EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6066 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Diophantine equation 4(7x)− py = z2 Kittipong Laipaporn1, Ratcharut Jankaew2, Poomipat Sae-Iu2, Adisak Karnbanjong1,∗ 1 Department of Mathematics and Statistics, Center of Excellence for Ecoinformatics, School of Science, Walailak University, Nakhon Si Thammarat , 80160, Thailand 2 Princess Chulabhorn Science High School, Nakhon Si Thammarat, 80330, Thailand Abstract. This paper determines all non-negative integer solutions to the Diophantine equation 4(7x) − py = z2, where p is a prime. Using modular arithmetic and congruence arguments, we classify all solutions as follows: a unique solution for p = 2, an infinite family of solutions for p = 3, no solutions for 5 ≤ p ≤ 17, and–for p ≥ 19–the existence of solutions requires that p ≡ 19 (mod 24) subject to specific modular constraints. Computational results support the conjecture that no further solutions exist beyond those identified. This work illustrates how modular techniques can fully resolve an exponential Diophantine equation and offers a framework for analyzing similar equations involving mixed exponential and polynomial terms. 2020 Mathematics Subject Classifications: 11A07 Key Words and Phrases: Exponential Diophantine Equation, Modulo 1. Introduction The study of Diophantine equations, named after the ancient Greek mathematician Diophantus of Alexandria (ca. 250 AD), forms a foundational aspect of number theory. Diophantus is often regarded as the “father of algebra” due to his pioneering work in using symbolic methods to represent equations in his famous treatise Arithmetica. Although originally consisting of thirteen books, only six have survived and contain around 130 problems with solutions. These problems demonstrate early algebraic reasoning and have inspired generations of mathematicians. A well-known anecdote regarding Diophantus’ life is his purported age at death, de- duced from a riddle composed by the poet Metrodorus. The riddle, when interpreted algebraically, leads to the equation x = 1 6 x+ 1 12 x+ 1 7 x+ 5 + 1 2 x+ 4, ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6066 Email addresses: lkittipo@wu.ac.th (K. Laipaporn), 6405825@pccnst.ac.th (R. Jankaew), 6405826@pccnst.ac.th (P. Sae-Iu), kadisak@mail.wu.ac.th (A. Karnbanjong) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 2 of 13 whose solution is x = 84, suggesting that Diophantus lived to be 84 years old. This kind of problem serves as an early example of a linear Diophantine equation, which is an equation of the form ax+ by = c where a, b, c ∈ Z, and the goal is to find integer solutions for x and y. A necessary and sufficient condition for such an equation to have solutions is that gcd (a, b)|c. Over the centuries, linear Diophantine equations have evolved into more intricate forms, including exponential and nonlinear variants. Table 1 provides a brief histori- cal overview of several notable Diophantine equations, ranging from Pell’s Equation to Fermat’s Last Theorem. Table 1: Historical Diophantine Equations. Equation Reference Equations Equation Names discovered in 1768 [1] x2 − ny2 = ±1 Pell’s Equation 1918 [2] w3 + x3 = y3 + z3 Hardy-Ramanujan Number 1948 [3] 4 n = 1 x + 1 y + 1 z Erdös-Straus Conjecture 1988 [4] x4 + y4 + z4 = w4 Euler’s Conjecture (disproved) 1995 [5] xn + yn = zn, n ≥ 3 Fermat’s Last Theorem Among these developments, Catalan’s conjecture—proposed by Eugène Catalan in 1844 and proved by P. Mihăilescu [6] in 2004 — stands out as a landmark result. It asserts that the equation ax − by = 1 has a unique solution in natural numbers when a, b, x, y ≥ 2, namely (a, b, x, y) = (3, 2, 2, 3). This result has had significant implications in the field of exponential Diophantine equations. In 2007, Acu [7] utilized Catalan’s theorem to analyze the equation 2x + 5y = z2, which has inspired the investigation of equations involving exponential terms equated to perfect squares. Over the past decade, many researchers have investigated equations of the general type a(px)± b(qy) = z2, where a, b ∈ Z+, and p, q are primes. Table 2 summarizes several recent contributions on exponential Diophantine equations of the form a(px)± b(qy) = z2, where p, q are primes. Some of these studies demonstrate the effectiveness of modular arithmetic and parametric forms in analyzing such equations, and they inform the approach adopted in this article. Motivated by previous studies on exponential Diophantine equations, we continue this line of inquiry by examining the equation 4(7x)− py = z2, where p is a prime and x, y, z ∈ Z0. By employing modular arithmetic and a detailed anal- ysis of congruence conditions, we classify all possible solutions and propose a conjecture on the nonexistence of further solutions beyond those explicitly identified. K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 3 of 13 Table 2: Examples of exponential Diophantine equations of the forms a(px)± b(qy) = z2 where a, b are positive integers and p, q are primes. Year Authors Equation 2018 J.F.T. Rabago [8] 4x − py = 3z2 2019 K. Laipaporn, et al. [9] 3x + p(5y) = z2 2020 A. Elshahed and H. Kamarulhaili [10] (4n)x − py = z2 2021 S. Thongnak, et al. [11] 7x − 5y = z2 2022 W. Tangjai, et al. [12] 7x + 5(py) = z2 2022 W. Orosram and A. Unchai [13] 22nx − py = z2 2022 M. Buosi, et al. [14] px − 2y = z2 2024 S. Thongnak, et al. [15] 11x − 17y = z2 2024 K. Laipaporn, et al. [16] ax ± ay = zn 2024 Y. Li, et al. [17] 2x ± (2kp)y = z2 and −2x + (2k3)y = z2 2024 J. Zhang and Y. Li [18] (−1)αpx + (−1)β(2k(2p− 1))y = z2 2025 K. Laipaporn, et al. [19] px + q2y = z2n 2. Main theorem We begin by analyzing the structure of the equation and presenting the main clas- sification result. To support the classification, we first introduce an auxiliary lemma to understand the behavior of power of p modulo 9. Lemma 1. Let p ≡ 6r+1 (mod 9) for some integer r ≥ 0. Then for all integers x, y ≥ 0, the congruence class of 4(7x) − py (mod 9) depends on the values of x (mod 3) and y (mod 3) as follows: 4(7x)− py ≡  1 (mod 9) if x ≡ 0 (mod 3) and y ≡ 0 (mod 3), 4− p (mod 9) if x ≡ 0 (mod 3) and y ≡ 1 (mod 3), 2 + p (mod 9) if x ≡ 0 (mod 3) and y ≡ 2 (mod 3), 0 (mod 9) if x ≡ 1 (mod 3) and y ≡ 0 (mod 3), 1− p (mod 9) if x ≡ 1 (mod 3) and y ≡ 1 (mod 3), p− 1 (mod 9) if x ≡ 1 (mod 3) and y ≡ 2 (mod 3), 6 (mod 9) if x ≡ 2 (mod 3) and y ≡ 0 (mod 3), 7− p (mod 9) if x ≡ 2 (mod 3) and y ≡ 1 (mod 3), 5 + p (mod 9) if x ≡ 2 (mod 3) and y ≡ 2 (mod 3). Proof. Since p ≡ 6r+1 (mod 9), we derive p3 ≡ 216r3+108r2+18r+1 ≡ 1 (mod 9). Therefore, for all integers t, y ≥ 0, py ≡ pt (mod 9) where y ≡ t (mod 3). It follows that py ≡  1 (mod 9) if y ≡ 