On the Existence of \(\{p,q\}\)-Diophantine Quadruples and Quintuples

#A92 INTEGERS 25 (2025) ON THE EXISTENCE OF {p, q}-DIOPHANTINE QUADRUPLES AND QUINTUPLES Volker Ziegler Department of Mathematics, University of Salzburg, Salzburg, Austria [email protected] Received: 1/24/25, Accepted: 9/25/25, Published: 11/5/25 Abstract Let Sbe a set of primes. We call an m-tuple (a1, . . . , am) of distinct positive integers S-Diophantine if, for all i=j, the integers si,j := aiaj+ 1 have only prime divisors coming from the set S, i.e., if all si,j are S-units. In this paper we prove that for S={p, q}, no S-Diophantine quadruple exists if p≡3 mod 4 or q≡3 mod 4. We also prove that for S={p, q}no S-Diophantine quintuple exists. 1. Introduction Let Abe a set of kdistinct, positive integers. Gy˝ory, S´ark¨ozy, and Stewart [2] considered the product Π = Y a,b∈A a=b ab + 1, and found lower bounds for the number ωof prime factors of Π in terms of k. In particular, they showed that ω≫log k. They also conjectured that the largest prime factor of (ab + 1)(ac + 1)(bc + 1) tends to infinity as max{a, b, c}→∞. A weaker form, namely that the largest prime factor of (ab + 1)(ac + 1)(bd + 1)(cd + 1) tends to infinity as max{a, b, c, d}→∞, was proved by Stewart and Tijdeman [5], and the full conjecture was proved independently by Corvaja and Zannier [1] and Hern´andez and Luca [3]. We want to emphasize that the results of Stewart and Tijdeman [5] are effective, while the results of Corvaja and Zannier [1] and Hern´andez and Luca [3] are ineffective. In order to obtain sharp lower bounds for ωfor small kwe introduce the notion of S-Diophantine m-tuples. Let Sbe a set of primes. We call an m-tuple (a1, . . . , am) DOI: 10.5281/zenodo.17535199 INTEGERS: 25 (2025) 2 of distinct, positive integers S-Diophantine if, for all i=j, the integers si,j := aiaj+ 1 have only prime divisors coming from the set S, i.e., if all si,j are S-units. We are interested in finding lower bounds for min terms of |S|. Indeed the result of Gy˝ory, S`ark¨ozy, and Stewart [2] implies that for m≥exp(C|S|), where Cis an effectively computable constant, no S-Diophantine m-tuple exists. This bound seems to be far from the truth. However, it seems to be a difficult problem to find sharp bounds for min terms of |S|even for small values of |S|, as for |S|= 2 or |S|= 3. Let us note that the case that |S|= 3 has been studied by the author in [10]. In particular, all {p, q, r}-Diophantine quadruples with 2 ≤p<q<r≤100 have been found. Let us note that in the context of S-Diophantine tuples, the results of Corvaja, Zannier, Hern´andez, and Luca [1, 3] imply that for a fixed, finite set of primes Sonly finitely many S-Diophantine triples exist. Moreover, the result due to Stewart and Tijdeman [5] implies that for a fixed set of primes Sall S-Diophantine quadruples can be effectively computed. However, the following conjecture is still open. Conjecture 1. Let S={p, q}be a set of primes with p<q. Then no SDiophantine quadruple exists. Conjecture 