THE NUMBER OF DIOPHANTINE QUINTUPLES. Yasutsugu Fujita College of Industrial Technology, Nihon University, Japan

GLASNIK MATEMATIČKI Vol. 45(65)(010), 15 9 THE NUMBER OF DIOPHANTINE QUINTUPLES Yasutsugu Fujita College of Industrial Technology, Nihon University, Japan Abstract. A set a 1,..., a m} of m distinct positive integers is called a Diophantine m-tuple if a i a j + 1 is a perfect square for all i, j with 1 i < j m. It is known that there does not exist a Diophantine sextuple and that there exist only finitely many Diophantine quintuples. In this paper, we first show that for a fixed Diophantine triple a, b, c} with a < b < c, the number of Diophantine quintuples a, b, c, d, e} with c < d < e is at most four. Using this result, we further show that the number of Diophantine quintuples is less than 10 76, which improves the bound 10 1930 due to Dujella. 1. Introduction Diophantus posed the problem of finding a set of four (positive rational) numbers which has the property that the product of any two numbers in the set increased by one is a square, and found such a set 1/16, 33/16, 68/16, 105/16}. A set a 1,...,a m } of m distinct positive integers is called a Diophantine m-tuple if a i a j + 1 is a perfect square for all i, j with 1 i < j m. Fermat found the first example of a Diophantine quadruple, which was the set 1, 3, 8, 10}. Euler found that for any Diophantine pair a, b} the set a, b, a+b+r} is a Diophantine triple, where r = ab + 1. We call such a triple regular. In 1979, Arkin-Hoggatt-Strauss ([1]) found that for any Diophantine triple a, b, c} the set a, b, c, d + } is a Diophantine quadruple, where d + = a+b+c+abc+rst and r = ab + 1, s = ac + 1, t = bc + 1, and conjectured the following. Conjecture 1.1. If a, b, c, d} is a Diophantine quadruple with a < b < c < d, then d = d +. 010 Mathematics Subject Classification. 11D09, 11J68, 11J86. Key words and phrases. Simultaneous Diophantine equations, Diophantine tuples. 15

16 Y. FUJITA We say that a Diophantine quadruple a, b, c, d} with a < b < c < d is regular if d = d +. Conjecture 1.1 immediately implies a folklore conjecture, which says that there does not exist a Diophantine quintuple. It is known that there does not exist a Diophantine sextuple and that there does not exist only finitely many Diophantine quintuples ([8]). Recently, we ([1]) showed that if a, b, c, d, e} is a Diophantine quintuple with a < b < c < d < e, then d = d + (note that all the quadruples contained in the quintuple, other than a, b, c, d}, are irregular; see Section 1 in [1]). The first result supporting Conjecture 1.1 is due to Baker and Davenport ([]), which asserts that if 1, 3, 8, d} is a Diophantine quadruple, then d = 10. This result has been generalized and it is known that the Diophantine quadruples containing the pairs k 1, k + 1} with k (cf. [4, 5, 10, 11]) or the triples F k, F k+, F k+4 } with k 1 (cf. [6]), where F ν is the ν- th Fibonacci number, are always regular. Besides them, the Diophantine quadruples containing the following triples are regular: 1, 8, 15}, 1, 8, 10}, 1, 15, 4}, 1, 4, 35},, 1, 4} (by Kedlaya [13]), 4, 1, 30} (by Dujella [8, p. 13]). Thus, one may easily check that if a, b, c, d, e} is a Diophantine quintuple with a < b < c < d < e, then b 8, d 6440 and ad 6888 (coming from 1, 35, 48, 6888}), bd 16 6440 (coming from 3, 16, 33, 6440}). The last two lower bounds play important roles in proving Theorem 1.. We are interested in bounding the number of Diophantine quintuples. We first consider the number of Diophantine quintuples a, b, c, d, e} with a < b < c < d < e for a fixed triple a, b, c}. Since the above-mentioned result ([1]) implies that d is unique, it suffices to bound the number of the e s. The essential tool to do this is an exponential gap principle between the solutions due to Okazaki (cf. [3, Lemma.]), which we may apply if there exist five such e s. Thus, we obtain the following. Theorem 1.. Fix a Diophantine triple a, b, c} with a < b < c. Then the number of Diophantine quintuples a, b, c, d, e} with c < d < e is at most four. Using Theorem 1., one may easily improve the bound 10 1930 due to Dujella ([9]) for the number of Diophantine quintuples. In fact, we reduce it to about the one-seventh power further by improving the upper bounds for b and d (in particular, for b). Theorem 1.3. The number of Diophantine quintuples is less than 10 76. This paper is organized as follows. In Section, we rephrase the assumption that a, b, c, d, e} is a Diophantine quintuple in terms of a system of three

