this axis in terms of the fraction of its radius. 2µ 2 shiftshift == hh//ρ == . k

SeeSee Figure 4. 2.3. It Ithas has values values in the in theinterval interval [0,1] [0,1] and it and measures it measures the degree thedegree the gasket the is gasket off isthe o ﬀaxialthe symmetry: axial symmetry: 0 for odd 0 for symmetryodd symmetry D1 and 1D for1 and even 1 symmetry for even symmetryD1. The fractionalD1. The fractionalvalue indicates value lack indicates of axial lacksymmetry, of axial except symmetry, the cases except of “accidental” the cases even* of “accidental” symmetry, even*when n symmetry, = k (this gives when B1 =n B=2)k (this gives B1 = B2).

TableTable A1 displays A1 displays first 24 ﬁrst integral 24 integral Apollonian Apollonian gaskets gasketsincluding including symmetry symmetrytype, principal type, principaldisk quintet, disk the quintet, parameters the parametersB, k, n, , andB, kthe, n ,shiftµ, and factor. the The shift first factor. even* The case ﬁrst is even*also caseincluded. is also included. Table A2 contains a longer list of integer Apollonian gaskets for the principal Table A2 contains a longer list of integer Apollonian gaskets for the principal curvatures curvaturesfrom 1 through from 32. 1 Curvature through 32.quintets Curvature (in brackets) quintets are followed (in brackets) by shift are factor. followed by shift factor.

symmetry Curvatures Four indices type -B0 B1 B2 B3 B4 B k n mu shift 1 1 odd 1 2 2 3 3 1 1 1 0 0 2 1 odd 2 3 6 7 7 2 1 4 0 0 3 1 odd 3 4 12 13 13 3 1 9 0 0 4 2 even 3 5 8 8 12 3 2 5 1 1 5 1 odd 4 5 20 21 21 4 1 16 0 0 6 2 even 4 8 9 9 17 4 4 5 2 1 7 1 odd 5 6 30 31 31 5 1 25 0 0 8 2 even 5 7 18 18 22 5 2 13 1 1 9 1 odd 6 7 42 43 43 6 1 36 0 0 10 2 odd 6 10 15 19 19 6 4 9 0 0 11 3 skew 6 11 14 15 23 6 5 8 2 4/5 12 1 odd 7 8 56 57 57 7 1 49 0 0 13 2 even 7 9 32 32 36 7 2 25 1 1 14 3 skew 7 12 17 20 24 7 5 10 1 2/5 15 1 odd 8 9 72 73 73 8 1 64 0 0 16 2 skew 8 13 21 24 28 8 5 13 1 2/5 17 3 even 8 12 25 25 33 8 4 17 2 1 18 1 odd 9 10 90 91 91 9 1 81 0 0 19 2 even 9 11 50 50 54 9 2 41 1 1 20 3 skew 9 14 26 27 35 9 5 17 2 4/5 21 4 skew 9 18 19 22 34 9 9 10 3 2/3 22 1 odd 10 11 110 111 111 10 1 100 0 0 23 2 odd 10 14 35 39 39 10 4 25 0 0 24 3 skew 10 18 23 27 35 10 8 13 2 1/2 … 34 6 even* 12 25 25 28 48 12 13 13 5 10/13

T able A1: Bend quintetsTable A1: for Bend Apollonian quintets integerfor Apollonian disk packingsinteger disk for packings principal for curvatures 1

through 15,principal including curvatures four indices1 through and 15, shifts. including four indices and shifts.

