There are only finitely many Diophantine quintuples


 Nathaniel Green
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1 There are only finitely many Diophantine quintuples Andrej Dujella Department of Mathematis, University of Zagreb, Bijenička esta Zagreb, Croatia Abstrat A set of m positive integers is alled a Diophantine mtuple if the produt of its any two distint elements inreased by 1 is a perfet square. Diophantus found a set of four positive rationals with the above property. The first Diophantine quadruple was found by Fermat (the set {1, 3, 8, 120}). Baker and Davenport proved that this partiular quadruple annot be extended to a Diophantine quintuple. In this paper, we prove that there does not exist a Diophantine sextuple and that there are only finitely many Diophantine quintuples. 1 Introdution A set of m positive integers {a 1, a 2,..., a m } is alled a Diophantine mtuple if a i a j + 1 is a perfet square for all 1 i < j m. Diophantus first studied the problem of finding four numbers suh that the produt of any two of them inreased by unity is a square. He found a set of four positive rationals with the above property: {1/16, 33/16, 17/4, 105/16}. However, the first Diophantine quadruple, {1, 3, 8, 120}, was found by Fermat. Euler was able to add the fifth positive rational, / , to the Fermat s set (see [6], pp ). Reently, Gibbs [13] found examples of sets of six positive rationals with the property of Diophantus. A folklore onjeture is that there does not exist a Diophantine quintuple. The first important result onerning this onjeture was proved in 1969 by Baker and Davenport [3]. They proved that if d is a positive integer suh that {1, 3, 8, d} forms a Diophantine quadruple, then d = 120. This problem was stated in 1967 by Gardner [12] (see also [17]). In 1979 Arkin, Hoggatt and Strauss [1] proved that every Diophantine triple an be extended to a Diophantine quadruple. More preisely, let {a, b, } be a Diophantine triple and ab + 1 = r 2, a + 1 = s 2, b + 1 = t 2, where r, s, t are positive integers. Define d + = a + b + + 2ab + 2rst. 0 Mathematis Subjet Classifiation (2000): Primary 11D09, 11D25, 11D48; Seondary 11B37, 11J68, 11J86. 1
2 There are only finitely many Diophantine quintuples 2 Then {a, b,, d + } is a Diophantine quadruple. Indeed, ad = (at + rs) 2, bd = (bs + rt) 2, d = (r + st) 2. There is a stronger version of the Diophantine quintuple onjeture. Conjeture 1 If {a, b,, d} is a Diophantine quadruple and d > max{a, b, }, then d = d +. Conjeture 1 was proved for ertain Diophantine triples [16, 20] and for some parametri families of Diophantine triples [7, 8, 10]. In partiular, in [10] it was proved that the pair {1, 3} annot be extended to a Diophantine quintuple. A Diophantine quadruple D = {a, b,, d}, where a < b < < d, is alled regular if d = d +. Equivalently, D is regular iff (1) (a + b d) 2 = 4(ab + 1)(d + 1) (see [14]). The equation (1) is a quadrati equation in d. One root of this equation is d +, and other root is d = a + b + + 2ab 2rst. It is easy to hek that all small Diophantine quadruples are regular; e.g. there are exatly 207 quadruples with max{a, b,, d} 10 6 and all of them are regular. Sine the number of integer points on an ellipti urve (2) y 2 = (ax + 1)(bx + 1)(x + 1) is finite, it follows that, for fixed a, b and, there does not exist an infinite set of positive integers d suh that a, b,, d is a Diophantine quadruple. However, bounds for the size [2] and for the number [19] of solutions of (2) depend on a, b, and aordingly they do not immediately yield an absolute bound for the size of suh set. The main result of the present paper is the following theorem. Theorem 1 There are only finitely many Diophantine quintuples. Moreover, this result is effetive. We will prove that all Diophantine quintuples Q satisfy max Q < Hene we almost ompletely solve the problem of the existene of Diophantine quintuples. Furthermore, we prove Theorem 2 There does not exist a Diophantine sextuple. Theorems 1 and 2 improve results from [9] where we proved that there does not exist a Diophantine 9tuple and that there are only finitely many Diophantine 8tuples. As in [9], we prove Conjeture 1 for a large lass of Diophantine triples satisfying some gap onditions. However, in the present paper these gap onditions are muh weaker than in [9]. Aordingly, the lass of Diophantine triples for whih we are able to prove Conjeture 1 is muh larger. In fat, in an arbitrary Diophantine quadruple, we may find a triple for whih we are able to prove Conjeture 1.