THE NUMBER OF DIOPHANTINE QUINTUPLES 17 Diophantine equations, which induces binary recurrent sequences. Then, we completely determine the initial terms of the sequences. In Section 3, using some congruence relations we give lower bounds for the solutions. In Section 4, using Baker s theory we give upper bounds for the solutions and for b and d. It is to be noted that its proof tells that any regular Diophantine triple a, b, c} with c = a+b+r > 10 50 cannot be extended to a Diophantine quintuple (cf. Corollary 4.5). In Section 5, we give an exponential gap principle between the solutions. Combining it with the upper bound for the solutions we prove Theorem 1.. Finally, in Section 6 using the bounds for b and d and Theorem 1., we prove Theorem 1.3 along the same lines as Theorem 4 in [9].. The fundamental solutions of the system of Diophantine equations In this section, we transform the problem into solving a system of three Diophantine equations, which induces binary recurrent sequences u l }, v m }, w n }. Then, we determine the initial terms of the sequences, which we call the fundamental solutions of the system. We also examine the relationship between l, m and n in the case of u l = v m = w n. Let a, b, c} be a Diophantine triple with a < b < c, and r, s, t positive integers such that ab + 1 = r, ac + 1 = s, bc + 1 = t. Furthermore, suppose that a, b, c, d, e} is a Diophantine quintuple with c < d < e, and put ad + 1 = x, bd + 1 = y, cd + 1 = z with positive integers x, y, z. Then there exist integers α, β, γ, δ such that ae + 1 = α, be + 1 = β, ce + 1 = γ, de + 1 = δ, from which, by eliminating e, we obtain the system of Diophantine equations. (.1) (.) (.3) aδ dα = a d, bδ dβ = b d, cδ dγ = c d. Note that by Theorem.1 in [1] we know that d = d + = a+b+c+abc+rst. Lemma.1. (cf. [8, Lemma 1],[1, Lemma.1]) Let (δ, α), (δ, β), (δ, γ) be positive solutions of (.1), (.), (.3), respectively. Then there exist solutions

18 Y. FUJITA (δ 0, α 0 ) of (.1), (δ 1, β 1 ) of (.) and (δ, γ ) of (.3) in the ranges x + 1 1 α 0 < 0.76 4 d d ad, 1 δ 0 a < 0.69d, y + 1 1 β 1 < 0.73 4 d d bd, 1 δ 1 b < 0.148d, z + 1 1 γ < 0.73 4 d d cd, 1 δ c < 0.148d such that (.4) (.5) (.6) δ a + α d = (δ 0 a + α0 d)(x + ad) l, δ b + β d = (δ 1 b + β1 d)(y + bd) m, δ c + γ d = (δ c + γ d)(z + cd) n for some integers l, m, n 0. Proof. This immediately follows from Lemma.1 in [1]. By (.4), (.5) and (.6), we may write δ as follows: δ = u l, where u 0 = δ 0, u 1 = xδ 0 + dα 0, u l+ = xu l+1 u l ; δ = v m, where v 0 = δ 1, v 1 = yδ 1 + dβ 1, v m+ = yv m+1 v m ; δ = w n, where w 0 = δ, w 1 = zδ + dγ, w n+ = zw n+1 w n. Lemma.. If δ = u l = v m = w n, then l m n 0 (mod ) and δ 0 = δ 1 = δ = ±1. Proof. If l is odd, then by Lemma 8 in [8] we have δ 0 = y and δ 0 = z, which are contradictions. Similarly, if m is odd, then δ 1 = x, δ 1 = z, and if n is odd, then δ = x, δ = y; in any case, we arrive at contradictions. It follows that l, m, n must be all even. Then Lemma 3 in [7] implies that δ 0 = δ 1 = δ. Put e 0 = (δ 0 1)/d. Then ae 0 + 1 = α, be 0 + 1 = β, ce 0 + 1 = γ, de 0 + 1 = δ 0. If δ 0 1, then a, b, c, d, e 0} is a Diophantine quintuple with e 0 < d, which contradicts d = d + (cf. Section 1). Therefore, we obtain δ 0 = 1. Lemma.3. If δ = u l = v m = w n, then 4 n m l n.

