ISOSCELES TRIANGLES IN Q 3. Matt Noble Department of Mathematics, Middle Georgia State University, Macon, Georgia

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1 #A9 INTEGERS 18 (2018) ISOSCELES TRIANGLES IN Q Matt Noble Deartment of Mathematics, Middle Georgia State University, Macon, Georgia matthew.noble@mga.edu Received: 7/2/17, Acceted: 2//18, Published: 2/19/18 Abstract In this article, we give a characterization of isosceles triangles having as their vertices oints of the rational Euclidean sace Q. For ositive integers d and r where d and r are both realized as distances in Q and r a 2 + b 2 + c 2 for integers a, b, c, it is shown that the triangle with edges of length d, d, and r is embeddable in Q if and only if the Diohantine euation x 2 + ry 2 (4d r)(a 2 + b 2 )z 2 0 has a non-trivial integer solution. This condition is used to give new roofs of a few recent results concerning G(Q, r), the Euclidean distance grah with vertex set Q where adjacent vertices are at distance r aart. Originally shown by E. J. Ionascu, we rove that G(Q, r) contains K as a subgrah if and only if the suare-free art of r is even but has no odd factor congruent to 2 (mod ). We conclude by roving that each of those grahs G(Q, r) which contain K as a subgrah have chromatic number Introduction As is customary, the rings of real numbers, rational numbers, and integers are given by R, Q, and Z, resectively. For any S R, S n denotes the set of all n-tules whose coordinate entries are elements of S. Throughout this aer, the distance between oints of any S n will be calculated using the standard Euclidean distance metric. For a, b, c 2 R +, we will denote by T (a, b, c) the triangle having edges of length a, b, and c, resectively. For any S R n, define T (a, b, c) to be embeddable, or alternately reresentable, in S if the triangle can be oriented such that its vertices are oints of S. We will hrase a large ortion of our work in the language and notation of Euclidean distance grahs. For d > 0 and S R n, let G(S, d) denote the grah with vertex set S where any two vertices are adjacent if and only if they are a Euclidean distance d aart. Such grahs have been widely studied (see [7], [14], or [15]) with the foremost structural uestion being the determination of the chromatic number (S, d). In other words, (S, d) reresents the minimum number of colors

2 INTEGERS: 18 (2018) 2 needed to color the vertices of G(S, d) such that no two adjacent vertices receive the same color. Central to our work will also be a few fundamentals from classical number theory. We exect the majority of readers to be familiar with these concets, but in the event clarification is needed, one could consult virtually any textbook on elementary number theory, with our ersonal reference being [9]. For integers a, b, say that a is a uadratic residue of b if there exists an integer x such that x 2 a (mod b). Often the word uadratic is droed from rint, which will be the case here as well. For an integer a and rime, define the Legendre symbol a 8 < : 0 if a is a multile of 1 if a is a residue of 1 if a is not a residue of a as follows: An imortant multilicative roerty of the Legendre symbol is that for integers a, b and rime,. Additional roerties of the Legendre symbol ab a b relevant to our discussion are summarized by the law of uadratic recirocity which is given with its common sulements below. Theorem 1. Let, be odd, ositive rimes. Then each of the following is true. (i) If at least one of, is congruent to 1 (mod 4), then. (ii) If both of, are congruent to (mod 4), then (iii) (iv) 1 1 if and only if 1 (mod 4). 2 1 if and only if ±1 (mod 8).. The structure of the aer will be as follows. In Section 2, for ositive integers r and d, we give a necessary and su cient condition for T ( r, d, d) to be embeddable in Q. Suosing that r and d are both distances actually realized between oints of Q and r a 2 + b 2 + c 2 for a, b, c 2 Z with a, b not both eual to 0, it is shown that T ( r, d, d) is embeddable in Q if and only if the Diohantine euation x 2 + ry 2 (4d r)(a 2 + b 2 )z 2 0 has a non-trivial solution. Furthermore, we show that the solvability of this euation is indeendent of which articular a, b, c are chosen to satisfy the above conditions. In Section, we use this characterization to rove a variant of a result of Ionascu [4] concerning the ossible edge-lengths of euilateral triangles in Q. We address Euclidean distance grahs G(Q, r) in Section 4, giving a new roof of the author s recent result [12] that if G(Q, r) contains a triangle, then (Q, r) 4.