3 There are only finitely many Diophantine quintuples 3 In the proof of Conjeture 1 for a triple {a, b, } we first transform the problem into solving systems of simultaneous Pellian equations. This redues to finding intersetions of binary reurrene sequenes. In Setion 5 we almost ompletely determine initial terms of these sequenes, under assumption that they have nonempty intersetion whih indues a solution of our problem. This part is a onsiderable improvement of the orresponding part of [9]. This improvement is due to new gap priniples developed in Setion 4. These gap priniples follow from the areful analysis of the elements of the binary reurrene sequenes with small indies. Let us mention that in a joint paper with A. Pethő [10] we were able to determine initial terms, in a speial ase of triples {1, 3, }, using an indutive argument. Applying some ongruene relations we get lower bounds for solutions. In obtaining these bounds we need to assume that our triple satisfies some gap onditions like b > 4a and > b 2.5. Let us note that these onditions are muh weaker then onditions used in [9], and this is due to more preise determination of the initial terms. Comparing these lower bounds with upper bounds obtained from the Baker s theory on linear forms in logarithms of algebrai numbers (a theorem of Matveev [18]) we prove Theorem 1, and omparing them with upper bounds obtained from a theorem of Bennett [5] on simultaneous approximations of algebrai numbers we prove Theorem 2. In the final steps of the proofs, we use again the above mentioned gap priniples. 2 Systems of Pellian equations Let us fix some notation. Let {a, b, } be a Diophantine triple and a < b <. Furthermore, let positive integers r, s, t be defined by ab + 1 = r 2, a + 1 = s 2, b + 1 = t 2. In order to extend {a, b, } to a Diophantine quadruple {a, b,, d}, we have to solve the system (3) ad + 1 = x 2, bd + 1 = y 2, d + 1 = z 2, with positive integers x, y, z. Eliminating d from (3) we get the following system of Pellian equations (4) (5) az 2 x 2 = a, bz 2 y 2 = b. In [9], Lemma 1, we proved the following lemma whih desribes the sets of solutions of the equations (4) and (5). Lemma 1 There exist positive integers i 0, j 0 and integers z (i) j = 1,..., j 0, with the following properties: (i) (z (i) 0, x(i) 0 ) and (z(j) 1, y(j) 1 0, x(i) 0, z(j) 1, y(j) ) are solutions of (4) and (5), respetively. 1, i = 1,..., i 0,
4 There are only finitely many Diophantine quintuples 4 (ii) (6) (7) (8) (9) z (i) 0, x(i) 0, z(j) 1, y(j) 1 satisfy the following inequalities a( a) s + 1 2(s 1) < 2 1 x (i) 0 < a, 1 z (i) (s 1)( a) 0 < 2a 2 a < 0.421, 1 y (j) 1 b( b) t + 1 2(t 1) < < b, 2 1 z (j) (t 1)( b) 1 < 2b 2 b < (iii) If (z, x) and (z, y) are positive integer solutions of (4) and (5) respetively, then there exist i {1,..., i 0 }, j {1,..., j 0 } and integers m, n 0 suh that (10) (11) z a + x = (z (i) 0 z b + y = (z (j) 1 (i) a + x 0 )(s + a) m, )(t + b) n. b + y (j) 1 Let (x, y, z) be a solution of the system (4) & (5). From (10) it follows that z = v (i) m for some index i and integer m 0, where (12) v (i) 0 = z (i) 0, v(i) 1 = sz (i) 0 + x (i) 0, v(i) m+2 = 2sv(i) m+1 v(i) m, and from (11) we onlude that z = w (j) n for some index j and integer n 0, where (13) w (j) 0 = z (i) 1, w(j) 1 = tz (j) 1 + y (j) 1, w(j) n+2 = 2tw(j) n+1 w(j) n. Let us onsider the sequenes (v m ) and (w n ) modulo 2. From (12) and (13) it is easily seen that (14) (15) v (i) 2m z(i) 0 (mod 2), v (i) 2m+1 sz(i) 0 + x (i) 0 (mod 2), w (j) 2n z(j) 1 (mod 2), w (j) 2n+1 tz(j) 1 + y (j) 1 (mod 2). We are searhing for solutions of the system (4) & (5) suh that d = (z 2 1)/ is an integer. Using (14) and (15), from z = v m = w n we obtain [z (i) 0 ]2 v 2 m z 2 1 (mod ), [z (j) 1 ]2 w 2 n z 2 1 (mod ). Therefore we are interested only in equations v m = w n satisfying [z (i) 0 ]2 [z (j) 1 ]2 1 (mod ).
5 There are only finitely many Diophantine quintuples 5 We will dedue later more preise information on the initial terms z (i) 0 and z (j) 1 (see Setion 5). Let us mention now a result of Jones [15], Theorem 8, whih says that if < 4b then z (j) 1 = 1. As a onsequene of Lemma 1 and the relations (14) and (15), we obtain the following lemma. From now on, we will omit the supersripts (i) and (j). Lemma 2 1) If the equation v 2m = w 2n has a solution, then z 0 = z 1. 2) If the equation v 2m+1 = w 2n has a solution, then z 0 z 1 < 0 and x 0 s z 0 = z 1. In partiular, if b > 4a and > 100a, then this equation has no solution. 3) If the equation v 2m = w 2n+1 has a solution, then z 0 z 1 < 0 and y 1 t z 1 = z 0. 4) If the equation v 2m+1 = w 2n+1 has a solution, then z 0 z 1 > 0 and x 0 s z 0 = y 1 t z 1. Proof. See [9], Lemma 3. 3 Relationships between m and n In [9], Setion 4, we proved that v m = w n implies m n if b > 4a, > 100b and n 3. We also proved that m 3 2n, provided {a, b, } satisfies some rather strong gap onditions. In this setion we will first prove an unonditional relationship between m and n, and then we will improve that result under various gap assumptions. Lemma 3 If v m = w n, then n 1 m 2n + 1. Proof. We have the following estimates for v 1 : v 1 = sz 0 + x 0 x 0 s z 0 = 2 a z 2 0 x 0 + s z 0 > a 2x 0 > 4x 0 > a, Hene (16) If > 4b, then v 1 < 2x 0 < a a (2s 1)m 1 < v m < a(2s) m 1 for m 1. w 1 2 b z 2 1 y 1 + b z 1 > 4y 1 > If < 4b then, by [15], Theorem 8, z 1 = ±1, y 1 = 1 and w 1 t = a + r = s > a > Furthermore, w 1 < 2y 1 < b and therefore (17) b b b (2t 1)n 1 < w n < b(2t) n 1 for n 1.