1 has been confirmed in the following special cases: •If p2∤qordp(q)−1, q2∤pordq(p)−1, and q < pξholds for some ξ > 1, then there exists an effectively computable constant C=C(ξ) such that for all such primes p, q > C no S-Diophantine quadruple exists (see [6]). •No S-Diophantine quadruple exists, if p≡q≡3 mod 4 (see [7]). •No S-Diophantine quadruple exists, if p= 2 or p= 3 (see [9]). •No S-Diophantine quadruple exists, if p≡3 mod 4 and vq(pq−1−1) ≥2,and vp(qp−1−1) ≥max 2,log q log p, where vp(x) and vq(x) denote the p-adic and q-adic valuation of x, respectively (see [9]). •No S-Diophantine quadruple exists, if p, q < 105(see [8]). •No S-Diophantine quadruple exists if q=p+ 2 is a prime and pis large enough (see [4]). Let us note that most of the results listed above use results on lower bounds for linear forms in complex and p-adic logarithms. However, in this paper we can prove by using only elementary methods (mainly divisibility properties) the following theorem. INTEGERS: 25 (2025) 3 Theorem 1. Assume that S={p, q}is a set of primes with at least one of por q congruent to 3modulo 4. Then there does not exist an S-Diophantine quadruple. In other words, to prove Conjecture 1 we only have to consider primes with p≡q≡1 mod 4. Although a complete proof of Conjecture 1 seems to be out of reach at the moment, at least we can prove that no S-Diophantine quintuple exists if |S|= 2. Theorem 2. Let p<qbe prime numbers and S={p, q}. Then no S-Diophantine quintuple exists. In the next section we list some useful results concerning S-Diophantine m-tuples, which will be frequently used in the proofs of Theorems 1 and 2. In Section 3 we prove Theorem 1 and in Section 4 we prove Theorem 2. 2. Preliminaries Let S={p, q}be a set of two primes with p<qand let (a, b, c, d) be a hypothetical S-Diophantine quadruple. Then we write ab + 1 =pα1qβ1, bc + 1 =pα4qβ4, ac + 1 =pα2qβ2, bd + 1 =pα5qβ5, ad + 1 =pα3qβ3, cd + 1 =pα6qβ6. Moreover, we denote by sithe S-unit pαiqβifor i= 1,...,6. If we compute abcd in different ways, we obtain abcd = (ab)(cd)=(s1−1)(s6−1) = s1s6−s1−s6+ 1 = (ac)(bd)=(s2−1)(s5−1) = s2s5−s2−s5+ 1 = (ad)(bc)=(s3−1)(s4−1) = s3s4−s3−s4+ 1. Therefore, we obtain the three non-linear S-unit equations s1s6−s1−s6=s2s5−s2−s5,(1) s3s4−s3−s4=s2s5−s2−s5,(2) s1s6−s1−s6=s3s4−s3−s4.(3) We start with the following simple divisibility condition which was proved in [6, Lemma 2.1]. Lemma 1 ([6]).Let (a, b, c)be an S-Diophantine triple, with a<b<c, then s∤t with s=ac + 1 and t=bc + 1. INTEGERS: 25 (2025) 4 An immediate consequence of Lemma 1 is that we can exclude the following relations between exponents: α2=α4, α3=α5, α3=α6, α5=α6, β2=β4, β3=β5, β3=β6, β5=β6. On the other hand, we have the following lemma (cf. [6, Proposition 1] or [8, Lemma 2.1]), which one obtains by taking p-adic and q-adic valuations applied to Equations (1), (2), and (3). Lemma 2 ([6]).The smallest two exponents occuring in each of the three quadruples (α2, α3, α4, α5),(α1, α2, α5, α6), and (α1, α3, α4, α6)coincide. The