THE NUMBER OF DIOPHANTINE QUINTUPLES 19 Proof. From Lemma 3 in [8] we see that n 1 m n + 1, n 1 l n + 1 and m 1 l m + 1, that is, n 1 m n + 1 and m 1 l n + 1, which together with Lemma. imply that n m l n. The first inequality n 4 follows from Lemmas.4 to.7 in [1] (see also [1, p. 7]). We need the following lemma in order to prove Theorem 1.. Lemma.4. If δ = u l = v m = w n, then m 6. Proof. It suffices to show that u 4 v 4 and u 6 > v 4. The latter immediately follows from Lemma.10 (1) in [1] and its proof, in view of δ 0 = δ 1 = ±1 and d > 4abc > b. Suppose that u 4 = v 4. By Lemma. we have ±a(ad + 1) + (ad + 1)x = ±b(bd + 1) + (bd + 1)y. Since (by ax)d + y x > 0, we have (b a) (b a)d + 1} = (by ax)d + y x. However, d > 4abc > b implies that (b a)d+1 < yd and by ax > (b a)y, which are contradictions. 3. Lower bounds for solutions In this section, we first introduce the notion of standard triples. Then, we give lower bounds for m in terms of d. Definition 3.1. Let a, b, c} be a Diophantine triple with a b 5 ; (ii) the second kind if b > 4a and c b ; (iii) the third kind if b > 1a and b 5/3 < c < b. A Diophantine triple is called standard if it is of the first, the second or the third kind. Remark 3.. The definition of standard triples in [8] differs from the one above in the following: The condition for the first kind in [8] is c > b 4.5 ; The second and the fourth kind in [8] correspond to the second kind above. We modified the first kind in order to get better upper bounds for b and d, and combined the second and the fourth kind because dividing those cases does not affect the bounds (see the proof of Proposition 4.3). Lemma 3.3. Any Diophantine quadruple contains a standard triple.

0 Y. FUJITA Proof. Let a, b, c, d} be a Diophantine quadruple. In view of the proof of Proposition in [8], we may assume that a, b, c, d} is regular and b < 4a. If c > b 3, then d > 4abc > b 5, whence a, b, d} is of the first kind. Assume that c b 3. Then, from the same argument as in the proof of Proposition in [8] we see that either (i) c = a + b + r or (ii) c > 4ab + a + b, and that in the case of (i), a, c, d} is of the second kind; in the case of (ii), d < c and d > 4abc > b c c 5/, whence a, c, d} is of the third kind. Remark 3.4. It follows from the proof of Lemma 3.3 that any Diophantine quadruple a, b, c, d} with a 10 100. (i) If a, b, c, d} contains a triple of the first kind, then m > d 0.05. (ii) If a, b, c, d} contains a triple of the second kind, then m > d 0.4. (iii) If a, b, c, d} contains a triple of the third kind, then m > d 0.19. Proof. Let A, B, C} be a standard triple with A C 0.05, (ii) k > C 0.4, (iii) k > C 0.19. (i) Suppose that k C 0.05. By Lemma 9 in [8] and Lemma., we have ±Aj + Sj ±Bk + Tk (mod 4C). Since Aj < C 0. 4C 0.05 < C, Sj < C 1. + 1 C 0.05 < C, Bk < C 0. C 0.05 < C, Tk < C 1. + 1C 0.05 < C, we have (3.1) ±Aj + Sj = ±Bk + Tk. Squaring (3.1) twice, we have (Aj Bk ) j k } 4j k (mod C). (3.) Since k C 0.05 yields (Aj Bk ) j k } < (Bk ) 4 < C 0.8 C 0. = C, 4j k 16k 4 16C 0.1 < C, the congruence (3.) is in fact an equation, that is, (3.3) Aj Bk = (j + k).

