EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 2, 2023, 724-735 ISSN 1307-5543 – ejpam.com Published by New York Business Global Non-existence of Positive Integer Solutions of the Diophantine Equation px + (p+ 2q)y = z2, where p, q and p+ 2q are Prime Numbers Suton Tadee1, Apirat Siraworakun1,∗ 1 Department of Mathematics, Faculty of Science and Technology, Thepsatri Rajabhat University, Lopburi 15000, Thailand Abstract. The Diophantine equation px + (p + 2q)y = z2, where p, q and p + 2q are prime numbers, is studied widely. Many authors give q as an explicit prime number and investigate the positive integer solutions and some conditions for non-existence of positive integer solutions. In this work, we gather some conditions for odd prime numbers p and q for showing that the Diophantine equation px + (p + 2q)y = z2 has no positive integer solution. Moreover, many examples of Diophantine equations with no positive integer solution are illustrated. 2020 Mathematics Subject Classifications: 11D61 Key Words and Phrases: Diophantine Equation, Legendre Symbol, The Chinese Remainder Theorem 1. Introduction Studying non-negative integer solutions of the Diophantine equation px + qy = z2, where p and q are prime numbers, has been done in numerous ways. One of them is that p and q are given as explicit prime numbers. For example, in [4] and [5], Kumar, Gupta and Kishan showed that the Diophantine equations 61x+67y = z2, 67x+73y = z2, 31x + 41y = z2 and 61x + 71y = z2 have no non-negative integer solution and Burshtein [3] revealed that the Diophantine equations 2x + 11y = z2 and 19x + 29y = z2 have no positive integer solutions (x, y, z). Many researchers studied the Diophantine equation by considering q = p+k, where k is an even number. In [2], Burshtein investigated the solutions of the Diophantine equation px + (p + 6)y = z2, where p and p + 6 are primes and x + y = 2, 3, 4. Gupta, Kumar and Kishan [6] studied the Diophantine equation px + (p + 6)y = z2, where p and p + 6 are sexy primes with p = 6n + 1 and n is a natural number. Burshtein [1] showed that ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i2.4702 Email addresses: suton.t@lawasri.tru.ac.th (S. Tadee), apirat.si@lawasri.tru.ac.th (A. Siraworakun) https://www.ejpam.com 724 © 2023 EJPAM All rights reserved. S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 725 the Diophantine equation px + (p + 4)y = z2, where p > 3 and p + 4 are primes, has no positive integer solutions (x, y, z). In addition, Rao [10] studied the Diophantine equation 3x + 7y = z2. Neres [9] investigated the Diophantine equation px + (p + 8)y = z2, where p > 3 and p + 8 are primes. Moreover, Tadee ([13], [14]) has given the solutions of the Diophantine equations px + (p + 10)y = z2 and px + (p + 14)y = z2, where p, p + 10 and p+ 14 are primes. In [7], Mina and Bacani use the concepts of Legendre symbol and Jacobi symbol to find some condition for non-existence of solutions of the Diophantine equations of the form px + qy = z2n. Two years later, the solutions of the Diophantine equation px + (p+ 4k)y = z2, where k is a natural number and p, p+ 4k are prime numbers, were investigated [8]. The goal of this article is to give some conditions on primes p and q to show that the Diophantine equation px+(p+2q)y = z2, where p, q and p+2q are prime numbers, has no positive integer solution. Moreover, the forms of odd prime numbers p, when q is a prime number, are investigated and many examples of Diophantine equations with no positive integer solution are demonstrated. 2. Preliminaries First, we recall some elementary definitions and theorems in number theory. See [11] for instance. Definition 1. Let n be a positive integer. The Euler phi-function ϕ(n) is defined to be the number of positive integers not exceeding n that are relatively prime to n. Definition 2. Let a and n be relatively prime integers with a ̸= 0 and n > 0. The least positive integer x such that ax ≡ 1 (mod n) is called the order of a modulo n and is denoted by ordna. Theorem 1. (Fermat’s Little Theorem). If p is a prime number and a is an integer with p ∤ a, then ap−1 ≡ 1 (mod p). Definition 3. Let r and n be relatively prime integers with n > 0. The integer r is called a primitive root modulo n if ordnr = ϕ(n). Theorem 2. Every prime number has a primitive root. The concepts of quadratic residue and Legendre symbol have important roles in this paper. Definition 4. Let a and m be positive integers with (a,m) = 1. We say that a is a quadratic residue of m if the congruence x2 ≡ a (mod m) has a solution. Otherwise, a is a quadratic nonresidue of m. S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 726 Definition 5. Let p be an odd prime number and a be an integer with p ∤ a. The Legendre symbol (ap ) is defined by( a p ) = { 1 if a is a quadratic residue of p −1 if a is a quadratic nonresidue of p. Some properties of Legendre symbol are given in Theorems 3 and 4. Theorem 3. Let p be an odd prime number and a, b be integers with p ∤ a and p ∤ b. The following statements hold. (i) If a ≡ b (mod p), then ( a p ) = ( b p ) . (ii) ( ab p ) = ( a p )( b p ) ; (iii) ( a2 p ) = 1. Theorem 4. (The Law of Quadratic Reciprocity). Let p and q be distinct odd prime numbers. Then ( p q )( q p ) = (−1)( p−1 2 )( q−1 2 ). The following theorem shows the form of odd prime number p in the Legendre symbol( 2 p ) . Theorem 5. Let p be an odd prime number. Then( 2 p ) = { 1 if p ≡ ±1 (mod 8) −1 if p ≡ ±3 (mod 8). Theorem 6. Let p and q be distinct odd prime numbers. (i) If p ≡ 1 (mod 4) or q ≡ 1 (mod 4), then ( p q ) = ( q p ) . (ii) If p ≡ 3 (mod 4) and q ≡ 3 (mod 4), then ( p q ) = − ( q p ) . Theorem 7. Let a1, a2, a3, . . . , an be integers and m1,m2,m3, . . . ,mn be positive integers. Then, the system of congruences x ≡ a1 (mod m1) x ≡ a2 (mod m2) x ≡ a3 (mod m3) ... x ≡ an (mod mn) has a solution if and only if (mi,mj)|(ai − aj) for all pairs of integer (i, j). S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 727 In 