6 There are only finitely many Diophantine quintuples 6 We thus get (18) Sine (19) (2s 1) m 1 < ab 2 (2t) n 1. 2s 1 = 2 a > a and 2t = 2 b + 1 < b, it follows that (2s 1) 2 > 3.12a > 2t. It implies and m 2n + 1. On the other hand, we have (2s 1) m 1 < (2t) n < (2t) n+0.43 < (2s 1) 2n+0.86 (2t 1) n 1 < ab 2 (2s) m 1 < ab 2 (2t 1) m 1 < (2t 1) m and n m + 1. Thus we proved the lemma for m, n 0. It remains to hek that v 0 < w 2 and w 0 < v 2. Indeed, 2t w 2 = 2tw 1 w 0 > b ( 4 ) b 2 b > > > v 0, v 2 = 2sv 1 v 0 > 2s a 2 b ( 4 ) a 2 a > > > w 0. 2 a Lemma 4 Assume that > If v m = w n and m, n 2, then 1) > b 4.5 = m 11 9 n + 7 9, 2) > b 2.5 = m 7 5 n + 3 5, 3) > b 2 = m 3 2 n + 1 2, 4) > b 5/3 = m 8 5 n Proof. As in the proof of Lemma 3, assuming > max{b 5/3, }, we have v m > x 0 (2s 1) m 1 > w n < b(2t) n 1, a (2s 1)m 1, for m, n 1. Hene (2s 1) m 1 < ab 2 (2t) n 1 and (20) m 1 a (m 1)/2 (m 1)/2 < n b (n 1)/2+1/4 (n 1)/2+1/4 a 1/4. If m 2 and > b ε, then (20) shows that at least one of the following two inequalities holds: (21) m 1 < ( n ) 1 ε + n 2, (22) m 1 < 1.004n. The inequality (22) implies all statements of the lemma. For ε = 4.5, (21) implies m < 11 9 n and m 11 9 n Similar arguments apply to all other statements of the lemma.
7 There are only finitely many Diophantine quintuples 7 4 Gap priniples In [9], Lemma 14, we proved that if {a, b,, d} is a Diophantine quadruple and a < b < < d, then d 4b. The proof was based on the fat, proved by Jones [15], Lemma 4, that = a + b + 2r or 4ab + a + b. In this setion we will develop a stronger and more preise gap priniple by examining the equality v m = w n for small values of m and n. Let us note that sine (4ab + 1) < d + < 4(ab + 1), we may expet an essential improvement only if we assume that d d +. Lemma 5 Let v m = w n and define d = (v 2 m 1)/. If {0, 1, 2} {m, n}, then d < or d = d +. Proof. From the proof of Lemma 3 we have: v 0 < w 2, w 0 < v 2, v 1 < w 3, w 1 < v 4, v 2 < w 4, w 2 < v 6. Therefore, the ondition {0, 1, 2} {m, n} implies (m, n) {(0, 0), (0, 1), (1, 0), (1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (2, 3), (3, 2), (4, 2), (5, 2)}. If 0 {m, n}, then d <. If (m, n) = (1, 1), then d < for z 0 < 0, and d = d + for z 0 > 0 (see [9], proof of Theorem 3). Assume that (m, n) = (1, 2). We have v 1 = sz 0 + x 0, w 2 = z 1 + 2(bz 1 + ty 1 ). By Lemma 2, if z 1 > 0 then z 0 < 0 and x 0 + sz 0 = z 1. Hene w 2 > v 1 = w 0. If z 1 < 0 then z 0 > 0 and (23) x 0 sz 0 = z 1. Inserting (23) into the relation v 1 = w 2 we obtain (24) bz 1 + ty 1 = x 0. From (23), (24) and the system (4) & (5), we obtain (b a)t = z0 2 (b a). Therefore, z 0 = t, x 0 = r, z 1 = st r, y 1 = rt bs. It implies v 1 = st + r and d = v2 1 1 = 1 (ab2 + a + b rst + ab ) = d +. Assume now that (m, n) = (2, 1). In the same manner as in the ase (m, n) = (1, 2), we obtain z 1 = s, y 1 = r, z 0 = st r, x 0 = rs at and d = d +. Let (m, n) = (2, 2). We have v 2 = z 0 + 2(az 0 + sx 0 ), w 2 = z 1 + 2(bz 1 + ty 1 ), and sine z 0 = z 1, we obtain az 0 + sx 0 = bz 0 + ty 1 and (b a)(y 2 1 x b a) = ax x 2 0 by y 2 1 2stx 0 y 1.
