EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 3, Article Number 6593 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Diophantine Equation px + (p+ 5k)y = z2 Merlyn C. Avenilla1, Jerico B. Bacani1,∗ 1 Department of Mathematics and Computer Science, College of Science, University of the Philippines Baguio, Baguio City 2600, Benguet, Philippines Abstract. With the use of modular arithmetic and other fundamental number theoretic methods, as well as the concepts of the floor function and the principle of mathematical induction, this study searches for possible nonnegative integer solutions of exponential Diophantine equations of the form px + (p+ 5k)y = z2, where k ∈ N. Results are obtained for the following cases: a) when p = 2; or b) when p and p+ 5k are prime pairs. In addition, the study is limited only to solutions where x and y are not both greater than 1. 2020 Mathematics Subject Classifications: 11D61, 11D72, 11A41 Key Words and Phrases: Exponential Diophantine equation, Prime pairs, Mihăilescu’s theorem 1. Introduction A prime number is a natural number greater than 1 whose divisors are 1 and itself. The difference between two prime numbers is called prime gap, and the two prime numbers are called prime pair. The prime pairs that have two gaps are called twin primes, those with four and six gaps are called cousin primes and sexy primes, respectively. In this present study, we incorporate prime bases and prime pairs with gaps of multiples of five. They are part of the study of exponential Diophantine equations. Diophantine equations are any equations that seek rational solutions, but usually integers. These equations are believed to have been introduced by Diophantus of Alexandria. There are two types of Diophantine equation, namely, linear and nonlinear. Each type could have no solution, unique solution, finitely many solutions, or infinitely many solutions. The Linear Diophantine Equation (LDE) is the famous one. In n unknowns, it is of the form a1x1 + a2x2 + . . .+ anxn = c, (1) ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i3.6593 Email addresses: mcavenilla@up.edu.ph (M. C. Avenilla), jbbacani@up.edu.ph (J. B. Bacani) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 2 of 24 where x1, x2, . . . , xn are the unknowns, and a1, a2, . . . , an, c are fixed integers, and not all coefficients are zero. The simplest LDE is the one with two unknowns, say x and y. An equation that is not of the form (1) is considered Nonlinear Diophantine Equa- tion (NDE). Examples of this type include the Pythagorean equation (in 3 unknowns) x2 + y2 = z2, Pell equation (in 2 unknowns) x2 − Dy2 = 1, Nagell-Ljunggren equation (in 4 unknowns) xn − 1 x− 1 = yq, and the Catalan equation (in 4 unknowns) ax − by = 1. The last equation is also known as an Exponential Diophantine Equation (EDE). It is an equation that involves variable exponents. In the past few years, many researchers have studied EDEs of the form px + qy = z2 (2) for various fixed values of p and q. In 2011, Singta et al. [1] proved that there are no solutions in the set N0 of non-negative integers when p = 4 and q = 7 or 11. In 2012, Sroysang [2] showed that (1, 0, 2) is the only solution in N0 for the case p = 3 and q = 5. In the same year, he also established in [3] that (1, 0, 3) is the only solution in N0 when