15 Jerzy Kocik Diophantine equation of Apollonian gakets 17

1 … (0, 0, 1, 1, 1) 1 odd (17, 18, 306, 307, 307) 0 1 odd (26, 27, 702, 703, 703) 0 2 even (17, 19, 162, 162, 166) 1 2 odd (26, 30, 195, 199, 199) 0 1 even (1, 2, 2, 3, 3) 0 3 skew (17, 22, 75, 78, 82) 2/5 3 skew (26, 31, 162, 163, 171) 4/5 1 odd (2, 3, 6, 7, 7) 0 4 skew (17, 27, 46, 54, 58) 1/5 4 skew (26, 34, 111, 115, 123) 1/2 5 skew (17, 30, 42, 43, 67) 12/13 5 skew (26, 43, 66, 79, 87) 4/17 1 odd (3, 4, 12, 13, 13) 0 6 skew (26, 51, 55, 66, 94) 14/25 2 even (3, 5, 8, 8, 12) 1 1 odd (18, 19, 342, 343, 343) 0 7 skew (26, 46, 63, 67, 99) 4/5 2 odd (18, 22, 99, 103, 103) 0 1 odd (4, 5, 20, 21, 21) 0 3 skew (18, 23, 83, 86, 90) 2/5 1 odd (27, 28, 756, 757, 757) 0 2 even (4, 8, 9, 9, 17) 1 4 skew (18, 31, 43, 54, 58) 2/13 2 even (27, 29, 392, 392, 396) 1 1 odd (5, 6, 30, 31, 31) 0 5 skew (18, 26, 59, 63, 71) 1/2 3 skew (27, 32, 173, 176, 180) 2/5 2 even (5, 7, 18, 18, 22) 1 6 skew (18, 27, 55, 58, 70) 2/3 4 skew (27, 37, 100, 108, 112) 1/5 7 skew (18, 35, 38, 47, 63) 8/17 5 skew (27, 36, 109, 112, 124) 2/3 1 odd (6, 7, 42, 43, 43) 0 6 skew (27, 45, 68, 80, 92) 1/3 2 odd (6, 10, 15, 19, 19) 0 1 odd (19, 20, 380, 381, 381) 0 7 skew (27, 40, 85, 88, 108) 10/13 3 skew (6, 11, 14, 15, 23) 4/5 2 even (19, 21, 200, 200, 204) 1 8 skew (27, 53, 56, 72, 92) 5/13 3 skew (19, 24, 92, 93, 101) 4/5 1 odd (7, 8, 56, 57, 57) 0 9 skew (27, 44, 72, 77, 101) 12/17 4 skew (19, 29, 56, 60, 72) 2/5 2 even (7, 9, 32, 32, 36) 1 10 skew (27, 52, 61, 64, 108) 22/25 5 skew (19, 32, 48, 53, 69) 8/13 3 skew (7, 12, 17, 20, 24) 2/5 6 skew (19, 36, 44, 45, 77) 16/17 1 odd (28, 29, 812, 813, 813) 0 1 odd (8, 9, 72, 73, 73) 0 2 odd (28, 44, 77, 93, 93) 0 1 odd (20, 21, 420, 421, 421) 0 2 skew (8, 13, 21, 24, 28) 2/5 3 skew (28, 33, 185, 188, 192) 2/5 2 odd (20, 36, 45, 61, 61) 0 3 even (8, 12, 25, 25, 33) 1 4 even (28, 32, 225, 225, 233) 1 3 even (20, 24, 121, 121, 129) 1 5 skew (28, 41, 89, 96, 108) 6/13 1 odd (9, 10, 90, 91, 91) 0 4 skew (20, 33, 52, 57, 73) 8/13 6 skew (28, 53, 60, 77, 93) 8/25 2 even (9, 11, 50, 50, 54) 1 5 skew (20, 37, 45, 52, 72) 10/17 7 skew (28, 48, 69, 77, 101) 3/5 3 skew (9, 14, 26, 27, 35) 4/5 6 even (20, 36, 49, 49, 81) 1 8 skew (28, 45, 77, 80, 108) 14/17 4 skew (9, 18, 19, 22, 34) 2/3 1 odd (21, 22, 462, 463, 463) 0 9 even (28, 44, 81, 81, 113) 1 1 odd (10, 11, 110, 111, 111) 0 2 odd (21, 30, 70, 79, 79) 0 10 skew (28, 57, 60, 65, 113) 24/29 2 odd (10, 14, 35, 39, 39) 0 3 even (21, 23, 242, 242, 246) 1 1 odd (29, 30, 870, 871, 871) 0 3 skew (10, 18, 23, 27, 35) 1/2 4 skew (21, 34, 55, 66, 70) 2/13 2 even (29, 31, 450, 450, 454) 1 5 skew (21, 38, 47, 62, 66) 2/17 1 odd (11, 12, 132, 133, 133) 0 3 skew (29, 34, 198, 199, 207) 4/5 6 skew (21, 26, 110, 111, 119) 4/5 2 even (11, 13, 72, 72, 76) 1 4 skew (29, 42, 94, 103, 111) 4/13 7 skew (21, 30, 71, 74, 86) 2/3 3 skew (11, 16, 36, 37, 45) 4/5 5 skew (29, 39, 114, 118, 130) 2/5 8 skew (21, 31, 66, 70, 82) 3/5 4 skew (11, 21, 24, 28, 40) 2/3 6 skew (29, 46, 79, 90, 102) 6/17 9 skew (21, 39, 46, 58, 70) 1/3 7 skew (29, 54, 63, 82, 94) 6/25 1 odd (12, 13, 156, 157, 157) 0 10 even (21, 39, 50, 50, 86) 1 8 skew (29, 55, 66, 70, 114) 11/13 2 odd (12, 21, 28, 37, 37) 0 1 odd (22, 23, 506, 507, 507) 0 3 skew (12, 17, 41, 44, 48) 2/5 1 odd (30, 31, 930, 931, 931) 0 2 odd (22, 26, 143, 147, 147) 0 4 even (12, 16, 49, 49, 57) 1 2 odd (30, 34, 255, 259, 259) 0 3 skew (22, 27, 119, 122, 126) 2/5 5 skew (12, 21, 29, 32, 44) 2/3 3 odd (30, 39, 130, 139, 139) 0 4 skew (22, 30, 83, 87, 95) 1/2 6 skew (12, 25, 25, 28, 48) 10/13 4 odd (30, 55, 66, 91, 91) 0 5 skew (22, 39, 51, 62, 74) 6/17 5 skew (30, 47, 83, 98, 102) 2/17 1 odd (13, 14, 182, 183, 183) 0 6 skew (22, 42, 47, 59, 75) 2/5 6 skew (30, 38, 143, 147, 155) 1/2 2 even (13, 15, 98, 98, 102) 1 7 skew (22, 35, 62, 63, 87) 12/13 7 skew (30, 39, 131, 134, 146) 2/3 3 skew (13, 18, 47, 50, 54) 2/5 1 odd (23, 24, 552, 553, 553) 0 8 skew (30, 55, 67, 82, 102) 2/5 4 skew (13, 23, 30, 38, 42) 1/5 2 even (23, 25, 288, 288, 292) 1 9 skew (30, 43, 102, 103, 127) 12/13