8 There are only finitely many Diophantine quintuples 8 Therefore (b a) 2 = (sy 1 tx 0 ) 2 and sine sy 1 = tx 0, we have (25) tx 0 sy 1 = b a. Furthermore, and (a + 1)(bx a b) = a(bx x (b a) 2 2t(b a)x 0 ) (b a)(x atx 0 + a 2 t 2 ) = (b a)(ab + 1)(a + 1) = (b a)r 2 s 2. Finally, x 0 = rs at, y 1 = rt bs, z 0 = st r. Now we obtain and d = d +. v 2 = st r + 2(ast ar + rs 2 ast) = st + r Let (m, n) = (3, 1). We may assume that {a, b, } {1, 3, 8}. Then a 15 and b 48. It implies 2s 1 > a and 2t < b. From (18) we obtain whih implies 6, a ontradition. (1.807 a) 2 < ab 2 < a 3, Let (m, n) = (3, 2). Assume that z 0 > 0, z 1 < 0. We have v 1 > 2sz 0, v 2 > (4s 2 1)z 0, v 3 > (4s 2 1)(2s 1)z 0 > 7(a) 3/2 z 0, and w 1 < 2 2t z 1, w 2 < 2tw 1 < 2 z 1. We have also z 1 > 4x 0 and therefore w 2 < 4x 0. Sine x 0 <, we obtain w 2 < 4 3/2 < 7(a) 3/2 z 0 < v 3. Assume now that z 0 < 0, z 1 > 0. Then z 1 = sz 0 + x 0 and the ondition v 3 = w 2 implies x 0 + 2az 1 = bz 1 + ty 1 > 2bz 1 and x 0 > 4z 1 > x 0, a ontradition. Let (m, n) = (2, 3). If z 0 > 0, z 1 < 0, then v 2 = w 3 implies y 1 + 2bz 0 = az 0 + sx 0 < 2az 0 + z 0. Therefore z0 2 < 4. On the other hand, z2 0 1 (mod ). Hene, z 0 = 1. But, if > 4b then z 0 > 4y 1 > 1. If < 4b then z 1 = 1, y 1 = 1, and Lemma 2 implies = b + 2, whih ontradits the fat that = a + b + 2r. Assume that z 0 < 0, z 1 > 0. As in the ase (m, n) = (3, 2), we have v 2 < 2 z 0 and w 3 > 7.3(b) 3/2 z 1. If > 4b then z 0 > and sine y 1 <, we obtain 4y 1 v 2 < 4 3/2 < 7.3(b) 3/2 z 1 < w 3. If < 4b then w 3 > 7.3b > > 2 > v 2. Let (m, n) = (4, 2). We have v 4 = z 0 + 4(2az 0 + sx 0 ) + 8a 2 (az 0 + sx 0 ), and the relation v 4 = w 2 implies (26) bz 0 + ty 1 = 4az 0 + 2sx 0 + 4a(az 0 + sx 0 ).
9 There are only finitely many Diophantine quintuples 9 It holds ty 1 b z 0 < z 0. Hene, if z 0 > 0 then the left hand side of (26) is 2bz 0 + z 0 < 3z 0, while the right hand side is > 4a 2 z 0 > 3z 0. If z 0 < 0, then the left hand side of (26) is z 0, while the right hand side is 4a 4a z 0 > a. Let (m, n) = (5, 2). It follows from (18) that and 10.66a 2 2 < 10.65b, a ontradition. (1.807 a) 4 < b b We now an prove the following gap prinipal, whih we will improve again in Proposition 1 below. Lemma 6 If {a, b,, d} is a Diophantine quadruple and a < b < < d, then d = d + or d b 1.5. Proof. By Lemma 5, if d d + then m 3 and n 3. From (17) it follows that and w 3 > (2t 1) b > 81 4 b d 1.161b > b 1.5. Using the gap priniple from Lemma 6 we an prove the following lemma. Lemma 7 Under the notation from above, we have v 3 w 3. Proof. If z 0, z 1 > 0 then define z := x 0 sz 0 = y 1 tz 1, and if z 0, z 1 < 0 then define z := x 0 + sz 0 = y 1 + tz 1. Define also d 0 = (z 2 1)/. Then d 0 is an integer. Furthermore, d = z 2, ad = 1 (a2 x 2 0 2asx 0 z 0 + a 2 z az 2 0 a + ) = (ax 2 0 2asx 0 z 0 + a 2 z x 2 0) = (sx 0 az 0 ) 2, bd = 1 (b2 y 2 1 2bty 1 z 1 + b 2 z 2 1 b + ) = (ty 1 bz 1 ) 2. We have Therefore z = 2 a z 2 0 x 0 + s z 0 > d 0 > a a and z <. > > 0, a
10 There are only finitely many Diophantine quintuples 10 and thus the set {a, b,, d 0 } forms a Diophantine quadruple. Sine d 0 <, by Lemma 6, we have two possibilities: the quadruple {a, b,, d 0 } is regular or (27) > 1.16d b 1.5. Assume that {a, b,, d 0 } is regular, i.e. d 0 = d. Then z = r st. From (x 0 r) = s( z 0 t) and gd(, s) = 1 it follows that z 0 t (mod ), and sine z 0 <, t <, we onlude that z 0 = t and x 0 = r. We an proeed analogously to prove that z 1 = s and y 1 = r. The ondition v 3 = w 3 implies sz 0 + 3x 0 + 4a(x 0 + sz 0 ) = tz 1 + 3y 1 + 4b(y 1 + tz 1 ) and we obtain a = b, a ontradition. Hene we may assume that the quadruple {a, b,, d 0 } is not regular. But it means that > 10 6, whih implies and z = 2 a z 2 0 x 0 + s z 0 > x 0 > a > a d 0 > a 1 > a Thus from (27) we obtain > a 1.25 b 1.5 > a 0.25 and a , whih ontradits the assumption that > Now we an prove the following strong gap priniple, whih is the main improvement to [9] and whih we will use several times later. Observe that espeially the dependene on is muh better than in the gap priniple in [9]. Proposition 1 If {a, b,, d} is a Diophantine quadruple and a < b < < d, then d = d + or d > a 2.5. Proof. From Lemmas 5 and 7 it follows that m 4 or n 4. By (16), we have v 4 Therefore, if m 4 then a (2s 1) a 5 5. Similarly, from (17) it follows w 4 and for n 4 we obtain d 2.696a b (2t 1)3 d 3.919b > a b 5 5, > b 2.5 > a 2.5.