0 (mod 3), p (mod 9) if y ≡ 1 (mod 3), p2 (mod 9) if y ≡ 2 (mod 3). K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 4 of 13 Next, we compute p2 (mod 9) using the given congruence for p: p2 ≡ 36r2 + 12r + 1 ≡ −6r + 1 ≡ 2− p (mod 9). Therefore, py ≡  1 (mod 9) if y ≡ 0 (mod 3), p (mod 9) if y ≡ 1 (mod 3), 2− p (mod 9) if y ≡ 2 (mod 3). Hence, 4(7x) ≡  4(1) ≡ 4 (mod 9) if x ≡ 0 (mod 3), 4(7) ≡ 1 (mod 9) if x ≡ 1 (mod 3), 4(4) ≡ 7 (mod 9) if x ≡ 2 (mod 3). Combining the congruences for 4(7x) and py, we obtain the desired result. We now proceed to a complete classification of the solutions based on the value of p: • Small primes p = 2, 3, where elementary computations can be used. • Intermediate primes 5 ≤ p ≤ 17, where contradiction via modular arguments can be established, • Large prime p ≥ 19, where structural congruence restrictions guide the solution forms. Theorem 1. Let p be any prime number. Then the solutions to the Diophantine equation 4(7x)− py = z2 where x, y and z are non-negative integers satisfy the following: (i) For the prime p = 2, the unique solution is (x, y, z, p) = (0, 2, 0, 2). (ii) For the prime p = 3, if a solution exists, then it must be of the following set (x, y, z, p) ∈ {(0, 1, 1, 3), (1, 1, 5, 3), (1, 3, 1, 3), (2, 3, 13, 3), (3, 1, 37, 3)} ∪ {(2k + 1, 4l + 1, 16m+ n, 3)| for any integers k ≥ 2, l,m ≥ 0 and n = 3, 5, 11, 13} ∪ {(2k + 1, 4l + 3, 16m+ n, 3)| for any integers k ≥ 2, l,m ≥ 0 and n = 1, 7, 9, 15}. (iii) There is no solution for any prime p with 5 ≤ p ≤ 17. (iv) For the prime p ≥ 19, a necessary condition for the existence of solutions is that p ≡ 19 (mod 24). In such cases, all solutions must satisfy: (x, y, z, p) ∈ {(2k+1, 2l+1, 24m+n, 24r+19)| for any integers k, l,m, r ≥ 0 and n = 3, 9, 15, 21}. Proof. We begin by examining the degenerate cases x = 0 and y ≥ 0. In this case, direct computation shows that (x, y, z, p) = (0, 1, 1, 3) and (0, 2, 0, 2) are valid solutions, as the expression 4(1)−py = z2 yields a non-negative perfect square. Next, suppose x ≥ 1 and y = 0. Then the equation becomes 4(7x)−1 = z2, which is impossible modulo 4 since z2 ≡ 0 or 1 (mod 4), but 4(7x)− 1 ≡ 0− 1 ≡ 3 (mod 4). Hence, no solution exists in this case. We now focus on the remaining case where both x and y are positive. To proceed, we consider three cases based on the value of the prime p: K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 5 of 13 Case 1 p = 2. Then 4(7x)− 2y ≡  0− 1 ≡ 6 (mod 7) if y ≡ 0 (mod 3), 0− 2 ≡ 5 (mod 7) if y ≡ 1 (mod 3), 0− 4 ≡ 3 (mod 7) if y ≡ 2 (mod 3). However, for any integer z, z2 ≡ 0, 1, 2, 4 (mod 7) so no solution is possible in this case. Case 2 p = 3. Since z is odd and not divisible by 3, it follows that z2 ≡ 1 (mod 24). Observe that for any x > 0, we have 4(7x)− 3y ≡ { 4− 3 ≡ 1 (mod 24) if y is odd, 4− 9 ≡ 19 (mod 24) if y is even. This contradicts the fact z2 ≡ 1 (mod 24) when y is even, so we only consider the case when y is an odd number. Subcase 2.1 x = 2k for some k ≥ 1. In this case, the equation 4(7x)− py = z2 becomes 3y = (2(7k)− z)(2(7k) + z), which is factorization of 3y into two positive integers. Let 3u = 2(7k) − z and 3y−u = 2(7k) + z for some 0 ≤ u < y. Adding these two expressions yields 4(7k) = 3(3u−1 + 3y−u−1). If 0 < u < y , then the right-hand side is divisible by 3 while the left-hand side is not, giving a contradiction. Hence, the