3 skew (23, 28, 129, 132, 136) 2/5 10 skew (30, 59, 66, 71, 119) 24/29 1 odd (14, 15, 210, 211, 211) 0 4 skew (23, 33, 76, 84, 88) 1/5 2 odd (14, 18, 63, 67, 67) 0 1 odd (31, 32, 992, 993, 993) 0 5 skew (23, 36, 64, 73, 81) 4/13 3 skew (14, 19, 54, 55, 63) 4/5 2 even (31, 33, 512, 512, 516) 1 6 skew (23, 40, 57, 60, 88) 14/17 4 skew (14, 22, 39, 43, 51) 1/2 3 skew (31, 44, 105, 116, 120) 2/13 7 skew (23, 48, 49, 52, 96) 22/25 5 skew (14, 27, 31, 34, 54) 10/13 4 skew (31, 57, 68, 92, 96) 1/13 1 odd (24, 25, 600, 601, 601) 0 5 skew (31, 36, 224, 225, 233) 4/5 1 odd (15, 16, 240, 241, 241) 0 2 odd (24, 33, 88, 97, 97) 0 6 skew (31, 41, 128, 132, 144) 2/5 2 odd (15, 24, 40, 49, 49) 0 3 even (24, 28, 169, 169, 177) 1 7 skew, 31, 48, 89, 96 ) ft= 10/17 3 even (15, 17, 128, 128, 132) 1 4 skew (24, 29, 140, 141, 149) 4/5 8 skew (31, 60, 65, 84, 104) 10/29 4 skew (15, 24, 41, 44, 56) 2/3 5 skew (24, 44, 53, 69, 77) 1/5 9 skew (31, 56, 72, 81, 113) 16/25 5 skew (15, 28, 33, 40, 52) 6/13 6 skew (24, 33, 89, 92, 104) 2/3 6 skew (15, 32, 32, 33, 65) 16/17 1 odd (32, 33, 1056, 1057, 1057) 0 7 skew (24, 37, 69, 76, 88) 6/13 1 odd (16, 17, 272, 273, 273) 0 2 skew (32, 37, 237, 240, 244) 2/5 8 skew (24, 40, 61, 69, 85) 1/2 2 even (16, 20, 81, 81, 89) 1 3 skew (32, 57, 73, 96, 100) 2/25 9 skew (24, 41, 60, 65, 89) 12/17 3 skew (16, 21, 68, 69, 77) 4/5 4 even (32, 36, 289, 289, 297) 1 10 skew (24, 49, 49, 60, 88) 14/25 4 skew (16, 29, 36, 45, 53) 4/13 5 skew (32, 45, 112, 117, 133) 8/13 5 skew (16, 32, 33, 41, 57) 1/2 1 odd (25, 26, 650, 651, 651) 0 6 skew (32, 48, 97, 105, 121) 1/2 2 even (25, 27, 338, 338, 342) 1 7 skew (32, 52, 85, 93, 117) 2/3

3 skew (25, 42, 62, 75, 83) 4/17 8 skew (32, 61, 69, 84, 112) 14/29

4 skew (25, 38, 75, 78, 98) 10/13 9 skew (32, 49, 96, 97, 129) 16/17

5 skew (25, 50, 51, 66, 86) 2/5

6 skew (25, 50, 54, 59, 99) 4/5

Table A2: Bend quintetsTable A2:for Bend Apollonian quintets for integerApollonian disk integer packings disk packings for principal curvatures 1 throughfor 32. principal curvatures 1 through 32. 16 Jerzy Kocik Diophantine equation of Apollonian gakets 18

References

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[11] J. B. Wilker, Inversive Geometry, in: The Geometric Vein, (C. Davis, B. Grün- baum, F.A. Sherk, Eds.), Spring-Verlag: New York 1981, pp. 379–442.