11 There are only finitely many Diophantine quintuples 11 Corollary 1 If {a, b,, d, e} is a Diophantine quintuple and a < b < < d < e, then e > d 3.5 b 2.5. Proof. Assume that {b,, d, e} is a regular Diophantine quadruple. Then e 4d(b + 1) < d 3. The quadruple {a,, d, e} is not regular and, by Proposition 1, we have e > 2.695d 3.5 a 2.5 > d 3. Therefore the quadruple {b,, d, e} is not regular and hene e > 2.695d 3.5 b Determination of the initial terms Using the gap priniples developed in the previous setion we will improve Lemma 2 and obtain more speifi information on the initial terms of the sequenes (v m ) and (w n ). Lemma 8 1) If the equation v 2m = w 2n has a solution, then z 0 = z 1. Furthermore, z 0 = 1 or z 0 = r st or z 0 < min{0.869 a 5/14 9/14, b }. 2) If the equation v 2m+1 = w 2n has a solution, then z 0 = t, z 1 = r st and z 0 z 1 < 0. 3) If the equation v 2m = w 2n+1 has a solution, then z 0 = r st, z 1 = s and z 0 z 1 < 0. 4) If the equation v 2m+1 = w 2n+1 has a solution, then z 0 = t, z 1 = s and z 0 z 1 > 0. Proof. 1) From Lemma 2 we have z 0 = z 1. Define d 0 = (z 2 0 1)/. Then d 0 is an integer and d = z 2 0, ad = 1 (az2 0 a + ) = x 2 0, bd = 1 (bz2 1 b + ) = y 2 1. Hene, we have three possibilities: d 0 = 0 or {a, b,, d} is a regular Diophantine quadruple or {a, b,, d} is an irregular Diophantine quadruple. If d 0 = 0 then z 0 = 1. If the quadruple {a, b,, d} is regular, then (sine d 0 < ) we have d 0 = d and z 0 = r st. Otherwise we may apply Proposition 1 to obtain 2.695d a2.5. Sine z 0 1, we have z We may assume that > 106 and therefore d 0 = z2 0 1 z2 ( ) > z Hene we obtain 4.5 > z 7 0 a2.5 and z 0 < 0.869a 5/14 9/14. Analogously, from Lemma 6, we obtain z 0 < 0.972b ) Let z = z 1 = x 0 + sz 0 if z 1 > 0, and z = z 1 = x 0 sz 0 if z 1 < 0. Define d 0 = (z 2 1)/. Then d 0 is an integer and d = z 2, ad = (sx 0 ± az 0 ) 2, bd = y 2 1. In the proof of Lemma 7 it is shown that d 0 > 0 and d 0 <. Therefore {a, b,, d} is a Diophantine quadruple, and by the proof of Lemma 7, it must be regular. It means that d 0 = d and z = r st. It implies z 1 = r st and z 0 = t.
12 There are only finitely many Diophantine quintuples 12 3) Let z = z 0 = y 1 + tz 1 if z 0 > 0, and z = z 0 = y 1 tz 1 if z 0 < 0, and define d 0 = (z 2 1)/. Then d = z 2, ad = x 2 0, bd = (ty 1 ± bz 1 ) 2 and 0 < d 0 <. If the quadruple {a, b, d 0, } is not regular, then from Lemma 6 we have (28) We an assume that > If > 4b then and if < 4b then Therefore, z = 2 b z 2 1 y 1 + t z d b 1.5. > y 1 > b > b, z > a > b. d 0 > b 1 > b Now (28) implies > /4 b 1/4 and b , a ontradition. It follows that the quadruple {a, b,, d} is regular, i.e. d 0 = d and z = r st. Hene z 0 = r st and (y 1 r) = t( z 1 s). Sine gd(t, ) = 1 we have z 1 s (mod ), whih implies z 1 = s. 4) Let z = x 0 s z 0 = y 1 t z 1 and d 0 = (z 2 1)/. In the proof of Lemma 7 we have shown that {a, b, d 0, } is a regular Diophantine quadruple, and that this fat implies z 0 = t and z 1 = s. 6 Standard Diophantine triples In [9] we proved Conjeture 1 for triples satisfying some gap onditions like b > 4a and > max{b 13, }. In Setion 7 we will prove Conjeture 1 under ertain weaker assumptions. This results will suffie for proving Theorem 1 sine we will show that every Diophantine quadruple ontains a triple whih satisfies some of our gap assumptions. Definition 1 Let {a, b, } be a Diophantine triple and a < b <. Diophantine triple of We all {a, b, } a the first kind if > b 4.5, the seond kind if b > 4a and > b 2.5, the third kind if b > 12a and b 5/3 < < b 2, the fourth kind if b > 4a and b 2 < 6ab 2. A triple {a, b, } is alled standard if it is a Diophantine triple of the first, seond, third or fourth kind.
13 There are only finitely many Diophantine quintuples 13 The Diophantine triples of the first and the seond kind appear naturally when we try to modify results from [9] using Lemma 8. They orrespond to triples with properties b > 4a, > b 13 and b > 4a, > b 5, onsidered in [9]. On the other hand, triples of the third and the fourth kind ome from the analysis what kind of triples a regular Diophantine quadruple may ontain. Note also that these four ases are not mutually exlusive. We now use the improved gap priniple (Proposition 1) again to show that the set of all standard Diophantine triples is large. Proposition 2 Every Diophantine quadruple ontains a standard triple. Proof. Let {a, b,, d} be a Diophantine quadruple. If it is not regular, then by Proposition 1 and [9], Lemma 14, we have d > 3.5 and > 4a. Hene {a,, d} is a triple of the seond kind. Assume that {a, b,, d} is a regular quadruple. Then (4ab + 1) < d < 4(ab + 1). If b > 4a and b 1.5, then d > b 2.5 and we see that {a, b, d} is a triple of the seond kind. If b > 4a and < b 1.5, then we have two possibilities: if = a + b + 2r then < 4b, 2 < d < 4b 2 and therefore {b,, d} is a triple of the fourth kind; if 4ab + a + b then d < 2, d > b > 5/3 and it follows that {a,, d} is a triple of the third kind. We may now assume that b < 4a. By [15], Theorem 8, we have = k (k 1) or = k (k 2), where the sequenes ( k ) and ( k ) are defined by 0 = 0, 1 = a + b + 2r, k = (4ab + 2) k 1 k 2 + 2(a + b), 0 = 0, 1 = a + b 2r, k = (4ab + 2) k 1 k 2 + 2(a + b). If > b 2.5 then d > 4ab > b 4.5 and we see that {a, b, d} is a triple of the first kind. Sine 2 = 4r(a + r)(b + r) > 4ab 2 > b 3, the ondition b 2.5 implies = 1 or = 2. If = 1 then a + b ab = ab ( a b + b a ) ab( ) < 4.81 ab and 4r. Hene d > 4ab > > 2 and d < 4r 2 2 r < 2a 2. Therefore the triple {a,, d} is of the fourth kind. If b 2.5 and = 2 4ab + 2a + 2b (we may assume b > a + 2, sine otherwise 2 = 1 ), then d < 2 and d > 4ab > b 2 > 1.8 > 5/3. Therefore the triple {a,, d} is of the third kind. 7 Lower bounds for solutions The main tool in obtaining lower bounds for m and n satisfying v m = w n (m, n > 2) is the ongruene method introdued in the joint paper of the author with A. Pethő [10].