THE NUMBER OF DIOPHANTINE QUINTUPLES 1 By (3.1) and (3.3), we have j(s 1) = k(t ± 1), which together with (3.1) implies that ( ) T ± 1 S(T ± 1) ± A B} k = T S 1 S 1. It follows from C = d > 10 100 that (S + 1)(T + S) k = (AT + BS A + B) > AC B(S + 1) > 0.49C0.3 > C 0.05, which contradicts the assumption. Therefore, we obtain k > C 0.05. (ii) Note that in this case Lemma 4 in [8] and k 4 together imply that j 3k/ + 1/ 1.65k. Suppose that k C 0.4. Since Aj < 1 4 C0.5 ( 1.65C 0.4) < C, Sj < 1 4 C1.5 + 1 1.65C 0.4 < C, Bk < C ( 0.5 C 0.4) < C, Tk < C 1.5 + 1C 0.4 < C, we have the equation (3.1). We now have Aj S < j A < 1.65C0.4 0.5C 0.5 C C 0.5 = 0.815C 0.01 < 0.0815, Bk T < k B < C0.4 C 0.5 C C 0.5 = C 0.01 < 0.1. It follows from (3.1) that 1.0815Sj > 0.9Tk, that is, j k > 0.9 1.0815 T B S > 0.83 A > 1.664, which contradicts j 1.65k. Therefore, we obtain k > C 0.4. (iii) Note that in this case Lemma 4 in [8] and k 4 together imply that j 8k/5 + 3/5 1.75k. Suppose that k C 0.19. Since Aj < ( 1 1 C0.6 1.75C 0.19) < C, Sj < 1 1 C1.6 + 1 1.75C 0.19 < C, Bk < C 0.6 ( C 0.19) < C, Tk < C 1.6 + 1C 0.19 < C, we have the equation (3.1). In the same way as (ii), we see that Aj S < 0.0505, Bk T < 0.1, and that 1.0505Sj > 0.9Tk, that is, j/k >.965, which contradicts j 1.75k. Therefore, we obtain k > C 0.19. This completes the proof of Lemma 3.5. 4. Upper bounds for b and d In this section, combining Lemma 3.5 with a theorem of Matveev ([14]) we give upper bounds for b and d. In its proof, it emerges that if d > 10 100, then d > b 5, which implies that any regular Diophantine triple a, b, c} with c = a + b + r > 10 50 cannot be extended to a Diophantine quintuple.

Y. FUJITA We begin by quoting the theorem of Matveev. Theorem 4.1 ([14]). Let Λ be a linear form in logarithms of N multiplicatively independent totally real algebraic numbers α 1,..., α N with rational integer coefficients b 1,..., b N (b N 0). Let h(α j ) denote the absolute logarithmic height of α j for 1 j N. Define the numbers D, A j (1 j N) and E by D = [Q(α 1,...,α N ) : Q], A j = maxdh(α j ), log α j }, E = max 1, max b j A j /A N ; 1 j N}}. Then, log Λ > C(N)C 0 W 0 D Ω, where C(N) = 8 (N 1)! (N + )(N + 3)(4e(N + 1))N+1, C 0 = log(e 4.4N+7 N 5.5 D log(exd)), W 0 = log(1.5exed log(exd)), Ω = A 1 A N. Proposition 4.. Suppose that a, b, c, d, e} is a Diophantine quintuple with a B 5/3 and C 6440 > 10 3 (cf. Section 1). Put Λ = j log ξ k log η + log µ, where ξ = S + AC, η = T + BC, µ = B( C ± A) A( C ± B). Then, by (60) in [8] we know that (4.1) 0 < Λ < 8 3 ACξ j. We apply Theorem 4.1 with N = 3, D = 4, α 1 = ξ, α = η, α 3 = µ. We have A 1 = logα 1 < log( AC + 1) < log(.0001c 0.8 ) < 1.8007 logc, A = logα < log( BC + 1) < log(.0001c 0.8 ) < 1.8007 logc. Since the minimal polynomial of µ is A (C B) X 4 + 4A B(C B)X 3 + AB(3AB AC BC C )X + 4AB (C A)X + B (C A)