only possibility is u = 0, leading to 1 = 2(7k)− z and 3y = 2(7k) + z. Substituting the first equation into the second gives 3y = 4(7k)−1. We now consider whether k is even or odd. If k = 2l for some l ≥ 1. Then we obtain 3y = 4(72l) − 1 = (2(7l)− 1)(2(7l) + 1). Since l ̸= 0, both factors are greater than 1 and differ by 2, so both must be divisible by 3, which is impossible since 3 does not divide 2. Therefore, no solution exists in the case k is even. Next, we suppose k = 2h+1 for some h ≥ 0. Then we have 3y = 4(74h+2)−1 = (2(72h+1)−1)(2(72h+1)+1), which again leads to a contradiction unless h = 0, i.e., x = 2. Hence, the only solution in this case (x, y, z) = (2, 3, 13). Subcase 2.2 x = 2k + 1 for some k ≥ 0. For k = 0 or 1, direct computation shows that (x, y, z) = (1, 1, 5), (1, 3, 1) and (3, 1, 37) are valid solutions, as the expression 4(7x) − 3y = z2 yields a non- negative perfect square. For k ≥ 2, consider the equation with modulo 16, we have z2 ≡ 4(7x)− 3y ≡ { 4(7)− 3 ≡ 9 (mod 16) if y ≡ 1 (mod 4), 4(7)− 11 ≡ 1 (mod 16) if y ≡ 3 (mod 4). Since z ≡ { 3, 5, 11, 13 (mod 16) if z2 ≡ 9 (mod 16), 1, 7, 9, 15 (mod 16) if z2 ≡ 1 (mod 16), K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 6 of 13 it follows that any solution (x, y, z) with x = 2k + 1, k ≥ 2 and y ≡ 1 or 3 (mod 4) must satisfy: (x, y, z) ∈ {(2k+1, 4l+1, 16m+n)| for any integers k ≥ 2, l,m ≥ 0 and n = 3, 5, 11, 13}∪{(2k+1, 4l+3, 16m+n)| for any integers k ≥ 2, l,m ≥ 0 and n = 1, 7, 9, 15}. Case 3 p ≥ 5. Subcase 3.1 x = 2k for some k ≥ 1. Since z is odd, the quantity d = gcd (2(7k)− z, 2(7k) + z) is also odd. Note that the product py = (2(7k)−z)(2(7k)+z) consists of two factors whose sum is 4(7k), implying d|4(7k). Therefore, d = 1 or 7|d. If d = 1 then 2(7k) − z = 1, which gives py = 4(7k)− 1. Reducing both sides modulo 3 yields py ≡ 0 (mod 3), so p = 3, contradicting the assumption p ≥ 5. Hence, we must have 7|d, which implies p = 7. In this case, the equation becomes 7y = (2(7k) − z)(2(7k) + z), where the two factors are coprime and their product is a power of 7. Therefore, we can write 7u = 2(7k) − z, 7y−u = 2(7k) + z for some 1 ≤ u < y. Adding these two equations yields 4(7k) = 7u + 7y−u. Reducing modulo 6, we find 7u + 7y−u ≡ 2 (mod 6), while 4(7k) ≡ 4 (mod 6). Since both sides are not congruent modulo 6, the equality cannot hold. Hence there is no solution when x is even and p ≥ 5. Subcase 3.2 x = 2k + 1 for some k ≥ 0. Note that 72k+1 ≡ (49k)7 ≡ 7 (mod 24). Since p is an odd prime, we have p ≡ ±1,±5,±7,±11 or 13 (mod 24), so py ≡ { 1 (mod 24) if y is even, p (mod 24) if y is odd. Therefore, 4(7x)− py ≡ { 4(7)− 1 ≡ 3 (mod 24) if y is even, 4(7)− p ≡ 4− p (mod 24) if y is odd. Since z is odd, it square must satisfy z2 ≡ 1 or 9 (mod 24). But 4(7x)− py ≡ 3 (mod 24) where y is even, which is not a quadratic residue modulo 24. Now, suppose y is odd. Then 4 − p ≡ 4(7x) − py ≡ z2 ≡ 1 or 9 (mod 24). This implies p ≡ 3 or 19 (mod 24). However, gcd (p, 3) = 1 so p ̸≡ 3 (mod 24). Therefore, we must have p ≡ 19 (mod 24), and hence z2 ≡ 9 (mod 24). Finally, we can conclude that if 4(7x) − py = z2 has a solution with prime p ≥ 19 then both x and y must be odd, z ≡ 3, 9, 15 or 21 (mod 24), and p ≡ 19 (mod 24). Based on Theorem 1(iv), we refine the possible values of the prime p by analyzing the divisibility properties of z, particularly