3 INTEGERS: 18 (2018) 2. Isosceles Triangles in Q We begin with a warmu roblem in the lane. Theorem 2. Let r, d 2 Q + where r and d are both realized as distances between oints of Q 2. An isosceles triangle T ( r, d, d) is embeddable in Q 2 if and only if r(4d r) 2 Q +. Proof. Let T T ( r, d, d). As described in [1] (and robably elsewhere as well), for vectors v 1, v 2 2 Q 2, there exists a distance-scaling bijection ' : Q 2! Q 2 such that '(v 1 ) v 2. In other words, for any oints 1, 2 2 Q 2, '( 1 ) '( 2 ) v 2 v With this in mind, we have that T is embeddable in Q 2 if and only if T 0 d T (1, r, d r ) is embeddable in Q2. Furthermore, we may assume that the side of T 0 of length 1 lies on the x-axis, say with endoints (0, 0) and (1, 0). It follows that T 0 is embeddable in Q 2 if and only if there exists a oint ( 1 2, y) 2 Q2 such that ( 1 2 )2 +y 2 d r, or stated more simly, if and only if r(4d r) 2 Q +. In Q, however, the situation is murkier. We do not have access to a -dimensional analogue of the transformation ' described in the roof of Theorem 2, and instead, we must make do with the following two lemmas. Lemma 1 is immediate as multilication by any ositive rational number is a distance-scaling bijection from any Q n to Q n. Lemma 2 follows from Rodrigues rotation formula and is given with roof in [12]. Lemma 1. For any 2 Q +, there exists a bijection : Q n! Q n that scales distance by a factor of. Lemma 2. For vectors v 1, v 2 2 Q with v 1 v 2, there exists a distancereserving bijection : Q! Q such that (v 1 ) v 2. In attemting to determine if a triangle T ( r, d, d) is embeddable in Q, without loss of generality we may assume by Lemma 1 that r, d are both integers. Now suose r a 2 + b 2 + c 2 for a, b, c 2 Z. The uestion of whether or not T ( r, d, d) is embeddable in Q is euivalent to the existence of a oint (x, y, z) 2 Q simultaneously at distance d from the origin and the oint (a, b, c). And in light of Lemma 2, although r may have many reresentations as a sum of three suares, it does not matter how the a, b, c are chosen. Consider the isosceles triangle T with vertices (0, 0, 0), (a, b, c), and (x, y, z). We may extend T to a right triangle T 0 having vertices (0, 0, 0), (a, b, c), and (2x, 2y, 2z) as in Figure 1 below, and we have that T is embeddable in Q if and only if T 0 is embeddable in Q. The uestion of whether T 0 is embeddable in Q is euivalent to whether there exist orthogonal Q vectors v 1 and v 2 such that v 1 ha, b, ci and v 2 h,, i where d r. This brings us to the main theorem of this section.

4 INTEGERS: 18 (2018) 4 Figure 1: Theorem. Let r, d 2 Z + where r and d are both realized as distances in Q. Let r a 2 + b 2 + c 2 for a, b, c 2 Z with a, b not both eual to zero. Then the triangle T ( r, d, d) is embeddable in Q if and only if the Diohantine euation x 2 + ry 2 (4d r)(a 2 + b 2 )z 2 0 has a non-trivial integer solution. Proof. Let T T ( r, d, d), and assume a 6 0. With regards to the comments above, T is embeddable in Q if and only if there exist,, 2 Q such that d r and a + b + c 0. We first consider the case of c 0. Hence a + b 0. If b 0, we have that a 0 imlies 0, which is euivalent to 4d r being reresentable as a sum of two suares. Now considering the euation x 2 + ry 2 (4d r)(a 2 + b 2 )z 2 0, we simlify to x 2 + a 2 y 2 (4d r)(a 2 )z 2 0 which, through absorbing the a 2 coe cient, is solvable if and only if x 2 + y 2 (4d r)z 2 0 is solvable. However, the last euation can be rewritten further as ( x z )2 + ( y z )2 4d r which is solvable if and only if 4d r can be exressed as a sum of two suares. If b 6 0, we write a b and substitute to get 2 + ( a b ) d r. This simlifies to 2 (1 + a2 b ) + 2 4d r where we may then write xb 2 z and y z, and then substitute again to get ( x z )2 (a 2 + b 2 ) + ( y z )2 4d r. Multilying both sides by z 2 (a 2 + b 2 ) and then absorbing the (a 2 + b 2 ) coe cient into x 2 leaves us with x 2 + ry 2 (4d r)(a 2 + b 2 )z 2 0. We now assume c 6 0 and again consider the euation d r. Write a +b c and substitute to obtain 2 a + b d r (1) c which simlifies to Euation 2 below. (a 2 + c 2 ) 2 + (b 2 + c 2 ) 2 + 2ab (4d r)c 2 (2)

