EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 4, 2023, 2208-2212 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Diophantine Equation (p+ n)x + py = z2 where p and p+ n are prime numbers Wachirarak Orosram Department of Mathematics, Faculty of Science, Buriram Rajabhat University, Buriram 31000, Thailand Abstract. In this paper, we study the Diophantine equation (p + n)x + py = z2, where p, p + n are prime numbers and n is a positive integer such that n ≡ 0 (mod 4). In case p = 3 and n = 4, Rao [7] showed that the non-negative integer solutions are (x, y, z) = (0, 1, 2) and (1, 2, 4). In case p > 3 and p ≡ 3 (mod 4), if n − 1 is a prime number and 2n − 1 is not prime number, then the non-negative integer solution (x, y, z) is (0, 1, √ p+ 1) or (1, 0, √ p+ n+ 1). In case p ≡ 1 (mod 4), the non-negative integer solution (x, y, z) is also (0, 1, √ p+ 1) or (1, 0, √ p+ n+ 1). 2020 Mathematics Subject Classifications: 11D61 Key Words and Phrases: Diophantine equation, Catalan’s conjecture 1. Introduction Many mathematicians have been studying the Diophantine equations of the type (p+ n)x+py = z2 with a constant n and a specific condition of p, for example, in case that p is a prime number. In 2015, Tatong and Suvarnamani [10] found that (p, x, y, z) = (3, 1, 0, 2) is a unique non-negative integer solution of the Diophantine equation px+(p+1)y = z2 where p is an odd prime number. In 2018, Burshtein [1] showed that the Diophantine equation px+(p+4)y = z2 when p > 3, p+4 are primes has no positive integer solution (x, y, z). In the same year, Fernando [3] showed that px+(p+8)y = z2 has no positive integer solution, when p > 3 and p+ 8 are primes. In addition, Kumar, Gupta and Kishan [5] proved that the solution of px+(p+12)y = z2 has no non-negative integer solution where p and p+12 are prime numbers and p = 6n + 1 for some natural number n. In 2021, Dokchan and Pakapongpun [2] 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 Jr and Bacani [4] solved the Diophantine equation Mx p + (Mq +1)y = z2 where Mp and Mq are Mersenne primes. In 2022, Tadee [8] gave the solutions of equations px+(p+14)y = z2, where p and p+14 are primes. In 2023, Viriyapong and Viriyapong [11] studied the Diophantine equation ax + (a+ 2)y = z2, where a ≡ 5 (mod 21) and showed DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4822 Email address: Wachirarak.tc@bru.ac.th (W. Orosram) https://www.ejpam.com 2208 © 2023 EJPAM All rights reserved. W. Orosram / Eur. J. Pure Appl. Math, 16 (4) (2023), 2208-2212 2209 that the equation has no non-negative integer solution (x, y, z). In the same year, Tadee and Siraworakun [9] studied the Diophantine equation px+(p+2q)y = z2, where p, q, p+2q are prime numbers and showed that the equation has no positive integer solution. In this paper, we give a solution of the Diophantine equation (p+n)x+py = z2, where p, p+n are odd prime numbers such that n ≡ 0 (mod 4). To obtain our result, we consider two cases of p in modulo 4, i.e., the case that p ≡ 1 (mod 4) and the case that p ≡ 3 (mod 4). In case p ≡ 3 (mod 4), we first consider case p = 3 and n = 4. The considered equation in this case is 3x + 7y = z2 in which the solutions were given by Rao [7] that (x, y, z) = (0, 1, 2) or (x, y, z) = (1, 2, 4). Next, we consider p ≡ 3 (mod 4) such that p > 3 with a specific condition that n− 1 is a prime number and 2n− 1 is not a prime number. Then, our final case is when p ≡ 1 (mod 4) and n is a positive integer. 