2004, Siraworakun investigated the forms of odd prime numbers p in Legendre symbol ( q p ) , where q is an odd prime number, in his unplublished senoir project. We review some important results in Theorem 8 - 11. Theorem 8. Let p be an odd prime number. Then there is no a primitive root modulo p in the form n2, where n is a natural number with n < p. Proof. Assume that there is a primitive root n2 0 modulo p, where n0 is a natural number with n0 < p. Then (n2 0) p−1 ≡ 1 (mod p). So (np−1 0 )2 ≡ 1 (mod p). Thus, np−1 0 ≡ 1 (mod p) or np−1 0 ≡ −1 (mod p). If np−1 0 ≡ 1 (mod p), then (n2 0) p−1 2 ≡ 1 (mod p). It contradicts to the order of n2 0 modulo p. Hence, np−1 0 ≡ −1 (mod p). Since n0 < p, it contradicts to Fermat ’s Little Theorem. Therefore, there is no a primitive root modulo p in the form n2, where n is a natural number with n < p. Theorem 9. Let p be an odd prime number and r be a primitive root modulo p. Then r2, r4, r6, . . . , rp−1 are quadratic residues of p and r1, r3, r5, . . . , rp−2 are quadratic non- residues of p. Proof. It is obvious that r2, r4, r6, . . . , rp−1 are quadratic residues of p. Since r is a primitive root modulo p and Theorem 8, there is no a natural number n0 with n0 < p such that n2 0 ≡ r (mod p). Then ( r p ) = −1. By Theorem 3(ii), (iii), we have ( r3 p ) = ( r5 p ) = · · · = ( rp−2 p ) = −1 . Hence, r1, r3, r5, . . . , rp−2 are quadratic nonresidues of p. To find the forms of odd prime numbers p in Legendre symbol ( q p ) , where q is an odd prime number, we use the Chinese Remainder Theorem for solving the system of congruences. Theorem 10. Let p and q be distinct odd prime numbers with q ≡ 1 (mod 4). Then( q p ) = { 1 if p ≡ q + rS1q + rS1 (mod 2q) −1 if p ≡ q + rS2q + rS2 (mod 2q) , where S1 ∈ {2, 4, 6, . . . , q− 1}, S2 ∈ {1, 3, 5, . . . , q− 2} and r is a primitive root modulo q. Proof. Since q ≡ 1 (mod 4), we have ( q p ) = ( p q ) by Theorem 6(i). Let r be a primitive root modulo q. By Theorem 9, we obtain that( p q ) = { 1 if p ≡ rS1 (mod q) −1 if p ≡ rS2 (mod q) , where S1 ∈ {2, 4, 6, . . . , q − 1} and S2 ∈ {1, 3, 5, . . . , q − 2}. Case 1. ( q p ) = 1. Then ( p q ) = 1. Thus, p ≡ rS1 (mod q). Since p ≡ 1 (mod 2) and by the Chinese Remainder Theorem, we obtain that p ≡ q + rS1q + rS1 (mod 2q). S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 728 Case 2. ( q p ) = −1. Then ( p q ) = −1. Thus, p ≡ rS2 (mod q). Since p ≡ 1 (mod 2) and by the Chinese Remainder Theorem, we obtain that p ≡ q + rS2q + rS2 (mod 2q). Theorem 11. Let p and q be distinct odd prime numbers with q ≡ 3 (mod 4). Then( q p ) = { 1 if p ≡ 3q + 4n0r S1 , − 3q + 4n0r S2 (mod 4q) −1 if p ≡ 3q + 4n0r S2 , − 3q + 4n0r S1 (mod 4q) , where S1 ∈ {2, 4, 6, . . . , q− 1}, S2 ∈ {1, 3, 5, . . . , q− 2}, r is a primitive root modulo q and n0 = q + 1 4 . Proof. Let r be a primitive root modulo q. By Theorems 6 and 9, we obtain that ( q p ) =  ( p q ) if p ≡ 1 (mod 4) − ( p q ) if p ≡ 3 (mod 4) and ( p q ) = { 1 if p ≡ rS1 (mod q) −1 if p ≡ rS2 (mod q) , where