with respect to modulo 3. This leads to the following corollary. Corollary 1. Let p ≥ 19 be a prime. Then all non-negative integer solutions (x, y, z, p) to the equation 4(7x)− py = z2 are of the following form: (x, y, z, p) ∈ {(6k + 1, 6l + 3, 24m+ n, 24r + 19)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21} K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 7 of 13 ∪ {(6k + 1, 6l + 1, 24m+ n, 72r + 19)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21} ∪ {(6k + 1, 6l + 5, 24m+ n, 72r + 19)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21} ∪ {(6k + 3, 6l + 5, 24m+ n, 72r + 43)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21} ∪ {(6k + 5, 6l + 1, 24m+ n, 72r + 43)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21} ∪ {(6k + 3, 6l + 1, 24m+ n, 72r + 67)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21} ∪ {(6k + 5, 6l + 5, 24m+ n, 72r + 67)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21}. Remark 1. This corollary implies that for any prime p ≥ 19, the equation 4(7x)−py = z2 has no solution if x ≡ 3 or 5 (mod 6) and y ≡ 3 (mod 6). Proof. From Theorem 1(iv), we know that all solutions are of the form (x, y, z, p) ∈ {(2k + 1, 2l + 1, 24m+ n, 24r + 19)|k, l,m, r ∈ Z+ 0 and n = 3, 9, 15, 21}. We first consider z = 24m+ n where m ≥ 0 and n ∈ {3, 9, 15, 21}. Since every n is divisible by 3, it follows that z2 ≡ 0 (mod 9). Next, we observe that p = 24r + 19 for some non-negative integer r, which implies p ≡ 6r + 1 (mod 9). By Lemma 1 we have that py ≡  1 (mod 9) if y ≡ 0 (mod 3), p (mod 9) if y ≡ 1 (mod 3), 2− p (mod 9) if y ≡ 2 (mod 3). Next, we examine the behavior 4(7x) with mod 9 and we have 4(7x) ≡  4(1) ≡ 4 (mod 9) if x ≡ 0 (mod 3), 4(7) ≡ 1 (mod 9) if x ≡ 1 (mod 3), 4(7) ≡ 7 (mod 9) if x ≡ 2 (mod 3). Then we analyze the equation 4(7x) − py ≡ z2 ≡ 0 (mod 9). The resulting congruence modulo 9 for each pair (x mod 3, y mod 3) yields the following: no solution if x ≡ 0 (mod 3) and y ≡ 0 (mod 3), p ≡ 4 (mod 9) if x ≡ 0 (mod 3) and y ≡ 1 (mod 3), p ≡ 7 (mod 9) if x ≡ 0 (mod 3) and y ≡ 2 (mod 3), no additional information if x ≡ 1 (mod 3) and y ≡ 0 (mod 3), p ≡ 1 (mod 9) if x ≡ 1 (mod 3) and y ≡ 1 (mod 3), p ≡ 1 (mod 9) if x ≡ 1 (mod 3) and y ≡ 2 (mod 3), no solution if x ≡ 2 (mod 3) and y ≡ 0 (mod 3), p ≡ 7 (mod 9) if x ≡ 2 (mod 3) and y ≡ 1 (mod 3), p ≡ 4 (mod 9) if x ≡ 2 (mod 3) and y ≡ 2 (mod 3). Combining these observations with Theorem 1(iv) and applying the Chinese Remainder Theorem to parameters x, y and p, we derive the following refined conditions for a prime p ≥ 19: K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 8 of 13 (i) If x ≡ 3 or 5 (mod 6) and y ≡ 3 (mod 6) then the equation has no solution. (ii) Under the assumption that a solution exists: (a) If (x ≡ 3 (mod 6) and y ≡ 1 (mod 6)) or (x ≡ 5 (mod 6) and y ≡ 5 (mod 6)) then p ≡ 67 (mod 72). (b) If (x ≡ 3 (mod 6) and y ≡ 5 (mod 6)) or (x ≡ 5 (mod 6) and y ≡ 1 (mod 6)) then p ≡ 43 (mod 72). (c) If x ≡ 1 (mod 6) and y ≡ 1 or 5 (mod 6) then p ≡ 19 (mod 72). (d) If x ≡ 1 (mod 6) and y ≡ 3 (mod 6) then p ≡ 19 (mod 24). 