same statement also holds with αreplaced by β. The following lemma also proves to be useful and yields upper bounds for a, b, c, and d. A proof of it can be found in [6, Lemma 3]. Lemma 3 ([6]).We have a|gcd s2−s1 gcd(s2, s1),s3−s1 gcd(s3, s1),s3−s2 gcd(s3, s2), b|gcd s4−s1 gcd(s4, s1),s5−s1 gcd(s5, s1),s5−s4 gcd(s5, s4), c|gcd s4−s2 gcd(s4, s2),s6−s2 gcd(s6, s2),s6−s4 gcd(s6, s4), d|gcd s5−s3 gcd(s5, s3),s6−s3 gcd(s6, s3),s6−s5 gcd(s6, s5). We also use the following result which is the content of [7, Section 4]. Lemma 4 ([7]).Let p, q be odd primes and assume that (a, b, c, d)is a {p, q}- Diophantine quadruple. Then the following system of equations cannot hold: ab + 1 = qβ1, bc + 1 = pα4qβ4, ac + 1 = pα2, bd + 1 = pα5, ad + 1 = pα3qβ3, cd + 1 = qβ6. (4) 3. Non-Existence of S-Diophantine Quadruples This section is devoted to the proof of Theorem 1. A key result to establish this theorem is a result obtained by Ziegler [9, Proposition 5.1]. Lemma 5 ([9]).Assume that p≡3 mod 4. Then one of the four cases in Table 1 holds. INTEGERS: 25 (2025) 5 Case The αexponents The βexponents I 0 = α1=α6< α4=α5< α2< α3β1=β2=β3< β4< β5< β6 II 0 = α1=α6< α2=α5< α3, α4β3=β4< β1=β2< β5< β6 III 0 = α2=α5< α1=α3≤α4< α6β1=β6< β3=β4< β2< β5 IV 0 = α3=α4< α1=α2< α5< α6β1=β6< β2=β5< β3, β4 Table 1: Restrictions to the exponents Note that if necessary we can exchange the roles of pand qto ensure that p≡3 mod 4. By the result due to Szalay and the author [6] we may assume that q≡1 mod 4, i.e., q≥5, and due to the author’s result [9] we may assume that p= 3, i.e., p≥7. In order to prove Theorem 1 we have to show that the restrictions from Lemma 5, i.e., the cases from Table 1, yield contradictions. Before we prove Theorem 1 we prove two lemmas first. We start with the following lemma. Lemma 6. Assume that Case III of Table 1 holds. Then we have d < 1.03·q2(β′−β) and b < 1.03 ·qβ′−β. Proof. First, we note that due to Lemma 3 we have d s6−s4 gcd(s4, s6)=pα6qβ−pα′qβ′ pα′qβ=pα6−α′−qβ′−β and, in particular, we obtain d < pα6−α′. Moreover, we have d b=s3−1 s1−1=pαqβ′−1 pαqβ−1=qβ′−β·1−p−αq−β′ 1−p−αq−β. Since we may assume that α′> α > 0, β > 0, p≥7, and q≥5, we get qβ′−β<d b<1.03 ·qβ′−β. Also, note that s−1 3=p−αq−β′< s−1 1=p−αq−β. Similarly we obtain d b=s6−1 s4−1=pα6qβ−1 pα′qβ′−1=pα6−α′qβ−β′·1−p−α6q−β 1−p−α′q−β′> pα6−α′qβ−β′. Comparing lower and upper bounds for d/b yields pα6−α′qβ−β′<d b<1.03 ·qβ′−β, INTEGERS: 25 (2025) 6 and hence we get pα6−α′<1.03 ·q2(β′−β). Therefore, we obtain d<pα6−α′<1.03 ·q2(β′−β). Now, we obtain the inequality for bby the following simple calculation: b=d·b d<1.03 ·q2(β′−β) qβ′−β<1.03 ·qβ′−β. Next, we prove the following lemma. Lemma 7. Assume that Case IV of Table 1 holds. Then we have b < 1.042·qβ′−β. Proof. We use Lemma 3 to obtain d s6−s5 gcd(s6, s5)=pα6qβ−pα′qβ′ pα′qβ=pα6−α′−qβ′−β, which implies d<pα6−α′. Further let us compute c b=s2−1 s1−1=pαqβ′−1 pαqβ−1=qβ′−β·1−p−αq−β′ 1−p−αq−β<1.03 ·qβ′−β. The last inequality comes from the fact that we may assume that