S1 ∈ {2, 4, 6, . . . , q−1} and S2 ∈ {1, 3, 5, . . . , q−2}. Since q ≡ 3 (mod 4), we choose an integer n0 = q + 1 4 . Then q = 4n0 − 1. Thus 3q ≡ 1 (mod 4) and 4n0 ≡ 1 (mod q). In the following cases, the systems of congruences are solved by the Chinese Remainder Theorem. Case 1. ( q p ) = 1. Case 1.1 ( q p ) = ( p q ) and ( p q ) = 1. Then p ≡ 1 (mod 4) and p ≡ rS1 (mod q). Thus, p ≡ 3q + 4n0r S1 (mod 4q). Case 1.2 ( q p ) = − ( p q ) and ( p q ) = −1. Then p ≡ 3 (mod 4) and p ≡ rS2 (mod q). Thus, p ≡ −3q + 4n0r S2 (mod 4q). Case 2. ( q p ) = −1. Case 2.1 ( q p ) = ( p q ) and ( p q ) = −1. Then p ≡ 1 (mod 4) and p ≡ rS2 (mod q). Thus, p ≡ 3q + 4n0r S2 (mod 4q). Case 2.2 ( q p ) = − ( p q ) and ( p q ) = 1. Then p ≡ 3 (mod 4) and p ≡ rS1 (mod q) . Thus, p ≡ −3q + 4n0r S1 (mod 4q). Moreover, Siraworakun has given the forms of prime numbers p in Legendre symbol( 2q p ) , where q is a prime number. S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 729 Theorem 12. Let p and q be distinct odd prime numbers with q ≡ 1 (mod 4). Then( 2q p ) = 1 if p ≡ q2 + 8n1r S1 , − q2 + 8n1r S1 , 3q2 + 8n1r S2 ,−3q2 + 8n1r S2 (mod 8q),( 2q p ) = −1 if p ≡ q2 + 8n1r S2 ,−q2 + 8n1r S2 , 3q2 + 8n1r S1 ,−3q2 + 8n1r S1 (mod 8q), where S1 ∈ {2, 4, 6, . . . , q− 1}, S2 ∈ {1, 3, 5, . . . , q− 2}, r is a primitive root modulo q and if q − 1 4 is an even number, then n1 = −q + 1 8 , and if otherwise, then n1 = 3q + 1 8 . Proof. By Theorem 3(ii), we have ( 2q p ) = ( 2 p )( q p ) . Let r be a primitive root modulo q. By Theorems 5 and 10, we know that( 2 p ) = { 1 if p ≡ ±1 (mod 8) −1 if p ≡ ±3 (mod 8) and ( q p ) = { 1 if p ≡ q + rS1q + rS1 (mod 2q) −1 if p ≡ q + rS2q + rS2 (mod 2q) , where S1 ∈ {2, 4, 6, . . . , q − 1} and S2 ∈ {1, 3, 5, . . . , q − 2}. Since q ≡ 1 (mod 4), we have q = 4k + 1 for some integer k. Then q2 ≡ 1 (mod 8). If k is even, then k = −2l for some integer l. It leads to 8l − 1 = −q. Otherwise, q = 8s + 5 for some integer s, so 8(3s + 2) − 1 = 3q. Choose n1 = −q + 1 8 , when k is even and otherwise, n1 = 3q + 1 8 . Thus, 8n1 ≡ 1 (mod q). In the following cases, we use the Chinese Remainder Theorem for solving the systems of congruences. Case 1. ( 2q p ) = 1. Case 1.1 ( 2 p ) = 1 and ( q p ) = 1. Then p ≡ ±1 (mod 8) and p ≡ q + rS1q + rS1 (mod 2q). So p ≡ rS1 (mod q). It implies that p ≡ q2 + 8n1r S1 (mod 8q) or p ≡ −q2 + 8n1r S1 (mod 8q). Case 1.2 ( 2 p ) = −1 and ( q p ) = −1. Then p ≡ ±3 (mod 8) and p ≡ q + rS2q + rS2 (mod 2q). So p ≡ rS2 (mod q). It implies that p ≡ 3q2 + 8n1r S2 (mod 8q) or p ≡ −3q2 + 8n1r S2 (mod 8q). Case 2. ( 2q p ) = −1. Case 2.1 ( 2 p ) = 1 and ( q p ) = −1. Then p ≡ ±1 (mod 8) and p ≡ q + rS2q + rS2 (mod 2q). So p ≡ rS2 (mod q). It implies that p ≡ q2 + 8n1r S2 (mod 8q) or p ≡ −q2 + 8n1r S2 (mod 8q). Case 2.2 ( 2 p ) = −1 and ( q p ) = 1. Then p ≡ ±3 (mod 8) and p ≡ q + rS1q + rS1 (mod 2q). So p ≡ rS1 (mod q). It implies that p ≡ 3q2 + 8n1r S1 (mod 8q) or p ≡ −3q2 + 8n1r S1 (mod 8q). S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 