THE NUMBER OF DIOPHANTINE QUINTUPLES 3 up to a multiple of a constant, the leading coefficient a 0 of the minimal polynomial of µ satisfies ( ) 1 C 4 B 1 a 0 A (C B). Since we have B B( C ± A) B( C + B) A < < A( C B) A(C B) B(1 + B/C) B < < 1.6709 A(1 B/C) A, B B( C ± A) B( C B) A > > A( C + B) A(C B) ( ) B B B > 1 C A > 0.5607 A, A 3 = 4h(µ) < log ( 1.6709 B (C B) ) < 3.3489 logc, and noting that A, B, C} is a standard triple with C > 10 3 we have ( ) A 3 > log 0.5607 B(C B) 4A > log(c 1/5 ) = 0. logc. Since k j, we have E < ja 1 A 3 < 9.0035j. Hence we obtain the following: Since C(3) = 8 5 9(16e)4 < 6.4407 10 8, C 0 = log ( e 0. 3 5.5 16 log(4e) ) < 9.8847, W 0 = log (6exE log(4e)) < log(351j), Ω = A 1 A A 3 < 10.8583(logC) 3. log 8 3 ACξ j < log 8 3 C(4C) j < 1.006j log C, Theorem 4.1 and (4.1) together yield 1.006j log C < 6.4407 10 8 9.8847 16 10.8583(logC) 3 log(351j), whence we obtain j log(351j) <.786 101 (log C). Proposition 4. now follows from j m and C = d.

4 Y. FUJITA Proposition 4.3. Suppose that a, b, c, d, e} is a Diophantine quintuple with a 10 100, then d > b 5. Proof. We may assume that d > 10 100. In case a, b, c, d} contains a triple of the first kind, by Lemma 3.5 we have m > d 0.05, which together with Proposition 4. implies that m f(m) := log(351m)(logm) < 4.458 1015. Hence we have m < 5.45 10 0. It follows that d < m 40 < 10 830 and b < d 0. < m 8 < 10 166. In case a, b, c, d} contains a triple of the second or the third kind, m > d 0.19 and Proposition 4. together imply that f(m) < 7.718 10 13, which yields m < 7. 10 18. On the other hand, the assumption d > 10 100 implies that m > d 0.19 > 10 19, which is a contradiction. Therefore, we obtain d < 10 830 and b < 10 166. In addition, we have proved that a, b, c, d} cannot contain a triple of either the second or the third kind. Hence a, b, c, d} has to contain a triple of the first kind, which yields d > b 5. Corollary 4.4. Suppose that a, b, c, d, e} is a Diophantine quintuple with a exp.786 10 1 log(351m) > 10833, which contradicts Proposition 4.3. Corollary 4.5. Let a, b, c} be a Diophantine triple with c = a + b + r, where r = ab + 1. If c > 10 50, then a, b, c} cannot be extended to a Diophantine quintuple. Proof. Suppose that a, b, c, d, e} is a Diophantine quintuple. We may assume that c < d = d + < e by Theorem 1. in [1]. Since c > 10 50 means d > c > 10 100, we see from Proposition 4.3 that d > b 5. On the other hand, since d < 4b c and b 8, we have c > b 3 /4 16b, which contradicts c = a + b + r < 4b. 5. Proof of Theorem 1. Let A, B, C} be a Diophantine triple with A < B < C and let S = AC + 1, T = BC + 1. In this section, we first give an exponential gap principle for the solutions of (5.1) AZ CX = A C, BZ CY = B C,