α, β > 0 because of Lemma 4. Also, note that we may assume that p≥7 and q≥5. On the other hand we also have c b=s6−1 s5−1=pα6qβ−1 pα′qβ′−1=pα6−α′qβ−β′·1−p−α6q−β 1−p−α′q−β′> pα6−α′qβ−β′. The last inequality comes from the fact that s−1 6=p−α6q−β< p−α′q−β′=s−1 5 holds. Combining the inequalities for c/b we get pα6−α′<1.03 ·q2(β′−β). Therefore, we obtain from the previously found upper bound for dthe upper bound d<pα6−α′<1.03 ·q2(β′−β). Next, we note that ad =qβ3−1 = qβ3(1 −q−β3)≥0.992 ·qβ3, since β3≥3. Also, note that qβ3=s3> s2=pαqβ′. These two inequalities together with the upper bound for dyields a≥0.992 ·qβ3 d>0.992 ·qβ3+2β−2β′ 1.03 >0.96 ·pαq2β−β′. INTEGERS: 25 (2025) 7 Hence, we obtain b=s1−1 a<pαqβ 0.96 ·pαq2β−β′= 1.042 ·qβ′−β. Now, we start with the proof of Theorem 1. Proof of Theorem 1. We consider the four cases as shown in Table 1. Case I of Table 1. Due to Lemma 3 we have that d s5−s3 gcd(s5, s3)=pα5qβ5−pα3qβ3 pα5qβ3=qβ5−β3−pα3−α5. This implies d<qβ5−β3. Thus we obtain that b+ 1 = s5−1 d+ 1 >s5 d=pα5qβ3> qβ3. We also have β3=β1and α1= 0, which yields b+ 1 > qβ3=qβ1=s1=ab + 1 ≥b+ 1, which is a contradiction. Case II of Table 1. In this case we get  cd + 1 bd + 1 ·b c−1 = pα6qβ6·b−pα5qβ5·c pα5qβ5·c = N cpα5 = c−b bcd +c <1 bd + 1 =1 pα5qβ5, where Nis some integer unequal to 0. Note that N= 0 would imply b=c, which is excluded. This inequality yields 1≤ |N|<c qβ5 and therefore c > qβ5. By a similar argument, we also obtain a lower bound for b. Namely, we consider  bd + 1 ad + 1 ·a b−1 = pα5qβ5·a−pα3qβ3·b pα3qβ3·b = N bpα3−α5 = b−a abd +b <1 ad + 1 =1 pα3qβ3, INTEGERS: 25 (2025) 8 and deduce that 1≤ |N|<b pα5qβ3. Hence, we have b > pα5qβ3. Combining the lower bounds for band cwe obtain pα5qβ5< pα5qβ3+β5< bc < bd + 1 = pα5qβ5, which is an obvious contradiction. Case III of Table 1. Let us write α=α1=α3,α′=α4,β=β1=β6, and β′=β3=β4. With this notation we have ab + 1 = s1=pαqβ, ac + 1 = s2=qβ2, ad + 1 = s3=pαqβ′, bc + 1 = s4=pα′qβ′, bd + 1 = s5=qβ5, cd + 1 = s6=pα6qβ. Let us note that, due to Lemma 4, α, β = 0, that is, we have 0 < α ≤α′and 0< β < β′. Next, we note that by Lemma 3 we have that c s4−s2 gcd(s4, s2)=pα′qβ′−qβ2 qβ′=pα′−qβ2−β′, which implies c < pα′. Together with the bound for b, which we obtained in Lemma 6, we find pα′qβ′=bc + 1 <1.03 ·qβ′−βpα′+ 1 < pα′qβ′, which is a contradiction. Case IV of Table 1. Let us write α=α1=α2,α′=α5,β=β1=β6, and β′=β2=β5. With this notation we have ab + 1 = s1=pαqβ, ac + 1 = s2=pαqβ′, ad + 1 = s3=qβ3, bc + 1 = s4=qβ4, bd + 1 = s5=pα′qβ′, cd + 1 = s6=pα6qβ. Again we have α, β = 0 due to Lemma 4, that is, we have 0 < α ≤α′and 0 < β < β′. Next, let us find an upper bound for c. Indeed we find c s4−s2 gcd(s4, s2)=qβ4−pαqβ′ qβ′=qβ4−β′−pα, which yields c < qβ4−β′. With this upper bound for ccombined with the upper bound for bobtained in Lemma 7 we obtain qβ4=bc + 1 <1.042 ·qβ′−βqβ4−β′+ 1 = 1.042 ·qβ4−β+ 1. This inequality cannot hold unless β= 0. However, we can exclude the case that β= 0 due to Lemma 4. INTEGERS: 25 (2025) 9 