730 Theorem 13. Let p and q be distinct odd prime numbers with q ≡ 3 (mod 4). Then( 2q p ) = 1 if p ≡ q2 + 32n0n1r S1 ,−q2 + 32n0n1r S2 , 3q2 + 32n0n1r S1 ,−3q2 + 32n0n1r S2 (mod 8q),( 2q p ) = −1 if p ≡ q2 + 32n0n1r S2 ,−q2 + 32n0n1r S1 , 3q2 + 32n0n1r S2 ,−3q2 + 32n0n1r S1 (mod 8q), where S1 ∈ {2, 4, 6, . . . , q − 1}, S2 ∈ {1, 3, 5, . . . , q − 2}, r is a primitive root modulo q, n0 = q + 1 4 and if q − 3 4 is an even number, then n1 = 5q + 1 8 , and if otherwise, then n1 = q + 1 8 . Proof. By Theorem 3(ii), we have ( 2q p ) = ( 2 p )( q p ) . Let r be a primitive root modulo q. By Theorems 5 and 11, we know that( 2 p ) = { 1 if p ≡ ±1 (mod 8) −1 if p ≡ ±3 (mod 8) and ( q p ) = { 1 if p ≡ 3q + 4n0r S1 , − 3q + 4n0r S2 (mod 4q) −1 if p ≡ 3q + 4n0r S2 , − 3q + 4n0r S1 (mod 4q) , where S1 ∈ {2, 4, 6, . . . , q− 1}, S2 ∈ {1, 3, 5, . . . , q− 2}, r is a primitive root modulo q and n0 = q + 1 4 . Since q ≡ 3 (mod 4), we have q = 4k + 3 for some integer k. Then q2 ≡ 1 (mod 8). If k is even, then k = 2l for some integer l. It leads to 8(5l + 2) − 1 = 5q. Otherwise, q = 8s+ 7 for some integer s, so 8(s+ 1)− 1 = q. Choose n1 = 5q + 1 8 , when k is even and otherwise, n1 = q + 1 8 . Thus, 8n1 ≡ 1 (mod q). In the following cases, the systems of congruences are solved by the Chinese Remainder Theorem and Theorem 7. Case 1. ( 2q p ) = 1. Case 1.1 ( 2 p ) = 1 and ( q p ) = 1. Then p ≡ ±1 (mod 8) and p ≡ 3q + 4n0r S1 ,−3q + 4n0r S2 (mod 4q). So p ≡ 4n0r S1 , 4n0r S2 (mod q). Since (8, 4q) = 4 and q ≡ 3 (mod 4), we get that 4 ∤ (−3q+4n0r S2)−1 and 4 ∤ (3q+4n0r S1)+1. Hence, the system of congruences p ≡ 1 (mod 8) and p ≡ −3q + 4n0r S1 (mod 4q) and the system of congruences p ≡ −1 (mod 8) and p ≡ 3q + 4n0r S1 (mod 4q) have no solution. Thus, p ≡ q2 + 32n0n1r S1 (mod 8q) or p ≡ −q2 + 32n0n1r S2 (mod 8q). Case 1.2 ( 2 p ) = −1 and ( q p ) = −1. Then p ≡ ±3 (mod 8) and p ≡ 3q+4n0r S2 ,−3q+ 4n0r S1 (mod 4q). So p ≡ 4n0r S2 , 4n0r S1 (mod q). Since (8, 4q) = 4 and q ≡ 3 (mod 4), we get that 4 ∤ (3q+4n0r S2)−3 and 4 ∤ (−3q+4n0r S1)+3. Hence, the system of congruences p ≡ 3 (mod 8) and p ≡ 3q + 4n0r S2 (mod 4q) and the system of congruences p ≡ −3 (mod 8) and p ≡ −3q + 4n0r S1 (mod 4q) have no solution. Thus, p ≡ 3q2 + 32n0n1r S1 (mod 8q) or p ≡ −3q2 + 32n0n1r S2 (mod 8q). S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 731 Case 2. ( 2q p ) = −1. Case 2.1 ( 2 p ) = 1 and ( q p ) = −1. Then p ≡ ±1 (mod 8) and p ≡ 3q+4n0r S2 ,−3q+ 4n0r S1 (mod 4q). So p ≡ 4n0r S2 , 4n0r S1 (mod q). Since (8, 4q) = 4 and q ≡ 3 (mod 4), we get that 4 ∤ (−3q+4n0r S1)−1 and 4 ∤ (3q+4n0r S2)+1. Hence, the system of congruences p ≡ 1 (mod 8) and p ≡ −3q + 4n0r S1 (mod 4q) and the system of congruences p ≡ −1 (mod 8) and p ≡ 3q + 4n0r S1 (mod 4q) have no solution. Thus, p ≡ q2 + 32n0n1r S2 (mod 8q) or p ≡ −q2 + 32n0n1r S1 (mod 8q). Case 2.2 ( 2 p ) = −1 and ( q p ) = 1. Then p ≡ ±3 (mod 8) and p ≡ 3q+4n0r S1 ,−3q+ 4n0r S2 (mod 4q). So p ≡ 4n0r S1 , 4n0r S2 (mod q). Since (8, 4q) = 4 and q ≡ 3 (mod 4), we get that 4 ∤ (3q+4n0r S1)−3 and 4 ∤ (−3q+4n0r S1)+3. Hence, the system of congruences p ≡ 3 (mod 8) and p ≡ 3q + 4n0r S1 (mod 4q) and the system of congruences p ≡ −3 (mod 8) and p ≡ −3q + 4n0r S2 (mod 4q) have no solution. Thus, p ≡ 3q2 + 32n0n1r S2 (mod 8q) or p ≡ −3q2 + 32n0n1r S1 (mod 8q). 