5 INTEGERS: 18 (2018) 5 We now erform a linear transformation to eliminate the term. Let a b b a y 0 a 2 +b x 2 0 a 2 +b and y 2 0 a 2 +b + x 2 0 a 2 +b. After making this 2 substitution and executing the somewhat arduous rocess of simlifying, we are eventually left with Euation. c 2 x ry 0 2 (4d r)c 2 (a 2 + b 2 ) () Note that Euation 2 is solvable in rationals, if and only if Euation is solvable in rationals x 0, y 0. Now moving to homogeneous coordinates by letting x 0 x cz and y 0 y cz, we arrive at Euation 4. This comletes the roof of the theorem. x 2 + ry 2 (4d r)(a 2 + b 2 )z 2 0 (4) The advantage (at least as it aears to us) of Theorem is that the otential solvability of homogeneous uadratic Diohantine euations in three variables (like Euation 4) is addressed by a classical and well-known theorem of Legendre which we give as Theorem 4. For a roof, see [9]. Theorem 4. Let a, b, c be non-zero integers, not each ositive or each negative, and suose that abc is suare-free. Then the euation ax 2 + by 2 + cz 2 0 has a non-trivial integer solution (x, y, z) if and only if each of the following are satisfied: (i) (ii) ab is a uadratic residue of c ac is a uadratic residue of b (iii) bc is a uadratic residue of a. As an examle of the interlay between Theorems and 4, consider the triangle T ( 6, 10, 10). Writing , Euation 4 becomes x 2 + 6y 2 (4)(2)z 2 0. Absorbing suared terms into z, we have that the revious euation is solvable if and only if x 2 + 6y 2 17z 2 0 is solvable. Now alying Theorem 4 along with Theorem 1, we have that this euation is in fact not solvable as the Legendre symbol ( 2 ) 1 and thus 17 is not a residue of 6. So T ( 6, 10, 10) is not embeddable in Q. For another cavalierly-generated examle, consider T ( 14, 110, 110). Writing , Euation 4 becomes x y 2 (06)(1)z 2 0 which is solvable if and only if 2x y 2 (1)(17)z 2 0 is solvable. Maniulation of Legendre symbols in conjunction with Theorem 1 shows that each of the exressions

6 INTEGERS: 18 (2018) , , and are eual to 1. By Theorem, T ( 14, 110, 110) is embeddable in Q. And for those readers who remain sketical, we note that the oints (0, 0, 0), (9, 5, 2), and (2, 11, ) do indeed constitute the vertices of such a triangle. We close this section by remarking that Lemma 2 guarantees that for given r, d 2 Z +, either all airs of Q oints distance r aart form two vertices of the isosceles triangle T ( r, d, d) or no airs of Q oints distance r aart form two vertices of T ( r, d, d). However, when letting r a 2 + b 2 + c 2, it is not immediately obvious that the solvability of Euation 4 is indeendent of the selection of a and b. We allay those concerns with the following observation. Since r (a 2 + b 2 ) c 2, (a 2 + b 2 ) is always a residue of r and r is always a residue of (a 2 + b 2 ). Furthermore, by a well-known characterization of integers that are reresentable as a sum of two suares (again, see [9] for elaboration), we have that the suare-free art of (a 2 + b 2 ) can contain no rime factors congruent to (mod 4). By Theorem 1, r is a residue of the suare-free art of (a 2 + b 2 ) as well.. Euilateral Triangles in Q In a series of aers (see [4] and [5]), Ionascu develos a characterization of euilateral triangles whose vertices are oints of Z. Aearing in [4] is the following result which in the author s eyes has roven articularly useful to the study of Euclidean distance grahs G(Q, d). Theorem 5. Let r 2 Z +. An euilateral triangle of side length r is embeddable in Z if and only if the suare-free art of r is even but contains no odd factor congruent to 2 (mod ). Ionascu roves Theorem 5 by first establishing a few roositions concerning binary uadratic forms. However, in this section, we aim to rove a weaker version of his result using only Theorems and 4 along with uadratic recirocity. Note also that the embeddability of an euilateral triangle T ( r, r, r) in Q is euivalent to the existence of the comlete grah K as a subgrah of G(Q, r). Theorem 6. Let r 2 Z +. The distance grah G(Q, r) contains K as a subgrah if and only if the suare-free art of r is even but contains no odd factor congruent to 2 (mod ). Proof. Regarding Lemma 1, we may without loss of generality assume r to be suare-free. Writing r a 2 + b 2 + c 2 for integers a, b, c with a non-zero, and setting d r in Euation 4, we arrive at Euation 5 below. rx 2 + y 2 (a 2 + b 2 )z 2 0 (5)