THE NUMBER OF DIOPHANTINE QUINTUPLES 5 using Okazaki s idea (cf. [3, Lemma.]). Then we prove Theorem 1. combining the gap principle with Corollary 4.4. We may express the solutions of (5.1) as Z = V j = W k (j 0, k 0) with some binary recurrent sequences V j } and W k } (cf. Section ). Lemma 5.1. Suppose that V 0 = W 0 = ±1 and that there exist three positive solutions (X i, Y i, Z i ) (1 i 3) of (5.1) with Z 1 < Z < Z 3 that come from the same fundamental solution Z = V 0 = W 0. Put Z i = V ji (1 i 3) with j 1 < j < j 3. Then we have = W ki (5.) where j 3 j > 3 8AC ξj1 log η, ξ = S + AC, η = T + BC, = k k 1 k 3 k j j 1 j 3 j > 0. Proof. The proof proceeds along the same lines as that of Lemma. in [3]. Let ǫ = V 0 = W 0. By Z = V j = W k, we have Z = 1 ( C + ǫ A)ξ j ( C ǫ A)ξ j} A = 1 B whence we can find three points on the curve ( C + ǫ B)η k ( C ǫ B)η k}, (p i, q i ) = (j i log ξ, k i log η) (1 i 3) (5.3) Since F(p, q) =( C + ǫ B)e q ( C ǫ B)e q B ( C + ǫ A)e p ( C ǫ A)e p} = 0. A F q (p, q) = ( C + ǫ B)e q + ( C ǫ B)e q > 0 for all p and q, we may implicitly differentiate (5.3) to obtain (5.4) ( C + ǫ B)e q + ( C ǫ B)e q} dq dp B = ( C + ǫ A)e p + ( C ǫ A)e p}, A

6 Y. FUJITA which yields dq ( C + ǫ A)e dp = p ( C ǫ } A)e p + 4(C A) [ A/B ( C + ǫ B)e q ( C ǫ } ] B)e q + 4(C B) (5.5) ( C + ǫ A)e = p ( C ǫ } A)e p + 4(C A) ( C + ǫ A)e p ( C ǫ } > 1. A)e p + 4(AC/B A) Similarly, we may implicitly differentiate (5.4) to obtain ( C + ǫ B)e q + ( C ǫ B)e q} d q dp + ( C + ǫ B)e q ( C ǫ B)e q}( dq dr which yields (5.6) d q dp = 1 = ( C + ǫ B)e q ( C ǫ B)e q, ( ) } dq dp From (5.5) and (5.6) we see that (5.7) ) ( C + ǫ B)e q ( C ǫ B)e q ( C + ǫ B)e q + ( C ǫ < 0. B)e q 0 < q q 1 p p 1 q 3 q p 3 p < q q 1 p p 1 1. On the other hand, we may transform V ji = W ki (i = 1, ) into inequalities for linear forms in three logarithms of algebraic numbers 0 < p i q i + log µ < 8 3 ACξ ji (i = 1, ), where (cf. (4.1)). Hence we have µ = B( C + ǫ A) A( C + ǫ B) 0 < p q + log µ < p 1 q 1 + log µ < 8 3 ACξ j1, which yields (5.8) 0 < (q q 1 ) (p p 1 ) < 8 3 ACξ j1.