4. Non-Existence of S-Diophantine Quintuples This section is devoted to the proof of Theorem 2. Therefore, let (a, b, c, d, e) be a {p, q}-Diophantine quintuple with a<b<c<d<e. Then we write ab + 1 = s1=pα1qβ1, cd + 1 = s6=pα6qβ6, ac + 1 = s2=pα2qβ2, ae + 1 = s7=pα7qβ7, ad + 1 = s3=pα3qβ3, be + 1 = s8=pα8qβ8, bc + 1 = s4=pα4qβ4, ce + 1 = s9=pα9qβ9, bd + 1 = s5=pα5qβ5, de + 1 = s10 =pα10 qβ10 . Before we start our proof of Theorem 2 we prove restrictions for the exponents αiand βifor an S-Diophantine quadruple (a, b, c, d). Proposition 1. Assume that (a, b, c, d)is a {p, q}-Diophantine quadruple. Then one of the 14 cases listed in Table 2 holds. Case exponents Case exponents Iα1=α2=α3< α4, α6< α5VIII α1=α4=α5< α2, α6< α3 β1=β4=β5< β2, β6< β3β1=β2=β3< β4, β6< β5 II α1=α2=α3< α4< α5< α6IX α1=α6< α4=α5< α2< α3 β1=β6< β4=β5< β2< β3β1=β2=β3< β4< β5< β6 III α1=α6< α2=α3< α4< α5Xα1=α4=α5< α2< α3< α6 β1=β4=β5< β2< β3< β6β1=β6< β2=β3< β4< β5 IV α2=α5< α1=α3< α4< α6XI α1=α6< α3=α4< α2< α5 β1=β6< β3=β4< β2< β5β2=β5< β1=β3< β4< β6 Vα2=α5< α1=α4< α3< α6XII α1=α6< α3=α4< α2< α5 β1=β6< β3=β4< β2< β5β2=β5< β1=β4< β3< β6 VI α1=α6< α2=α5< α3, α4XIII α3=α4< α1=α2< α5< α6 β3=β4< β1=β2< β5< β6β1=β6< β2=β5< β3, β4 VII α2=α5< α3=α4< α1< α6XIV α1=α6< α3=α4< α2< α5 β1=β6< β3=β4< β2< β5β2=β5< β3=β4< β1< β6 Table 2: Restrictions on the exponents for the S-Diophantine quadruple (a, b, c, d) Let us note that the Cases I-VII of Table 2 are the same cases as the Cases VIII-XIV of Table 2 but with the roles of the α’s and β’s exchanged. This is easily explained by exchanging the roles of pand q. Proof of Proposition 1. We consider S-unit Equation (2) in view of Lemma 2 and obtain that one of the following four cases holds: •α2=α3< α4, α5, INTEGERS: 25 (2025) 16 References [1] P. Corvaja and U. Zannier, On the greatest prime factor of (ab + 1)(ac + 1), Proc. Amer. Math. Soc. 131 (6) (2003), 1705–1709. [2] K. Gy˝ory, A. S´ark¨ozy, and C. L. Stewart, On the number of prime factors of integers of the form ab + 1, Acta Arith. 74 (4) (1996), 365–385. [3] S. Hern´andez and F. Luca, On the largest prime factor of (ab + 1)(ac + 1)(bc + 1), Bol. Soc. Mat. Mexicana (3) 9(2) (2003), 235–244. [4] F. Luca, Diophantine S-quadruples with two primes which are twin, Acta Math. Hungar. 159 (2) (2019), 589–602. [5] C. L. Stewart and R. Tijdeman, On the greatest prime factor of (ab + 1)(ac + 1)(bc + 1), Acta Arith. 79 (1) (1997), 93–101. [6] L. Szalay and V. Ziegler, On an S-unit variant of Diophantine m-tuples, Publ. Math. Debrecen 83 (1-2) (2013), 97–121. [7] L. Szalay and V. Ziegler, S-Diophantine quadruples with two primes congruent to 3 modulo 4, Integers 13 (2013), #A80, 9 pp. [8] L. Szalay and V. Ziegler, S-Diophantine quadruples with S={2, q},Int. J. Number Theory 11 (3) (2015), 849–868. [9] V. Ziegler, On the existence of S-Diophantine quadruples, Glas. Mat., III. Ser. 54 (2) (2019), 279–319. [10] V. Ziegler, Finding all S-diophantine quadruples for a fixed set of primes S,Monatsh. Math. 196 (3) (2021), 617–641.