3. Conclusion In this article, we have determined all non-negative integer solutions to the exponential Diophantine equation 4(7x)−py = z2, where p is a prime number. Through a careful case- by-case analysis based on the value of the prime p, we find: • A unique solution for p = 2 , namely (x, y, z, p) = (0, 2, 0, 2). • A parametrized family of possible solutions satisfying specific congruence conditions for p = 3 , where all solutions require x and y to be odd, except for the cases (x, y, z, p) = (0, 1, 1, 3) and (2, 3, 13, 3), in which x is even. • No solutions exist for 5 ≤ p ≤ 17. • For p ≥ 19, a necessary condition for the existence of the solutions is that p ≡ 19 (mod 24), and the values of x, y, z and p must satisfy precise congruence conditions derived using quadratic residues and the Chinese Remainder Theorem. Moreover, there is no solution when x ≡ 3 or 5 (mod 6) and y ≡ 3 (mod 6). In contrast to the previous studies summarized in Table 2, our analysis addresses a distinct class of exponential Diophantine equations in which the term py is subtracted from a fixed exponential base 7x, rather than involving variable coefficients or dual exponential terms. For instance, Rabago [8] analyzed 4x − py = 3z2, while Laipaporn et al. [9] examined 3x + p(5y) = z2, which is additive in nature. In contrast, our equation exhibits a more intricate balance between exponential growth and quadratic structure, requiring careful refinement through modular analysis. Interestingly, the Diophantine equation 4(7x) − py = z2 was also examined compu- tationally using Algorithm 1, which internally calls two functions based on Algorithm 2, and Algorithm 3. The complete procedure was implemented in Python code (accessible at https://github.com/kadisak/EJPAM6066.git). For the case p = 3, the computational search was performed over the range 0 ≤ x ≤ 100, 000 and 1 ≤ y ≤ ⌊logp (4(7x))⌋. Within this domain, exactly five solutions were obtained, namely (x, y, z) = (0, 1, 1), (1, 1, 5), (1, 3, 1), (2, 3, 13), and (3, 1, 37), see Table 3. K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 9 of 13 Table 3: Valid solutions to the equation 4(7x)− py = z2 for p = 3, 0 ≤ x ≤ 100, 000, and 1 ≤ y ≤ ⌊logp (4(7x))⌋. p x y z 3 0 1 1 3 1 1 5 3 1 3 1 3 2 3 13 3 3 1 37 For prime numbers p satisfying 2 ≤ p ≤ 100, 000 with p ̸= 3, and under the parameter constraints 1 ≤ x ≤ 10, 000 and 1 ≤ y ≤ ⌊logp (4(7x))⌋, a total of 31 solutions were found when p ≡ 19 (mod 24), summarized in Table 4. Table 4: All computed solutions to the equation 4(7x)−py = z2. The search was conducted over the range 2 ≤ p ≤ 100, 000, 1 ≤ x ≤ 10, 000, and 1 ≤ y ≤ ⌊logp (4(7x))⌋, for primes p ≡ 19 (mod 24), as related to Corollary 1. Congruence class p x y z p ≡ 19 (mod 72) 19 1 1 3 86491 7 1 1791 p ≡ 67 (mod 72) 283 3 1 33 643 3 1 27 1291 3 1 9 p ≡ 43 (mod 72) 2203 5 1 255 5227 5 1 249 8179 5 1 243 11059 5 1 237 16603 5 1 225 19267 5 1 219 21859 5 1 213 24379 5 1 207 33739 5 1 183 35899 5 1 177 37987 5 1 171 Congruence class p x y z p ≡ 43 (mod 72) 41947 5 1 159 49003 5 1 135 50587 5 1 129 54907 5 1 111 57427 5 1 99 58579 5 1 93 59659 5 1 87 61603 5 1 75 62467 5 1 69 64627 5 1 51 65203 5 1 45 65707 5 1 39 66499 5 1 27 67003 5 1 15 67219 5 1 3 Based on both our computational findings and Theorem 1, we propose the following conjectures: • If z ̸≡ 1, 5 and 13 (mod 16), then the equation 4(7x)− 3y = z2 has no solution. • For any prime p ≡ 19 (mod 24) and y ≥ 3, our computational results suggest that no solutions exist within the tested parameter range. This leads us to conjecture that such solutions are either extremely rare or do not exist at