THE NUMBER OF DIOPHANTINE QUINTUPLES 7 It follows from (5.7) and (5.8) that 0 < q q 1 p p 1 q 3 q p 3 p < (q q 1 ) (p p 1 ) p p 1 < Substituting p i = j i log ξ and q i = k i log η, we have Therefore, we obtain 0 < k k 1 k 3 k 8AC < j j 1 j 3 j 3(j j 1 )ξ j1 log η. 8AC 3(p p 1 )ξ j1. as desired. j 3 j > 3 8AC ξj1 log η, We are now ready to prove Theorem 1.. Proof of Theorem 1.. Suppose that there exist five Diophantine quintuples a, b, c, d, e} (c < d < e) containing the fixed triple a, b, c}. Since d = d + by Theorem 1. in [1], we have five such e s, which yield five positive solutions δ = u li = v mi = w ni (1 i 5) with l 1 < l < l 3 < l 4 < l 5 of the system of Diophantine equations (.1), (.), (.3). Let A, B, C} be a standard triple with A < B < C contained in a, b, c, d} and Z = V j = W k the solutions of (5.1). Then, V 0 = W 0 = ±1 by Lemma. and we have three positive solutions (X i, Y i, Z i ) (1 i 3) of (5.1) that come from the same fundamental solution. Hence, putting Z i = V ji = W ki, we have the gap principle (5.). Since we may assume that C = d, we know that AC ad 6888 and BC bd 16 6440 (cf. Section 1), which together with Lemma.4 imply that ξ j1 AC > ( ad) m 1 1 (ad) 5 > 6.35 10, ad log η > log( bd) > 6.46. Moreover, since each of l i, m i and n i is even by Lemma., we have 0 (mod 4), which yields 4. It follows from (5.) that l 5 j 3 > 1.5 6.35 10 6.46 > 6.15 10 3, which contradicts Corollary 4.4. This completes the proof of Theorem 1..

8 Y. FUJITA 6. Proof of Theorem 1.3 In this section, we prove Theorem 1.3 along the same lines as Theorem 4 in [9]. Proof of Theorem 1.3. Suppose that a, b, c, d, e} is a Diophantine quintuple with a < b < c < d < e. We know by Proposition 4.3 that d < 10 830 and b < 10 166. We first bound the number of the a, b} s. If b 10 10, then the number is at most 10 04. Assume that 10 10 1 ω(b)log ω(b), where ω(b) denotes the number of distinct prime factors of b. If ω(b) b 0.3, then (6.1) implies that ω(b) < 101.6, which yields b < 10 10, a contradiction. Hence we have ω(b) < b 0.3, and the number of corresponding a, b} s is less than the following (see the proof of Theorem 1 in [9]): 10 166 1 b=10 10 +1 ω(b)+1 < 10 166 1 b=10 10 +1 10 166 b 0.3 < b 0.3 db < 10 16. 10 10 Therefore, the number of the a, b} s is less than 10 16. Secondly, for a fixed a, b} the number c such that a, b, c} is a Diophantine triple belongs to the union of finitely many binary recurrent sequences, and the number of the sequences is less than or equal to the number of solutions of the congruence z0 1 (mod b) with 0.71b0.75 < z 0 < 0.71b 0.75 (cf. [7, Lemma 1]). If b 10 58, then the number of the sequences is 0.71 10 58 0.75 < 5 10 43. Assume that 10 58 < b < 10 166. We see from (6.1) that ω(b) < b 1/3 in the same way as above. Hence, the number of the sequences is less than ω(b)+1 < 4b 1/3 < 9 10 55 (cf. [9, Lemma 1]). Moreover, the number of elements contained in each of the sequences is less than log 3 (10 830 ) < 600 (see the proof of Theorem 4 in [9]). Therefore, the number of the c s is less than 9 10 55 600 < 6 10 58. Consequently, we see from Theorem 1. that the number of Diophantine quintuples is less than 10 16 6 10 58 4 < 10 76, which completes the proof of Theorem 1.3. Acknowledgements. We would like to thank Professor Andrej Dujella and Professor Alan Filipin for pointing out some errors in the original draft of this manuscript. Thanks also go to the referee for the helpful suggestions. References [1] J. Arkin, V. E. Hoggatt and E. G. Strauss, On Euler s solution of a problem of Diophantus, Fibonacci Quart. 17 (1979), 333 339.

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