7 INTEGERS: 18 (2018) 7 Let w be the suare-free art of (a 2 + b 2 ) and note that we may select the a, b so that w is odd. Furthermore, note that does not divide w as w is a suare-free sum of two suares. Let g gcd(r, w), and set r 0 r g and w 0 w g. Euation 6 now fits the reuired form of Theorem 4 and is solvable if and only if Euation 5 is solvable. r 0 x 2 + gy 2 w 0 z 2 0 (6) For su ciency, we must verify three things i) gw 0 is a residue of r 0, ii) gr 0 is a residue of w 0, and iii) w 0 r 0 is a residue of g. We first consider the case of r not divisible by. For i), we have gw 0 w. Since r (a 2 + b 2 ) c 2, and (a 2 + b 2 ) euals w times a suare, we have that w is a residue of r (and thus of r 0 ). We now need to show that is a residue of r 0 as well. Consider an odd rime dividing r 0 where 1 by hyothesis 1 (mod ). If 1 (mod 4), we have (1) 1. If (mod 4), we have ( 1)( 1) 1. For ii), we have gr 0 r. Since r (a 2 + b 2 ) c 2 and every rime factor of w 0 is congruent to 1 (mod 4), we have that r is a residue of w 0. Since r 2 (mod ), r 1 (mod ), and we are left with r being a residue of. For iii), let be a rime dividing g. Since r and w, we have that 1 (mod 12). Thus 1 1. Again considering r (a 2 + b 2 ) c 2, we rewrite this exression as r 0 g w 0 d 2 g c 2 for some integer d. This imlies c 0 (mod g), so write c c 0 g and then simlify to r 0 w 0 d 2 gc 2 0. This gives us r 0 w 0 d 2 (mod g), and it follows that r 0, w 0 are either both residues of g or both non-residues of g. In either case, we have that r0w 0 1. We are left with w 0 r 0 being a residue of g. In the case of r being divisible by, write r 0 r 1. Euation 6 is solvable if and only if the following Euation 7 is solvable. 1 r 1 x 2 + gy 2 w 0 z 2 0 (7) Regarding Theorem 4, to show Euation 7 is solvable, the only thing we need aart from the work already done above is to establish that r 1 w 0 is a residue of. As r r 1 2 (mod ), we have that 1 1. As r a 2 + b 2 + c 2 is congruent to 0 (mod ) but not 0 (mod 9), it follows that each of a, b, c 6 0 (mod ). Thus (a 2 + b 2 ) 2 (mod ) and wd 2 2 (mod ) imlies w 2 (mod ). Since each factor of g is congruent to 1 (mod ), we are left with w 0 2 (mod ) as well. So w0 1 and we have shown that the roduct r 1 w 0 is a residue of. This concludes the su ciency art of the argument. For necessity, we consider searately the ossibility of r being odd and the ossibility of r having some odd rime factor that is congruent to 2 (mod ). In each case we will show that Euation 6 is not solvable.