14 There are only finitely many Diophantine quintuples 14 Lemma 9 1) v 2m z 0 + 2(az 0 m 2 + sx 0 m) (mod 8 2 ) 2) v 2m+1 sz 0 + [2asz 0 m(m + 1) + x 0 (2m + 1)] (mod 4 2 ) 3) w 2n z 1 + 2(bz 1 n 2 + ty 1 n) (mod 8 2 ) 4) w 2n+1 tz 1 + [2btz 1 n(n + 1) + y 1 (2n + 1)] (mod 4 2 ) Proof. See [9], Lemma 4. If v m = w n then, of ourse, v m w n (mod 4 2 ) and we an use Lemma 9 to obtain some ongruenes modulo. However, if a, b, m and n are small ompared with, then these ongruenes are atually equations. It should be possible to prove that these new equations are in ontradition with the starting equations v m = w n. This will imply that m and n annot be too small. We will prove a lower bound for n (and therefore also for m by Lemma 3) depending on and we will do this separately for Diophantine triples of the first, seond, third and fourth kind in the following four lemmas. Lemma 10 Let {a, b, } be a Diophantine triple of the first kind and > If v m = w n and n > 2, then n > Proof. Assume that n By Lemma 17, we have max{ m/2 +1, n/2 +1} < n Aording to Lemma 8, we will onsider six ases. 1.1) v 2m = w 2n, z 0 = 1 From Lemma 9 we have (29) ±am 2 + sm ±bn 2 + tn (mod 4). Sine > b 4.5, we have am 2 < <, sm < <, bn 2 < <, tn < <. Therefore we may replae by = in (29): (30) ±am 2 + sm = ±bn 2 + tn. From (30), squaring twie, we obtain [(am 2 bn 2 ) 2 m 2 n 2 ] 2 4m 2 n 2 (mod ). Sine 4m 2 n 2 < < and [(am 2 bn 2 ) 2 m 2 n 2 ] 2 < <, we therefore have [(am 2 bn 2 ) 2 m 2 n 2 ] 2 = 4m 2 n 2, and (31) am 2 bn 2 = ±m ± n. From (30) and (31) we obtain (32) m(s ± 1) = n(t ± 1). Inserting (32) into (30) we obtain (33) n = (s ± 1)[t(s ± 1) (t ± 1)s] ±[a(t ± 1) 2 b(s ± 1) 2 ] = (s ± 1)(±t s) ±(±2at + 2a ± 2bs 2b).
15 There are only finitely many Diophantine quintuples 15 Sine (s ± 1)(±t s) (s 1)(t s) = (s 1)(b a) t + s > 2(s 1) 2 b a > 2 b and ± 2at + 2a 2bs 2b 4bs + 4b < 6b a, (33) implies that n > ontradition. 1.2) v 2m = w 2n, z 0 = r st We may assume that b 4. Then we have 12 b > 0.377, a z 0 = z 1 = 2 a b 1 r + st > r > ab, and from z 1 < 2 it follows < b 5.4a2 b < 5.4b 3 < b 4.5, a ontradition. 1.3) v 2m = w 2n, z 0 1, r st From Lemma 9 we obtain (34) az 0 m 2 + sx 0 m bz 0 n 2 + ty 1 n (mod 4). By Lemma 8, we have Therefore, (35) az 0 m 2 < <, sx 0 m < (a z 0 + z 0 )m < 0.8 <, bz 0 n 2 < <, ty 1 n < (b z 1 + z 1 )n < 0.87 <. az 0 m 2 + sx 0 m = bz 0 n 2 + ty 1 n. Assume now that b > 4a. Sine z 0 1, we have z0 2 max{ + 1, 3/a}. This implies 0 sx 0 a z 0 1 = x2 0 + a a2 a z 0 (sx 0 + a z 0 ) 1.001a 2a 2 z 2 0 < , 0 ty 1 b z 1 1 = y2 1 + b b2 b z 1 (ty 1 + b z 1 ) 1.001b 2b 2 z 2 0 < If z 0 > 1 then, by (35), we have az 0 m(m ) bz 0 n(n + 1), and sine m, n 2, we obtain a m 2 > bn 2 and m n But now Lemma 4 implies n 1, a ontradition. If z 0 < 1 then (35) implies a z 0 m(m 1)) b z 0 n(n ). Hene, by Lemma 4, we onlude that 4n(n ) < ( 11 9 n + 7 )( n 11 ) < n n , 18 whih implies n n < 0 and n 1, a ontradition.