3. Main Results In this section, we study the Diophantine equation px+(p+2q)y = z2, where p, q and p+ 2q are prime numbers. Thus, p is an odd prime number with (p, q) = 1. For the case q = 2, Burshtein [1] showed that the Diophantine equation px+(p+4)y = z2, where p > 3 and p + 4 are primes, has no non-negative solution. Moreover, Rao [10] investigated the same equation, when q = 2 and p = 3. From now on, x, y are positive integers and p, q are distinct odd prime numbers. Lemma 1. Let x be an even number. If the Diophantine equation px+(p+2q)y = z2 has a positive integer solution, then 2q ≡ 1 (mod p). Proof. Assume that the Diophantine equation px + (p + 2q)y = z2 has a positive integer solution. Since x is even, there exists a positive integer k such that x = 2k. Thus, (p + 2q)y = z2 − p2k = (z − pk)(z + pk). Since p + 2q is a prime number, we have z− pk = (p+2q)u and z+ pk = (p+2q)y−u, where u is a non-negative integer. So y > 2u and 2pk = (p+ 2q)u((p+ 2q)y−2u − 1). Since p and p+ 2q are prime numbers, we obtain that u = 0 and so 2pk = (p+ 2q)y − 1 = (p+ 2q − 1)((p+ 2q)y−1 + (p+ 2q)y−2 + · · ·+ 1). Hence p|(p+ 2q − 1). Therefore 2q ≡ 1 (mod p). Lemma 2. Let x be an odd number. If the Diophantine equation px + (p+ 2q)y = z2 has a positive integer solution, then ( 2q p ) = 1. Proof. Assume that the Diophantine equation px + (p + 2q)y = z2 has a positive integer solution. Then px ≡ z2 (mod p + 2q). By Division Algorithm, we can write x = (p + 2q − 1)m + l, where m and l are integers with 0 ≤ l < p + 2q − 1. Since x is odd, we obtain l is odd. By Theorem 1, we obtain that pp+2q−1 ≡ 1 (mod p + 2q). So p(p+2q−1)m+l ≡ pl (mod p+2q). Then px ≡ pl (mod p+2q). Thus, z2 ≡ pl (mod p+2q). S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 732 Hence, ( pl p+2q ) = 1. Since l is an odd number, we obtain ( p p+2q ) = 1 by Theorem 3(ii). By Theorem 4, we obtain ( p p+2q )( p+2q p ) = (−1)( p−1 2 )( p+2q−1 2 ) = 1. Thus, ( p+2q p ) = 1. By Theorem 3(i), we have ( 2q p ) = ( p+2q p ) . Then ( 2q p ) = 1. From above lemmas, we have the following result. Theorem 14. Let p and q be distinct prime numbers with 2q ̸≡ 1 (mod p) and ( 2q p ) = −1. Then the Diophantine equation px + (p+ 2q)y = z2 has no positive integer solution. By applying Theorems 12 and 14, the forms of odd prime number p are identified, when q ≡ 1 (mod 4). Theorem 15. Let q be a prime number such that q ≡ 1 (mod 4). If p is a prime number with 2q ̸≡ 1 (mod p) and satisfies any of the following conditions: (i) p ≡ q2 + 8n1r S2 (mod 8q), (ii) p ≡ − q2 + 8n1r S2 (mod 8q), (iii) p ≡ 3q2 + 8n1r S1 (mod 8q), or (iv) p ≡ − 3q2 + 8n1r S1 (mod 8q), where S1 ∈ {2, 4, 6, . . . , q−1}, S2 ∈ {1, 3, 5, . . . , q−2}, r is