8 INTEGERS: 18 (2018) 8 First, suose that 2 (mod ) where is an odd rime and r. For solvability, we must have gw 0 w being a residue of of. As w is a residue of r, it is in turn a residue of, and we are left with reuiring 1. If 1 1 (mod 4), 2 1. If (mod 4), we have ( 1)( 1) 2 1. Thus Euation 6 or 7 (whichever 1 is alicable) is not solvable. Now assume that r is odd, and first consider the case of r not divisible by. Let be any rime dividing r 0. For solvability of Euation 6, we need w to be a residue of. By the argument in the receding aragrah, we have that 1 (mod ). However, for solvability of Euation 6, we also need r to be a residue of. This r gives a contradiction as 2 1. Considering the case of dividing r, we shift our attention to Euation 7 and note that for solvability, we need r 1 w 0 to be a residue of and w to be a residue of r 1. By revious arguments, w being a residue of r 1 imlies that is a residue of r 1 and thus every rime factor of r 1 is congruent to 1 (mod ). As r 1 is then congruent to 1 (mod ), for r 1 w 0 to be a residue of, we must have w 0 1 (mod ). This cannot haen because of the reviously mentioned reason of r a 2 + b 2 + c 2 being congruent to 0 (mod ) but not 0 (mod 9) results in (a 2 + b 2 ) 2 (mod ). It follows that w 0 2 (mod ) as well. This comletes the necessity ortion of the argument. 4. Euclidean Distance Grahs in Q The holy grail of Euclidean distance grah coloring has always been the determination of (R 2, 1), the fabled chromatic number of the lane. Those essimists among us may say that not only has it always been, it very likely will always be. It is known that 4 ale (R 2, 1) ale 7 and frustratingly, no imrovement has been made to these bounds in over sixty years of e ort. Interestingly enough, both bounds can be obtained through elementary methods, with the simlest derivation of the lower bound given by Leo and Willy Moser in [11] where they observe that the grah in Figure 2 is a subgrah of G(R 2, 1). This grah is commonly referred to as the Moser sindle and is easily seen to have chromatic number 4. For a mostly u-to-date reference on the status of coloring roblems in the rational sace Q n, see Johnson s comendium [7] on the subject. We will not go nearly as in-deth here, but still, we should relay a few facts of this case. Although Woodall was the first to consider coloring Q n, showing in a 1972 article [16] that (Q 2, 1) 2, the most influential work was done by Benda and Perles in [1]. In their manuscrit [1], they obtain a number of results on coloring Q n for small values of n and also ose a few still oen roblems including the following:

9 INTEGERS: 18 (2018) 9 Figure 2: Does there exist r 2 Z + such that G(Q, r)? Regarding Lemma 1, we may assume r to be suare-free. It is shown in [6] that for odd r, the grah G(Q, r) can be roerly 2-colored. For even r, it is shown in [2] that ale (Q, r) and in [8] that (Q, r) ale 4. So to comletely resolve Benda and Perles uestion, we are left with having to decide for each even suarefree r whether (Q, r) or (Q, r) 4. In [12], the author gives a artial answer with the following theorem given as a corollary to a few stronger results concerning colorings of the grah G(Z, r). Theorem 7. Let r 2 Z + where the suare-free art of r is even but contains no odd factor congruent to 2 (mod ). Then (Q, r) 4. The method of roof is straightforward. For each r it is shown that for some m, the grah in Figure is a subgrah of G(Q, r). We will refer to this grah as a generalized sindle, and it too is easily seen to have chromatic number 4. Figure : As a -dimensional reresentation of the generalized sindle is highly non-rigid, articularly at the vertices v 2,..., v m 1, the mechanics of the roof of Theorem 7 just consist of showing that the grah can be drawn in Q with v 1 and v m distance r aart. Now, this line of roof is satisfactory is establishing the theorem, but

