EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 15, No. 4, 2022, 1593-1596 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Diophantine Equation (p+ 4n)x + py = z2 Wachirarak Orosram1,∗, Kitsanuphong Makonwattana1, Saichon Khongsawat1 1Department of Mathematics, Faculty of Science, Buriram Rajabhat University, Buriram 31000, Thailand Abstract. In this paper, we study the Diophantine equation (p + 4n)x + py = z2, where n is a non-negative integer and p, p+4n are prime numbers such that p ≡ 7 (mod 12). We show that the non-negative integer solutions of such equation are (x, y, z) ∈ {(0, 1, √ p+ 1)}∪{(1, 0, 2 √ n+ p+1 4 )}, where √ p+ 1 and √ n+ p+1 4 are integers. 2020 Mathematics Subject Classifications: 11D61 Key Words and Phrases: exponential Diophantine equation, Catalan’s conjecture 1. Introduction A problem related to the Diophantine equation has been investigated by many re- searchers. It is considered one of the significant problems in elementary number theory. The proving method mainly uses a property in the integer system and algebraic number theory. Some of which appear in a higher system of the integer called the ring of integers. In 2011, Suvarnamani [10] considered a Diophantine equation 2x + py = z2 when p > 2 and p is a prime number. The result showed that (x, y, z) = (3, 0, 3) is a solution of the equation for all prime p > 2. If p = 3, then (x, y, z) = (4, 2, 5) is also a solution of the equation. If p = 1 + 2k+1 for some non-negative integer k, then (x, y, z) = (2k, 1, 1 + 2k). In 2012, the Diophantine equation 4x + py = z2, where x, y and z are non-negative in- tegers and p is a positive prime number was studied by Chotchaisthit [2]. The study revealed that the equation has no non-negative integer solution. In 2014, Suvarnamani [11] proved that the equation px+(p+1)y = z2 has a unique non-negative integer solution (p, x, y, z) = (3, 1, 0, 2) when p is an odd prime number. In 2016, Hoque [6] proved that there are exactly two solutions to (Mpq) x + (Mpq + 1)y = z2, where p, q ∈ Z such that p > 0, q > 1 and Mpq = pq − 1. In 2018, Kumar et al. [7] showed that the non-linear diophantine equation px + (p+ 6)y = z2 has no solution. Moreover, Fernando [4] showed ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v15i4.4508 Email addresses: Wachirarak.tc@bru.ac.th (W. Orosram), 620112210001@bru.ac.th (K. Makonwattana)620112210034@bru.ac.th (S. Khongsawat) https://www.ejpam.com 1593 © 2022 EJPAM All rights reserved. W. Orosram, K. Makonwattana, S. Khongsawat / Eur. J. Pure Appl. Math, 15 (4) (2022), 1593-1596 1594 that a Diophantine equation px + (p + 8)y = z2 has no positive-integer solution, when p, p + 8 are primes such that p > 3. In 2019, Kumar et al. [8] proved that the solution of an exponential Diophantine equation px + (p + 12)y = z2 has no non-negative integer solution, when p and p+12 are prime numbers such that p is in the form of 6n+1. In 2020, Burshtein [1] proved that a Diophantine equation px+(p+12)y = z2 has no positive integer solution (x, y, z), when p is a prime number such that p+ 5 = 22u. In 2021, Dokchan and Pakapongpun [3] studied a Diophantine equation px + (p+ 20)y = z2, when p and p+ 20 are primes and showed that the equation has no positive integer solution (x, y, z). In the same year, Gayo and Bacani [5] solved the Diophantine equation Mx p + (Mq + 1)y = z2 when Mp and Mq are Mersenne primes . In this work, we give solutions of the Diophantine equations 1+by = z2, 1+(d+4t)x = z2 where b, t, d are positive integers. Then, we extend to the solutions of the Diophantine equation (p+4n)x+py = z2 where p, p+4n are prime numbers such that p ≡ 7 (mod 12) and n is a positive integer such that n ≡ 0, 1 (mod 3). 