a primitive root modulo q, and if q − 1 4 is an even number, then n1 = −q + 1 8 , and if otherwise, then n1 = 3q + 1 8 . Then, the Diophantine equation px + (p+ 2q)y = z2 has no positive integer solution. Example 1. Let q = 17 and r = 3. Then r is a primitive root of q and n1 = −17 + 1 8 = −2 since 17− 1 4 is an even number. Consider a prime number p that satisfies any of the following congruences: (i) p ≡ 289 + (−16) · (3)S2 (mod 136), (ii) p ≡ − 289 + (−16) · (3)S2 (mod 136), (iii) p ≡ 867 + (−16) · (3)S1 (mod 136), or (iv) p ≡ − 867 + (−16) · (3)S1 (mod 136), where S1 ∈ {2, 4, 6, 8, 10, 12, 14, 16} and S2 ∈ {1, 3, 5, 7, 9, 11, 13, 15}. Thus, p ≡ ±7,±13,±19,±21,±23,±31,±35,±39,±41,±43,±53,±57,±59,±63,±65,±67 (mod 136). By Theorem 15, we obtain that the Diophantine equation px + (p + 34)y = z2 has no positive integer solution. For example, 7x + 41y = z2, 13x + 47y = z2, 19x + 47y = z2, 53x + 87y = z2 and 67x + 101y = z2. S. Tadee, A. Siraworakun / Eur. J. Pure Appl. Math, 16 (2) (2023), 724-735 733 Example 2. Let q = 5 and r = 2. Then r is a primitive root of q and n1 = 3(5) + 1 8 = 2 since 5− 1 4 is an odd number. Consider a prime number p that satisfies any of the following congruences: (i) p ≡ 25 + (16) · (2)S2 (mod 40), (ii) p ≡ − 25 + (16) · (2)S2 (mod 40), (iii) p ≡ 75 + (16) · (2)S1 (mod 40), or (iv) p ≡ − 75 + (16) · (2)S1 (mod 40), where S1 ∈ {2, 4} and S2 ∈ {1, 3}. Therefore, p ≡ ±7,±11,±17,±19 (mod 40). By Theorem 15, we obtain that the Diophantine equation px + (p+ 10)y = z2 has no positive integer solution. For example, 7x+17y = z2, 19x+29y = z2 (Burshtein [3]), 61x+71y = z2 (Kumar [5]), 73x + 83y = z2 and 97x + 107y = z2. From Theorems 13 and 14, the forms of odd prime number p can be obtained, when q ≡ 3 (mod 4). Theorem 16. Let p and q be distinct prime numbers such that q ≡ 3 (mod 4) and 2q ̸≡ 1 (mod p). If p satisfies any of the following conditions: (i) p ≡ q2 + 32n0n1r S2 (mod 8q), (ii) p ≡ − q2 + 32n0n1r S1 (mod 8q), (iii) p ≡ 3q2 + 32n0n1r S2 (mod 8q), or (iv) p ≡ − 3q2 + 32n0n1r S1 (mod 8q), where S1 ∈ {2, 4, 6, . . . , q − 1}, S2 ∈ {1, 3, 5, . . . , q − 2}, r is a primitive root modulo q, n0 = q + 1 4 and if q − 3 4 is an even number, then n1 = 5q + 1 8 , and if otherwise, then n1 = q + 1 8 . Then, the Diophantine equation px + (p + 2q)y = z2 has no positive integer solution. Example 3. Let q = 3 and r = 2. Then r is a primitive root of q, S1 = 2, S2 = 1, n0 = 3 + 1 4 = 1 and n1 = 5(3) + 1 8 = 2 since 3− 3 4 is an even number. Consider a prime number p that satisfies any of the following congruences: (i) p ≡ 9 + (64) · (2)1 (mod 24), (ii) p ≡ − 9 + (64) · (2)2 (mod 24), (iii) p ≡ 27 + (64) · (2)1 (mod 24), or REFERENCES 734 (iv) p ≡ − 27 + (64) · (2)2 (mod 24). Thus, p ≡ ±7,±11 (mod 24). By Theorem 16, we obtain that the Diophantine equa- tion px + (p + 6)y = z2 has no positive integer solution. For example, 7x + 13y = z2, 11x + 17y = z2, 13x + 19y = z2, 17x + 23y = z2 and 61x + 67y = z2(Kumar [4]). Example 4. Let q = 7 and r = 3. Then