16 There are only finitely many Diophantine quintuples 16 (36) Assume now that b < 4a. Squaring twie the relation (35) we obtain By Proposition 1, we have [(am 2 bn 2 ) 2 x 2 0m 2 y 2 1n 2 ] 2 4x 2 0y 2 1m 2 n 2 (mod ). y 2 1 < b z < 4a a 5/7 9/ < 3.018(a) 2/7 < This implies that the both sides of the ongruene (36) are less than. Indeed, the left hand side is bounded above by max{ , , } < <, while 4x 2 0 y2 1 m2 n 2 < = <. Therefore we have an equality in (36), and this implies (37) am 2 bn 2 = ±x 0 m ± y 1 n. From (35) and (37) we obtain x 0 m(s ± 1) = y 1 n(t ± 1) and n = A/B, where A = x 2 0y 1 (s ± 1)(±t s), B = z 0 [ab(y 2 1 x 2 0) + 2(a b) ± 2aty 2 1 2bsx 2 0]. We have the following estimates These estimates yield A x 2 0y 1 (s + 1)(t + s) < 2.005x 2 0y 1 ab, B > z 0 [ab(2y 1 1) 2b 4aty1 2 2s(b a)] [ > z 0 y 1 ab ty 1 y 1 ay 1 b 2s ] > z 0 y 1 ab. ay 1 n < 2.005x2 0 y 1 ab z 0 y 1 ab < x 2 0 z 0 ab < az z 0 ab < 1.28 z 0 < 1/4 < 1, a ontradition. 2) v 2m+1 = w 2n The impossibility of this ase is proven in 1.2). 3) v 2m = w 2n+1 By Lemma 8 and 1.2), z 0 > 5.4b 3 < b 4.5, a ontradition. 4) v 2m+1 = w 2n+1 From Lemmas 8 and 9 we obtain (38) 2.155, and from z ab 0 < 2 a it follows < 5.4ab2 < ±astm(m + 1) + rm ±bstn(n + 1) + rn (mod 2), where m = m, n = n if z 0 < 0, and m = m + 1, n = n + 1 if z 0 > 0. Multiplying (38) by s and t, respetively, we obtain (39) (40) ±atm(m + 1) + rsm ±btn(n + 1) + rsn (mod 2), ±asm(m + 1) + rtm ±bsn(n + 1) + rtn (mod 2). We have btn(n + 1) < <, rtn < <. Therefore we have equalities in (39) and (40). This implies rm = rn and am(m + 1) = bn(n + 1), and finally m = n = 0.
17 There are only finitely many Diophantine quintuples 17 Lemma 11 Let {a, b, } be a Diophantine triple of the seond kind and > v m = w n and n > 2, then n > If Proof. Assume that n Then max{ m/2 + 1, n/2 + 1} < n ) v 2m = w 2n, z 0 = 1 Sine am 2 < 0.408, sm < 0.701, bn 2 < and tn < 0.701, (29) implies (30). Furthermore am 2 sm < m a < 0.06 < 0.001, bn 2 tn < n b < 0.06 < Hene m n t s > b a > 1.99, ontrary to Lemma ) v 2m = w 2n, z 0 = r st We have z 0 = z 1 = 2 a b 1 > r + st 2r > ab, 2 b it follows < 5.25a2 b < 1.32ab 2. Hene {a, b, } is a Diophantine and from z 1 < triple of the fourth kind, and this ase will be treated in Lemma ) v 2m = w 2n, z 0 1, r st By Proposition 1, we have > b 3.5, and Lemma 8 implies az 0 m 2 < <, sx 0 m < <, bz 0 n 2 < 0.98 <, ty 1 n < <. Therefore, equation (35) holds again. As in Lemma 10, we obtain a ontradition with Lemma 4. 2) v 2m+1 = w 2n This ase is impossible by Lemma 2. Namely, if 100a then a and a 1.03, a ontradition. 3) v 2m = w 2n+1 As in 1.2), we obtain < 5.25ab 2 and again {a, b, } is a triple of the fourth kind. 4) v 2m+1 = w 2n+1 Relation (38) implies [am(m + 1) bn(n + 1)] 2 r 2 (n m) 2 (mod 2). Sine a 2 m 2 (m + 1) 2 < 0.96, b 2 n 2 (n + 1) 2 < 0.96, and r 2 m 2 < 0.48, we obtain (41) bn(n + 1) am(m + 1) = ±r(m n), whih implies (42) 2n + 1 a 2m + 1 b = ±4r(m n) + b a [(2n + 1) b + (2m + 1) a] b(2m + 1).