2. Main results Proposition 1. (Catalan’s conjecture) (a, b, x, y) = (3, 2, 2, 3) is the unique solution of the Diophantine equation ax − by = 1, where a, b, x and y are integers such that min{a, b, x, y} > 1. This proposition was proved in 2004 by Mihailescu [9]. Lemma 1. Let b be a positive integer. The non-negative integer solutions to the Diophan- tine equation 1 + by = z2 is (y, z) = (1, √ b+ 1) if √ b+ 1) is a positive integer. Proof. Let b be a positive integer. We have z2−by = 1. By proposition 1, it is sufficient to consider the case b = 1 , z ≤ 1 or y ≤ 1. Hence, it remains to consider the following cases of b, y and z. If b = 1, then we have z2 = 2, which is impossible. If z = 0 or z = 1, then there is no solution. If y = 0, then we have z2 = 2 which is impossible. If y = 1, then we have z2 = b+ 1 or z = √ b+ 1. Thus, we have (y, z) = (1, √ b+ 1). Corollary 1. Let p be a prime number such that p ≡ 7 (mod 12). The non-negative integer solutions to the Diophantine equation 1+py = z2 is (y, z) = (1, √ p+ 1) if √ p+ 1) is a positive integer. Lemma 2. Let t and d be positive integers. The non-negative integer solutions of the Diophantine equation 1 + (d + 4t)x = z2 is (x, z) = ( 1, 2 √ t+ d+1 4 ) if √ t+ d+1 4 is a positive integer. Proof. Let t, d be positive integers such that √ t+ d+1 4 is a positive integer. We have z2 − (d + 4t)x = 1. By proposition 1, it is sufficient to consider only the case that z ≤ 1 or x ≤ 1. Hence, we consider the following cases of z and x. For z = 0 and z = 1, there is no solution. If x = 0, then we have z2 = 2, which is impossible. If x = 1, then we have W. Orosram, K. Makonwattana, S. Khongsawat / Eur. J. Pure Appl. Math, 15 (4) (2022), 1593-1596 1595 z2 = 4t + d + 1. Thus z = 2 √ t+ d+1 4 where √ t+ d+1 4 is a positive integer. Therefore, (x, z) = (1, 2 √ t+ d+1 4 ). Corollary 2. Let n be a positive integer and p, (p+4n) be prime numbers such that n ≡ 0, 1 (mod 3) and p ≡ 7 (mod 12). The non-negative integer solutions of the Diophantine equation 1+ (p+4n)x = z2 is (x, z) = ( 1, 2 √ n+ p+1 4 ) if √ n+ p+1 4 is a positive integer. Theorem 1. Let n be a positive integer such that n ≡ 0, 1 (mod 3), p ≡ 7 (mod 12). If √ p+ 1 and √ n+ p+1 4 are also integers, then all of the non-negative integer solutions to the Diophantine equation (p + 4n)x + py = z2 are given by (x, y, z) ∈ {(0, 1, √ p+ 1)} ∪ {(1, 0, 2 √ n+ p+1 4 )}, where p and p+ 4n are prime number. Proof. Since p is a prime number such that p ≡ 7 (mod 12), it is clear that p ≡ 3 (mod 4) and p ≡ 1 (mod 3). Let (x, y, z) be a non-negative integer solution of (p+4n)x+ py = z2. If x = 0 or y = 0, then (x, y, z) = (0, 1, √ p+ 1) or (x, y, z) = ( 1, 0, 2 √ n+ p+1 4 ) . Suppose x > 0 and y > 0. We consider the following cases. Case 1. x and y are even numbers. Since (p + 4n)x + py = z2, it follows that z is even. So z2 ≡ 0 (mod 4). Note that (p + 4n)x ≡ 1 (mod 4) and py ≡ 1 (mod 4). Thus (p+ 4n)x + py ≡ 2 (mod 4) which contradicts with z2 ≡ 0 (mod 4). Case 2. x and y are odd numbers. Since (p+4n)x ≡ 3 (mod 4) and py ≡ 3 (mod 4), it follows that (p+ 4n)x + py ≡ 2 (mod 4) which contradicts with z2 ≡ 0 (mod 4). Case 3. x is an even number and y is an odd number. Let x = 2k, k ≥ 1 and y = 2s+1, s ≥ 0. We have (p+4n)2k+p2s+1 = z2, or equivalently p2s+1 = z2−(p+4n)2k = [z + (p + 4n)k][z − (p + 4n)k]. Thus, there exist non-negative integers α, β such that pα = z+(p+4n)k and pβ = z− (p+4n)k, where α > β and α+β = 2s+1. Then, we have 2(p + 4n)k = pβ(pα−β − 1). This implies that β = 0. We have 2(p + 4n)k = (p2s+1 − 1), which is impossible because 2(p+ 4n)k ≡ 1, 2 (mod 3) but (p2s+1 − 1) ≡ 0 (mod 3). Case 4. x is an odd number and y is an even number. Let x = 2k + 1, k ≥ 0 and y = 2s, s ≥ 1. We have (p+ 4n)2k+1 + p2s = z2, or equivalently (p+ 4n)2k+1 = z2 − p2s = (z + ps)(z − ps). 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