all. K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 10 of 13 Overall, this work contributes to the classification program of exponential Diophantine equations of the form a(px)−b(qy) = z2, especially when one exponential base is fixed. By combining classical number-theoretic tools with modern computation, we not only resolve the given equation but also establish a pathway for studying similar equations involving asymmetrical exponential and quadratic forms. We hope these results will inspire fur- ther theoretical generalizations and computational techniques for analyzing Diophantine equations of this type. Algorithm 1 Pseudocode for solution search and verification. Input: pmin: minimum prime considered; pmax: maximum prime considered; xmax: max- imum value investigated for x. Output: Verification Results: Diophantine Equation Solutions. for p← pmin to pmax do for x← 1 to xmax do ymax ← ⌊logp 4(7x)⌋; // the maximum value of y for each pair (p, x) for y ← 1 to ymax do if 4(7x)− py is perfect square then z ← √ 4(7x)− py if (x, y, z, p) satisfies Theorem 1 (verified by Algorithm 2) then Output: ”The solution (x, y, z, p) satisfies the Theorem 1” else Output: ”The solution (x, y, z, p) does not satisfy the Theorem 1” end if p ≥ 19 then if (x, y, z, p) satisfies Corollary 1 (verified by Algorithm 3) then Output: ”The solution (x, y, z, p) satisfies the Corollary 1” else Output: ”The solution (x, y, z, p) does not satisfy the Corollary 1” end end end Output: ”(x, y, z, p) is not a solution to the equation” end end end K. Laipaporn et al. / Eur. J. Pure Appl. Math, 18 (3) (2025), 6066 11 of 13 Algorithm 2 Pseudocode for verifying a solution (x, y, z, p) to the Theorem 1. Input: (x, y, z, p): solution of the Diophantine equation Output: 1: the solution satisfies the Theorem, 0: the solution does not satisfy Theorem function satisfyMainTheorem(x,y,z,p) if (x, y, z, p) ∈ {(0, 2, 0, 2)} then return 1 else if p = 3 then if (x ≥ 5 and x mod 2 = 1 and y mod 4 = 1 and z mod 16 ∈ {3, 5, 11, 13}) or (x ≥ 5 and x mod 2 = 1 and y mod 4 = 3 and z mod 16 ∈ {1, 7, 9, 15}) or ((x, y, z) ∈ {(0, 1, 1), (1, 1, 5), (1, 3, 1), (2, 3, 13), (3, 1, 37)}) then return 1 end else if p ≥ 19 and x mod 2 = 1 and y mod 2 = 1 and z mod 24 ∈ {3, 9, 15, 21} then return 1 return 0 Algorithm 3 Pseudocode for verifying a solution (x, y, z, p) to the Corollary 1. Input: (x, y, z, p): solution of the Diophantine equation Output: 1: the solution satisfies the Corollary, 0: the solution does not satisfy Corollary function satisfyCorollary(x,y,z,p) if z mod 24 ∈ {3, 9, 15, 21} then if x mod 6 = 1 then if y mod 6 = 3 and p mod 24 = 19 then return 1 ; // Congruence class: p ≡ 19 (mod 24) else if y mod 6 = 1 and p mod 72 = 19 then return 1 ; // Congruence class: p ≡ 19 (mod 72) else if y mod 6 = 5 and p mod 72 = 19 then return 1 ; // Congruence class: p ≡ 19 (mod 72) else if x mod 6 = 3 then if y mod 6 = 5 and p mod 72 = 43 then return 1 ; // Congruence class: p ≡ 43 (mod 72) else if y mod 6 = 1 and p mod 72 = 67 then return 1 ; // Congruence class: p ≡ 67 (mod 72) else if x mod 6 = 5 then if y mod 6 = 5 and p mod 72 = 67 then return 1 ; // Congruence class: p ≡ 67 (mod 72) else if y mod 6 = 1 and p mod 72 = 43 then return 1 ; // Congruence class: p ≡ 43 (mod 72) end return 0 K. Laipaporn et al. / Eur. J. Pure Appl. 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