2. Preliminaries Proposition 1. (Catalan’s conjecture) The Diophantine equation ax − by = 1 where min{a, b, x, y} > 1 has a unique solution (a, b, x, y) = (3, 2, 2, 3). This proposition was proved in 2004 by Mihailescu [6]. Lemma 1. Let p be odd prime number. The non-negative integer solution to the Dio- phantine equation 1 + py = z2 is (y, z) = (1, √ p+ 1) if √ p+ 1 is a positive integer. Proof. Let (y, z) be a non-negative integer solution of 1 + py = z2. If y = 0, then z2 = 2, which is impossible. If y > 1, then z > 1. By Catalan’s conjecture, there is no non-negative integer solution. If y = 1, then z2 = p+ 1. So z = √ p+ 1. This means that√ p+ 1 is a non-negative integer as z is a non-negative integer. Lemma 2. Let n be a positive integer such that n ≡ 0 (mod 4) and let p, p+ n be prime numbers. The non-negative integer solutions of the Diophantine equation 1+(p+n)x = z2 is (x, z) = (1, √ p+ n+ 1) if √ p+ n+ 1 is a positive integer. Proof. Let (x, z) be a non-negative integer solution of 1+(p+n)x = z2. If x = 0, then z2 = 2, which is impossible. If x > 1, then z > 1. By Catalan’s conjecture, there is no of a non-negative integer solution. If x = 1, then z2 = p + n + 1. So z = √ p+ n+ 1. This means that √ p+ n+ 1 is a non-negative integer as z is a non-negative integer. Theorem 1. ([7]) Let n = 4 and p = 3. The non-negative integer solutions of the Dio- phantine equation (p+ n)x + py = z2 are (x, y, z) = (0, 1, 2) and (1, 2, 4). This theorem was proved in 2017 by Rao [7]. 3. Main results In Theorem 2, we use the same method of proof as appeared in [1, 2] to derive the result. W. Orosram / Eur. J. Pure Appl. Math, 16 (4) (2023), 2208-2212 2210 Theorem 2. Let n and p be positive integers where n ≡ 0 (mod 4), p ≡ 3 (mod 4) and p > 3 such that p, p+n, n−1 are prime numbers and 2n−1 is not a prime number. If √ p+ 1 and √ p+ n+ 1 are integers, then all of the non-negative integer solutions of the Diophan- tine equation (p+n)x+py = z2 are given by (x, y, z) ∈ {(0, 1, √ p+ 1), (1, 0, √ p+ n+ 1)}, where x, y and z are non-negative integer. Proof. Let (x, y, z) be a non-negative integer solution of (p + n)x + py = z2. If x = 0 or y = 0, then (x, y, z) = (0, 1, √ p+ 1) or (x, y, z) = ( 1, 0, √ p+ n+ 1 ) by Lemma 1 and Lemma 2. Now, we suppose x > 0 and y > 0. We consider the following cases. If x and y are even, then (p+n)x ≡ 1 (mod 4) and py ≡ 1 (mod 4). Thus (p+n)x+py ≡ 2 (mod 4) which is impossible since z2 ≡ 0, 1 (mod 4). If x and y are odd, then (p+n)x ≡ 3 (mod 4) and py ≡ 3 (mod 4). Thus (p+ n)x + py ≡ 2 (mod 4) which is impossible since z2 ≡ 0, 1 (mod 4). Now, there are two remaining cases to be considered. Case 1. x is even and y is odd. There exist k ≥ 1 and s ≥ 0 such that x = 2k, y = 2s+1. We have (p+n)2k+p2s+1 = z2, which can be rewritten as p2s+1 = z2 − (p+ n)2k = [ z − (p+ n)k ] [ z + (p+ n)k ] . Thus, there exist non-negative integers α, β that pα = z− (p+ n)k and pβ = z+(p+ n)k, where α < β and α+ β = 2s+ 1. Hence 2 (p+ n)k = pα [ pβ−α − 1 ] . If α ≥ 1, then p | (p + n) which is impossible since p and p + n are different primes. In Case α = 0, we have 2(p + n)k = p2s+1 − 1. If s = 0, then 2(p + n)k + 1 = p which is impossible. If s ≥ 1, then we have 2(p+ n)k = (p− 1) [ p2s + p2s−1 + · · ·+ p+ 1 ] . Since p−1 is even and p2s+p2s−1+· · ·+p+1 