p = 8 and q = 19, and posed an open problem for the case q = 17. This problem was later addressed by Rabago [4], who demonstrated that (2) has only finitely many non-negative integer solutions when p = 8 and q = 17, namely, (1, 0, 3), (1, 1, 5), (2, 1, 9), and (3, 1, 23). A year later, Sroysang [5] examined the same class of EDEs with p = 5 and q = 7, but found no solutions in N0. In 2013, Chotchaisthit [6] studied the case where p and q are two consecutive integers, but p must be a Mersenne prime. Other Diophantine equations that involve bases of primes and Mersenne primes are found in the works of Gayo, Mina and Bacani (cf. [7–11]). Most of these studies utilize Mihăilescu’s theorem (known originally as Catalan’s conjecture). Basically, this theorem states that the only solution to the Diophantine equation px − qy = 1 is (p, q, x, y) = (3, 2, 2, 3), with the assumption that all variables have values greater than 1 [12]. It is also observed that the EDE (2) has been studied for specific prime pairs (p, q). Some examples are found in Gupta’s and Sroysang’s papers [2, 5, 13]. In 2015, Bacani and Rabago [14] investigated the case where p and q are twin primes. In 2018, Burshtein ([15], [16]) published two articles, namely, when p and q are cousin primes and when they are sexy primes with the condition that x + y = 2, 3, 4. In the same year, Neres [17] proved the solvability of (2) when p and q have a prime gap of 8, and p > 3. Recently, Tadee [18] studied the case where the prime gap is 14. A more interesting case was what Mina and Bacani [19] published in 2021, where they studied the case where p and q have prime gaps of multiples of four. In 2022, Orosram, et al. [20] also studied the same prime pairs with an additional constraint that p ≡ 7 (mod 12). Motivated by the papers mentioned above, the present study searches for nonnegative integer solutions of the Diophantine equation of the form px + (p+ 5k)y = z2, (3) where k ∈ N, and p satisfies any of the following conditions: M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 3 of 24 a) p = 2; or b) p and p+ 5k are prime pairs. The principle of mathematical induction is also used in this study, as well as the concept of floor function, which is usually not seen in related literature. Furthermore, the results presented here extend those of Dokchan and Panngam [21], who investigated the same equation under the condition p ≡ 1 (mod 5) and sought solutions in N. 2. Main results The results are divided into two parts. First, we present interesting results for the case p = 2. Next, we discuss the solutions of the equation when p and p+ 5k are prime pairs. 2.1. On the Diophantine Equation 2x + (2 + 5k)y = z2 This subsection talks about the Diophantine equation (3), where p = 2, that is, 2x + (2 + 5k)y = z2. (4) It is further divided into two parts. The first part discusses the case when y = 1, and the second part contains all other claims when y ̸= 1. 