r is a primitive root of q, n0 = 7+1 4 = 2 and n1 = 7 + 1 8 = 1 since 7− 3 4 is an odd number. Consider a prime number p that satisfies any of the following congruences: (i) p ≡ 49 + 64(3S2) (mod 56), (ii) p ≡ − 49 + 64(3S1) (mod 56), (iii) p ≡ 147 + 64(3S2) (mod 56), or (iv) p ≡ − 147 + 64(3S1) (mod 56), where S1 ∈ {2, 4, 6} and S2 ∈ {1, 3, 5}. Thus, p ≡ ±3,±15,±17,±19,±23,±27 (mod 56). By Theorem 16, we obtain that the Diophantine equation px + (p + 14)y = z2 has no positive integer solution. For example, 3x + 17y = z2(Sroysang [12]), 17x + 31y = z2, 23x + 37y = z2, 29x + 43y = z2, and 53x + 67y = z2. Acknowledgements This work was supported by Research and Development Institute and Faculty of Sci- ence and Technology, Thepsatri Rajabhat University, Thailand. References [1] N Burshtein. The diophantine equation px + (p + 4)y = z2 when p > 3 and p + 4 are primes is insolvable in positive integers x, y, z. Annals of Pure and Applied Mathematics, 16(2):283–286, 2018. [2] N Burshtein. Solutions of the diophantine equation px+(p+6)y = z2 when p, p+6 are primes and x+ y = 2, 3, 4. Annals of Pure and Applied Mathematics, 17(1):101–106, 2018. [3] N Burshtein. The diophantine equations 2x+11y = z2 and 19x+29y = z2 are insolv- able in positive integers x, y, z. Annals of Pure and Applied Mathematics, 22(2):119– 123, 2020. [4] S Kumar S Gupta and H Kishan. On the non-linear diophantine equation 61x+67y = z2 and 67x + 73y = z2. Annals of Pure and Applied Mathematics, 18(1):91–94, 2018. [5] S Kumar S Gupta and H Kishan. On the non-linear diophantine equations 31x+41y = z2 and 61x+71y = z2. Annals of Pure and Applied Mathematics, 18(2):185–188, 2018. REFERENCES 735 [6] S Gupta S Kumar and H Kishan. On the non-linear diophantine equation px + (p+ 6)y = z2. Annals of Pure and Applied Mathematics, 18(1):125–128, 2018. [7] R J S Mina and J B Bacani. Non-existence of solutions of diophantine equations of the form px + qy = z2n. Mathematics and Statistics, 7(3):78–81, 2019. [8] R J S Mina and J B Bacani. On the solutions of the diophantine equation px + (p + 4k)y = z2 for prime pairs p and p + 4k. European Journal of Pure and Applied Mathematics, 14(2):471–479, 2021. [9] F Neres. On the solvability of the diophantine equation px+(p+8)y = z2 when p > 3 and p+ 8 are primes. Annals of Pure and Applied Mathematics, 18(1):9–13, 2018. [10] C G Rao. On the diophantine equation 3x + 7y = z2. EPRA International Journal of Research and Development, 3(6):93–95, 2018. [11] K H Rosen. Elementary Number theory. Clays Ltd, Great Britain, 2014. [12] B Sroysang. On the diphantine equation 3x+17y = z2. International Journal of Pure and Applied Mathematics, 89(1):111–114, 2013. [13] S Tadee. On the diophantine equation px+(p+10)y = z2 when p and p+10 are primes. Udon Thani Rajabhat University Journal of Sciences and Technology, 10(2):155–162, 2022. [14] S Tadee. On the diophantine equation px+(p+14)y = z2, where p, p+14 are primes. Annals of Pure and Applied Mathematics, 26(2):125–130, 2022.