is odd, it follows that p−1 is an even positive divisor of 2(p + n)k that is p − 1 = 2 (p+ n)j , for some integer j such that 0 ≤ j < k. If j = 0, then p = 3 which contradicts p > 3. If 1 ≤ j < k, then 2(p+ n)j + 1 = p which is also impossible. Case 2. x is odd and y is even. There exist k ≥ 0 and s ≥ 1 such that x = 2k + 1 and y = 2s. We now have (p+ n)2k+1 + (p)2s = z2, which can be rewritten as (p+ n)2k+1 = z2 − p2s = (z + ps) (z − ps) . Thus, there exist non-negative integers α, β such that (p+ n)α = z − ps and (p+ n)β = z + ps, where α < β and α+ β = 2k + 1. Then 2ps = (p+ n)α [ (p+ n)β−α − 1 ] . W. Orosram / Eur. J. Pure Appl. Math, 16 (4) (2023), 2208-2212 2211 If α ≥ 1, then (p+n) | p which is impossible. In Case α = 0, we have 2ps = (p+ n)2k+1−1. If k = 0, then 2ps−p = n−1. Hence p [ 2ps−1 − 1 ] = n−1. Since n−1 is prime, it follows that p = n− 1 this contradicts the fact that p+ n = 2n− 1 is not prime. If k ≥ 1, then 2ps = (p+ n− 1) [ (p+ n)2k + (p+ n)2k−1 + · · ·+ (p+ n) + 1 ] . Since p + n − 1 is even and (p+ n)2k + (p+ n)2k−1 + · · · + (p+ n) + 1 is odd, it follows that p + n − 1 is an even positive divisor of 2ps that is p + n − 1 = 2pl, for some integer l such that 0 ≤ l < s. If l = 0, then p+ n = 3, which is impossible since p > 3 and n are positive integer. If 1 ≤ l < s, then p [ 2(p)l−1 − 1 ] = n− 1 which is also impossible. Example 1. There are infinitely many n, p of the form n ≡ 0 (mod 4),p ≡ 3 (mod 4) where p > 3 such that p, p+n, n−1 are prime numbers and 2n−1 is not a prime number. Some Diophantine equations of particular values of n where n is between 1 to 70 with positive integers √ p+ 1 and √ p+ n+ 1 are given in the table below. n (p+ n)x + py = z2 (x, y, z) 8 (p+ 8)x + py = z2 {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 9)} 20 (p+ 20)x + py = z2 [2] {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 21)} 32 (p+ 32)x + py = z2 {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 33)} 44 (p+ 44)x + py = z2 {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 45)} 48 (p+ 48)x + py = z2 {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 49)} 60 (p+ 60)x + py = z2 {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 61)} 68 (p+ 68)x + py = z2 {(0, 1, √ p+ 1} ∪ {(1, 0, √ p+ 69)} Table 1: Diophantine equations satisfying the condition in Theorem 2. Theorem 3. Let p and n be a positive integer where n ≡ 0 (mod 4), p ≡ 1 (mod 4) such that p and p + n are prime numbers. If √ p+ 1 and √ p+ n+ 1 are integers, then the all of the non-negative integer solutions of (p + n)x + py = z2 are given by (x, y, z) ∈ {(0, 1, √ p+ 1), (1, 0, √ p+ n+ 1)}. Proof. Let (x, y, z) be a non-negative integer solution of (p + n)x + py = z2. If x = 0 or y = 0, then (x, y, z) = (0, 1, √ p+ 1) or (x, y, z) = ( 1, 0, 2 √ p+ n+ 1 ) by Lemma 1 and Lemma 2. If x > 0 and y > 0, then (p + n)x ≡ 1 (mod 4) and py ≡ 1 (mod 4). Thus (p+ n)x + py ≡ 2 (mod 4) which is impossible since z2 ≡ 0, 1 (mod 4). Acknowledgements The authors wish to thank the referees for their kind suggestions and comments to improve the article. REFERENCES 2212 References [1] Nechemia Burshtein. On the diophantine equation px+(p+4)y = z2 when p > 3, p+4 are primes is insolvable in positive integers. Annals of Pure and Applied Mathematics, 16(2):283–286, 2018. [2] Rakporn Dokchan and Apisit Pakapongpun. 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