2.1.1. Part I: The case where y = 1. Consider the equation 2x + (2 + 5k) = z2. (5) For this case, the following variables will be used: Variable Meaning xn the value of x at a specific value of n ∈ N0 z(0,xn) the first value of z at a specific value of xn z(m,xn) the (m+ 1)st value of z at a specific value of xn k(0,xn) the first value of k at a specific value of xn k(1,xn) the second value of k at a specific value of xn αxn the difference between the first and second value of k at a specific value of xn k(m,xn) the (m+ 1)st value of k at a specific value of xn Table 1: Variables Considered in Solving 2x + (2 + 5k) = z2 We derive k(0,xn) and k(m,xn) from (5), and define αxn as the difference of k(0,xn) and k(1,xn): k(0,xn) = z2(0,xn) − 2xn − 2 5 , (6a) M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 4 of 24 k(m,xn) = z2(m,xn) − 2xn − 2 5 , (6b) αxn = k(1,xn) − k(0,xn). (6c) Here, m ∈ N. We begin the discussion with the following lemmas. Lemma 1. Suppose the exponential Diophantine equation (5) has a solution. Then, x ≡ 1 (mod 4) if and only if z ≡ 2 (mod 5) or z ≡ 3 (mod 5). Proof. Let x ≡ 1 (mod 4) in the exponential Diophantine equation (5). This means that 2x ≡ 2 (mod 5) and 2 + 5k ≡ 2 (mod 5) for all k ∈ N. Thus, z2 ≡ 2x + (2 + 5k) ≡ 4 (mod 5). We claim that z ≡ 2 (mod 5) or z ≡ 3 (mod 5). Suppose on the contrary that z ≡ 0 (mod 5), z ≡ 1 (mod 5), or z ≡ 4 (mod 5). Then, z2 ≡ 0 (mod 5), z2 ≡ 1 (mod 5), or z2 ≡ 1 (mod 5), respectively. However, any of these is impossible since we already established that z2 ≡ 4 (mod 5). This implies that z ≡ 2 (mod 5) or z ≡ 3 (mod 5). Now, let z ≡ 2 (mod 5) or z ≡ 3 (mod 5) in equation (5). Then, z2 ≡ 4 (mod 5). Assume the contrary that x ̸≡ 1 (mod 4). We first consider x ≡ 2 (mod 4). Then, 2x ≡ 4 (mod 5) and the equation 2x + (2 + 5k) ≡ 4 + 2 ≡ 1 (mod 5). Next, we let x ≡ 3 (mod 4). Consequently, 2x ≡ 3 (mod 5) and 2x + (2 + 5k) ≡ 3 + 2 ≡ 0 (mod 5). Lastly, we take x ≡ 0 (mod 4). Then, 2x ≡ 1 (mod 5) and 2x + (2 + 5k) ≡ 1 + 2 ≡ 3 (mod 5). For all these cases, we have 2x + (2 + 5k) ̸≡ z2 (mod 5). Hence, x must be congruent to 1 (mod 4). Therefore, the equation 2x + (2 + 5k) = z2 has a solution when x ≡ 1 (mod 4) if and only if z ≡ 2 (mod 5) or z ≡ 3 (mod 5). From equation (5), we can first look at the smallest possible value of z and then add a multiple of five to get the other values of z. To get the smallest number of possible conditions for z(0,xn), we use ⌊√ 2x ⌋ because √ 2x will never be an integer since x ≡ 1 (mod 4). As a result, we obtain the following formulas: x := xn = 1 + 4n (7a) z(0,xn) =  ⌊√ 2xn ⌋ + 6 if x = 1,⌊√ 2xn ⌋ + 1 if ⌊√ 2xn ⌋ ≡ 1 (mod 5),⌊√ 2xn ⌋ + 5 if ⌊√ 2xn ⌋ ≡ 2 (mod 5),⌊√ 2xn ⌋ + 4 if ⌊√ 2xn ⌋ ≡ 3 (mod 5),⌊√ 2xn ⌋ + 3 if ⌊√ 2xn ⌋ ≡ 4 (mod 5),⌊√ 2xn ⌋ + 2 if ⌊√ 2xn ⌋ ≡ 0 (mod 5), (7b) M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 5 of 24 or z(0,xn) =  ⌊√ 2xn ⌋ + 2 if ⌊√ 2xn ⌋ ≡ 1 (mod 5),⌊√ 2xn ⌋ + 1 if ⌊√ 2xn ⌋ ≡ 2 (mod 5),⌊√ 2xn ⌋ + 5 if ⌊√ 2xn ⌋ ≡ 3 (mod 5),⌊√ 2xn ⌋ + 4 if ⌊√ 2xn ⌋ ≡ 4 (mod 5),⌊√ 2xn ⌋ + 3 if ⌊√ 2xn ⌋ ≡ 0 (mod 5), (7c) z(m,xn) = z(0,xn) + 5m, (7d) where n ∈ N0 and m ∈ N. The five-tuple (p, k, x, y, z) = (2, k(m,xn), xn, 1, z(m,xn)) is a solution of (3). As an example, we use our equations when x = xn = 1 + 4n and z ≡ 2 (mod 5). If n = 0, we have x = 1, z(0,1) = 7, and z(1,1) = 12 from (7a), (7b), and (7d), respectively. Note that when x = x0 = 1 and z is congruent to 2 (mod 5), the smallest value of z is ⌊ √ 2x0⌋ + 6 because we must have k > 0. From (6a) and (6b), we have that k(0,1) = 9 and k(1,1) = 28. Thus, the first two solutions when x = 1 and z ≡ 2 (mod 5) are (p, k, x, y, z) = (2, 9, 1, 1, 7) and (2, 28, 1, 1, 12). When z ≡ 3 (mod 5), we have x = 1, z(0,1) = 