10 INTEGERS: 18 (2018) 10 it is a little unsatisfying for the following reasons. First of all, not much is being said about the lacement of the vertices v 2,..., v m 1. A reresentation of the generalized sindle embedded as a Euclidean distance grah in Q would not at all resemble the way the grah is drawn in Figure. So, what would it look like? Additionally, nothing is being said about what values of m can be used to embed the grah in Q. With this in mind, we will use the remainder of this section to o er an alternate roof of Theorem 7, one that ossesses a more geometric flavor than that outlined above. An imortant layer in this new roof of Theorem 7 will be the Diohantine euation x 2 y 2 1 where is a ositive rime congruent to 11 (mod 12). Mollin gives Pell-tye Diohantine euations of this form an extensive treatment in [10], where the following theorem aears. Theorem 8. Let D > 2 with D not a erfect suare. Then l( D) is even if and only if one of the following holds: (i) There exists a factorization D ab with 1 < a < b such that ax 2 by 2 ±1 is solvable. (ii) There exists a factorization D ab with 1 ale a < b such that ax 2 by 2 ±2 is solvable. The arameter l( D) denotes the length of the eriod in the continued fraction exansion of D. A digression into the realm of continued fractions would take us far afield, but all that is really needed for our roof is observed in [], where it is shown that l( D) odd imlies that D has no rime factor congruent to (mod 4). Lemma. Let be a ositive rime congruent to 11 (mod 12). Then the Diohantine euation x 2 y 2 1 is solvable. Proof. Let D. Then l( D) is even and Theorem 8 imlies that one of the euations x 2 y 2 ±1, x 2 y 2 ±2, or x 2 y 2 ±2 is solvable. First, note that x 2 y 2 1 is not solvable, as considering both sides of the euation modulo, we see that a solution would reuire y 2 2 (mod ) which is imossible. Similarly, neither x 2 y 2 ±2 nor x 2 y 2 ±2 is solvable as either of the congruences x 2 y 2 2 (mod 4) or x 2 y 2 2 (mod 4) would contradict the fact that for any integer n, n 2 0 or 1 (mod 4). We are then left with the fact that x 2 y 2 1 must be solvable. Note that in any integer solution x, y of x 2 y 2 1, we must have x even, for if x was odd, x 2 (mod 4) imlies y 2 2 (mod 4) which in turn imlies that y 2 2 (mod 4) which is imossible. So writing x 2x 0, we substitute back into the original euation and arrive at the following corollary.

11 INTEGERS: 18 (2018) 11 Corollary 1. Let be a ositive rime congruent to 11 (mod 12). Then the Diohantine euation 12x 2 y 2 1 is solvable. The final tool we will need in our alternate roof of Theorem 7 is Dirichlet s well-known theorem on rimes in arithmetic rogressions. It is given below. Theorem 9. Let a, b 2 Z + where gcd(a, b) 1. Then the arithmetic rogression a, a + b, a + 2b,... contains infinitely many rimes. We are now ready for a roof of Theorem 7. Let r be a suare-free ositive integer that is even but contains no odd factor congruent to 2 (mod ). Our lan will be to use Theorem to show that for some n 2 Z, the isosceles triangle T ( r, n r, n r) is embeddable in Q. Such an embedding guarantees that (Q, r) 4 by the following rationale. First note that if T 1 and T 2 are euilateral triangles, each of side length r, that lie in the same lane and share exactly one edge, their unshared vertices must be distance r aart. Suose that T T ( r, n r, n r) is in fact embeddable in Q. Label the vertices of T as v, x n, y n where x n y n r. There are rational oints x 1,..., x n 1 that slit the edge from v to x n into n line segments, each of length r. As well, there are rational oints y 1,..., y n 1 that slit the edge from v to y n into n line segments, each of length r. Now designate by S the set of all two-element subsets { 1, 2 } Q such that 1 2 r. For each i 2 {1,..., n 1} and air of oints x i and x i+1, Theorems 2 and 6 together guarantee that there exists some Q 2 S such that the two oints of Q are at distance r from each of x i and x i+1. A similar claim can be made about the airs of oints y i and y i+1 for i 2 {1,..., n 1} and the airs of oints v, x 1 and v, y 1. So in summary, if we can use Theorem to show that T is embeddable in Q for some n, we have in e ect shown that a generalized sindle with m 2n is a subgrah of G(Q, r) and thus (Q, r) 4. We do that with the following theorem. Theorem 10. Let r be a suare-free, even ositive integer that contains no odd factor congruent to 2 (mod ). Then there exists n 2 Z + such that the triangle T ( r, n r, n r) is embeddable in Q. Proof. To show that T T ( r, n r, n r) is embeddable in Q, Theorem indicates that we must establish the solvability of the euation x 2 + ry 2 (12rn 2 r)(a 2 + b 2 )z 2 0 (8) where r a 2 + b 2 + c 2 for a, b, c 2 Z. Euation 8 is solvable if and only if Euation 9 below is solvable. rx 2 + y 2 (12n 2 1)(a 2 + b 2 )z 2 0 (9) Now suose r has rime factorization r where each i 1 (mod 4) and each j (mod 4). By the Chinese Remainder Theorem, there exists a ositive integer s satisfying the following system of linear congruences:

12 INTEGERS: 18 (2018) 12 (i) s 11 (mod 12). (ii) s (mod 8). (iii) For each i, s 1 (mod i ). (iv) For each j, s 1 (mod j ). Note that gcd(r, s) 1. By Theorem 9, the arithmetic rogression s, s + r, s + 2r,... contains infinitely many rimes. Let t be a rime in this seuence. Corollary 1 guarantees that the Diohantine euation 12n 2 tk 2 1 is solvable. So in Euation (9), we may relace the (12n 2 1) coe cient with (tk 2 ), then absorb the k 2 to arrive at the euation rx 2 + y 2 t(a 2 + b 2 )z 2 0, (10) whose solvability imlies the solvability of Euation (9). Regarding Theorem 4, to establish the solvability of Euation 10, we need to show that t(a 2 + b 2 ) is a residue of r and that r is a residue of both t and (a 2 + b 2 ). However, in Section 2 we already observed that ±r is a residue of (a 2 + b 2 ) and (a 2 +b 2 ) is a residue of r. So really, we only need that t is a residue of r and that t 1 1 r is a residue of t. For any i, we have i i i 1. For any j, we t 1 1 r have j j j 1. Thus t is a residue of r. We also have that t t t t t 1t t ( 1)( 1)(1) (1)(1) (1) 1. This shows that r is a residue of t, guaranteeing that Euation 10 is solvable and comleting the roof of the theorem. References [1] Miro Benda and Micha Perles, Colorings of metric saces, Geombinatorics 9 (January, 2000), Note: Although it did not aear in rint until the year 2000, this work was originally roduced in the mid-1970s. Coies of the unublished manuscrit circulated for many years, and its results were widely known. [2] T. Chow, Distances forbidden by two-colorings of Q and A n, Discrete Math. 115 (199), [] H. Davenort, The Higher Arithmetic, Cambridge University Press, Cambridge, 199. [4] E. J. Ionascu, A arametrization of euilateral triangles having integer coordinates, J. Integer Seuences 10 (2007), # [5] E. J. Ionascu, Counting all euilateral triangles in {0, 1,..., n}, Acta Math. Univ. Comenianae 77 (1) (2008),

13 INTEGERS: 18 (2018) 1 [6] Peter D. Johnson Jr., Two-colorings of a dense subgrou of Q n that forbid many distances, Discrete Math. 79 (1989/1990), [7] Peter D. Johnson, Jr., Euclidean distance grahs on the rational oints, Ramsey Theory: Yesterday, Today, and Tomorrow, Alexander Soifer, editor. Progress in Mathematics 285 (2010), Birkhauser, [8] Peter Johnson, Andrew Schneider, and Michael Tiemeyer, B 1 (Q ) 4, Geombinatorics 16 (Aril, 2007), [9] William J. LeVeue, Fundamentals of Number Theory, Addison Wesley, Reading, Massachusetts, [10] R. A. Mollin, A continued fraction aroach to the Diohantine euation ax 2 by 2 ±1, JP Jour. of Algebra, Number Theory, and Al. 4 (2004), [11] L. Moser and W. Moser, Solution to Problem 10, Can. Math. Bull. 4 (1961), [12] Matt Noble, On 4-chromatic subgrahs of G(Q, d), Australasian J. of Combinatorics 65 (1) (2016), [1] Matt Noble, An isomorhism roblem in Z 2, Theory and Alications of Grahs 2 (1) (2015), Article 1. [14] A. M. Raigorodskii, Coloring distance grahs and grahs of diameters, Thirty Essays on Geometric Grah Theory, J. Pach editor, Sringer, 201, [15] Alexander Soifer, The Mathematical Coloring Book, Sringer, [16] D. R. Woodall, Distances realized by sets covering the lane, J. Combin. Theory Ser. A 14 (197),