18 There are only finitely many Diophantine quintuples 18 By Lemma 4, the right hand side of (42) is 4r(m n) b(2n + 1)(2m + 1) + Therefore, (43) b b(2n + 1)(2m + 1) < 2n + 1 2m + 1 < < b 2 4n+5 10 b(2n + 1)(2m + 1) < = If n = 1 then m = 1 (by Lemma 4), ontrary to (43). If n = 2 then m = 2 or m = 3, whih both ontradit the relation (43). Hene we may assume that n 3, m 3, and now (42) implies 2n+1 2m+1 < On the other hand, from Lemma 4 we see that 2n+1 2m (2m+1) 0.653, a ontradition. Lemma 12 Let {a, b, } be a Diophantine triple of the third kind and > v m = w n and n > 2, then n > If Proof. Assume that n Then max{ m/2 + 1, n/2 + 1} < n ) v 2m = w 2n, z 0 = 1 Sine am 2 < 0.9, sm < 0.951, bn 2 < 0.9 and tn < <, the proof is idential to that of Lemma ) v 2m = w 2n, z 0 = r st From (34) we have (44) ±astm(m 1) + rm ±bstn(n 1) + rn (mod ). Multiplying (44) by 2st we obtain (45) ±2[am(m 1) bn(n 1)] 2rst(n m) (mod 2). Let α be the absolutely least residue of 2rst modulo 2, and let A = (2rst 2r 2 +)(st+r). Then α (st + r) A and A = 2ar + 2br + 2r + st 2 r < 2ar + 2br + 2r. Sine r st = 2 b a 1 r+st < 2 2 < b b < br, we obtain A < 2r(a+b+1) < 13 ab 6 br and α < 13br 12 < 1.15b. We have 2am(m 1) < ab 0.904, 2bn(n 1) < 0.904, α(m n) < Therefore ±2[am(m 1) bn(n ± 1)] = α(n m), and it implies (46) 2n 1 a 2m 1 b = ±2α(m n) + b a [(2n 1) b + (2m 1) a] b(2m 1). By Lemma 4, the right hand side of (46) is 1.15b 6n+3 5 b(2n 1)(2m 1) + 1 (2n 1)(2m 1) < 0.495,
19 There are only finitely many Diophantine quintuples 19 and therefore (47) 2n 1 2m 1 < If n = 2 then Lemma 4 implies m = 2 or 3. The ase m = 2 ontradits the inequality (47), while for m = 3 relation (46) implies 2n 1 2m 1 < But for n = 2, m = 3 we have 2n 1 2m 1 { 3 5, 5 7 }, a ontradition. Hene n 3, and sine n = m = 3 ontradits (47), we have also m 4. Now, from (46) we obtain 2n 1 2m+1 2m 1 < But for m 4 we have 2n+1 > 2n 1 2m 1 > (2m 1) > 0.517, a ontradition. 1.3) v 2m = w 2n, z 0 1, r st Proposition 1 implies > b 3.5 and therefore this ase is impossible. 2) v 2m+1 = w 2n The impossibility of this ase is shown in Lemma 11. 3) v 2m = w 2n+1 From Lemmas 8 and 9 it follows that and (48) ±2astm(m 1) + r(2m ± 1) ±2bstn(n + 1) + r(2n + 1) (mod 4) ±2[am(m 1) bn(n + 1)] 2rst(n m + δ) (mod 2), where δ {0, 1}. Let α be defined as in 1.2). As in 1.2), we obtain and (49) ±2[am(m 1) bn(n + 1)] = α(n m + δ) 2n + 1 a 2m 1 b = The right hand side of (49) is and therefore (50) ±2α(m n δ) + b a [(2n + 1) b + (2m 1) a] b(2m 1). 1.15b 6n+11 5 b(2n 1)(2m 1) + 1 (2n 1)(2m 1), 2n + 1 2m 1 < 0.861, 2n + 1 2m + 1 < 0.617, respetively. If n = 1, then by Lemmas 4 and 7 we have m = 2. This is learly impossible if we have 2n+1 2n+1 2m 1 on the left hand side of (49), while for 2m+1 we obtain 3 5 < 0.509, a ontradition. If n = 2 then, by Lemma 17, we have m = 2, 3 or 4, and (50) implies m = 4. Sine m = n = 3 is also impossible by (50), we onlude that n 2 and m 4. This implies 2n+1 2m 1 < 0.469, 2n+1 2m+1 < 0.429, respetively. On the other hand, we have 2n+1 2m 1 > 2n+1 2m m+1 > 0.513, a ontradition. 4) v 2m+1 = w 2n+1
20 There are only finitely many Diophantine quintuples 20 Let α be defined as in 1.2). From (38) we obtain whih yields ±2[am(m + 1) bn(n + 1)] = α(n m), (51) 2n + 1 a 2m + 1 b = ±2α(m n) + b a [(2n + 1) b + (2m + 1) a] b(2m + 1). From (51) it follows that 2n+1 2m+1 a ontradition. < But Lemma 4 implies 2n+1 2m (2m+1) 0.55, Lemma 13 Let {a, b, } be a Diophantine triple of the fourth kind and > v m = w n and n > 2, then n > 0.2. If Proof. Assume that n 0.2. Then max{ m/2 + 1, n/2 + 1} < n ) v 2m = w 2n, z 0 = 1 The proof is idential to that of Lemma ) v 2m = w 2n, z 0 = r st The first part of the proof is idential to that of Lemma 12. In partiular, the relation (45) is valid. Estimating r st, we get r st < 2 ab < 6ab2 2 ab < 3b ab < 3br. Therefore A < 2r(a + b + 1) < 2.5br and α < 2.5br 2 < 1.33b. Note that ab α 2 4r 2 s 2 t 2 4r 2 (mod 2), 4r 2 4.5ab < 1.125b 2 < 2 and α 2 < 1.77b 2 < 2. Hene, α 2 = 4r 2 and α = ±2r. Sine am(m + 1) < 0.9, bn(n + 1) < 0.9, r(m n) < 0.7, we have (52) bn(n + 1) am(m + 1) = ±r(m n). Inserting α = ±2r in (46) and using Lemma 4 and the estimate α = 2r < b < 1.001b, we obtain (53) 2n 1 2m 1 < n + 1 (2n 1)(2m 1) < If (m, n) = (3, 2), then (52) implies 6b 12a = ±r and (4b 9a)(9b 16a) = 1, whih is learly impossible for b > 4a. Hene n 3, m 4 and (53) yields 2n 1 2m 1 < On the other hand, 2n+1 2m+1 > 2n 1 2m (2m 1) > 0.619, a ontradition. 1.3) v 2m = w 2n, z 0 1, r st Proposition 1 implies > b 3.5, and we have < 6ab 2 < 1.5b 3. Therefore, this ase is impossible. 2) v 2m+1 = w 2n
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