3, and z(1,1) = 8 from (7a), (7c), and (7d), respectively, for n = 0. We also have k(0,1) = 1 and k(1,1) = 12 from (6a) and (6b). Thus, the first two solutions when x = 1 and z ≡ 3 (mod 5) are (p, k, x, y, z) = (2, 1, 1, 1, 3) and (2, 12, 1, 1, 8). Lemma 2. Suppose the exponential Diophantine equation (5) has a solution. Then, x ≡ 2 (mod 4) if and only if z ≡ 1 (mod 5) or z ≡ 4 (mod 5). Proof. Consider the Diophantine equation (5). Let x ≡ 2 (mod 4). This means that 2x ≡ 4 (mod 5) and 2 + 5k ≡ 2 (mod 5) for all k ∈ N. Thus, 2x + (2 + 5k) ≡ 1 (mod 5) ≡ z2. We claim that z ≡ 1 (mod 5) or z ≡ 4 (mod 5). Suppose otherwise that z ≡ 0 (mod 5), z ≡ 2 (mod 5), or z ≡ 3 (mod 5). Then, we have z2 ≡ 0 (mod 5), z2 ≡ 4 (mod 5), or z2 ≡ 4 (mod 5), respectively. Any of these is a contradiction since we already established that z2 ≡ 1 (mod 5). Now, let z ≡ 1 (mod 5) or z ≡ 4 (mod 5). Then, both will give z2 ≡ 1 (mod 5). Thus, when x ≡ 2 (mod 4), (5) only has a solution when z ≡ 1 (mod 5) or z ≡ 4 (mod 5) . Now, let z ≡ 1 (mod 5) or z ≡ 4 (mod 5) in (5). Then, z2 ≡ 1 (mod 5). Assume the contrary that x ̸≡ 2 (mod 4). If x ≡ 1 (mod 4), then 2x ≡ 2 (mod 5) and 2x+(2+5k) ≡ 4 (mod 5). If x ≡ 3 (mod 4), then 2x ≡ 3 (mod 5) and 2x + (2 + 5k) ≡ 0 (mod 5). Lastly, if x ≡ 0 (mod 4), then 2x ≡ 1 (mod 5) and 2x + (2 + 5k) ≡ 3 (mod 5). In any case, we have 2x + (2 + 5k) ̸≡ z2. Hence, when z ≡ 1 (mod 5) or z ≡ 4 (mod 5), (5) only has a solution when x ≡ 2 (mod 4). Therefore, the equation 2x + (2 + 5k) = z2 has a solution when x ≡ 2 (mod 4) if and only if z ≡ 1 (mod 5) or z ≡ 4 (mod 5). We note that for this case, √ 2x is always an integer. To reduce the number of possible conditions to be considered for the smallest possible value of z, we use √ 2x. With this, M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 6 of 24 we have the following formulas. x := xn = 2 + 4n (8a) z(0,xn) = {√ 2xn + 4 if n is even,√ 2xn + 3 if n is odd, (8b) or z(0,xn) = {√ 2xn + 2 if n is even,√ 2xn + 1 if n is odd, (8c) z(m,xn) = z(0,xn) + 5m, (8d) where n ∈ N0 and m ∈ N. The five-tuple (p, k, x, y, z) = (2, k(m,xn), xn, 1, z(m,xn)) is a solution of (3). As an example, we use our equations when x = xn = 2 + 4n and z ≡ 1 (mod 4). If n = 0, we have x = 2, z(0,2) = 6, and z(1,2) = 11 from (8a), (8b), and (8d), respectively. From (6a) and (6b), we have that k(0,2) = 6 and k(1,2) = 23. Thus, the first two solutions when x = 2 and z ≡ 1 (mod 5) are (p, k, x, y, z) = (2, 6, 2, 1, 6) and (2, 23, 2, 1, 11). When z ≡ 4 (mod 5), we have x = 2, z(0,2) = 4, and z(1,2) = 9 from (8a), (8c), and (8d), respectively, for n = 0. We also have k(0,2) = 2 and k(1,2) = 15 from (6a) and (6b). Thus, the first two solutions when x = 2 and z ≡ 4 (mod 5) are (p, k, x, y, z) = (2, 2, 2, 1, 4) and (2, 15, 2, 1, 9). Lemma 3. Suppose the exponential Diophantine equation (5) has a solution. Then, x ≡ 3 (mod 4) if and only if z ≡ 0 (mod 5). The proof is omitted as it follows the same reasoning as in the proof of Lemma 1. Similar to the case when x ≡ 1 (mod 4), we use ⌊√ 2x ⌋ because √ 2x will never be an integer since x ≡ 3 (mod 4). Thus, we will have the smallest number of possible condition for z(0,xn). Hence, we have these formulas: x = xn = 3 + 4n, (9a) z(0,xn) =  ⌊√ 2xn ⌋ + 4 if ⌊√ 2xn ⌋ ≡ 1 (mod 5),⌊√ 2xn ⌋ + 3 if ⌊√ 2xn ⌋ ≡ 2 (mod 5),⌊√ 2xn ⌋ + 2 if ⌊√ 2xn ⌋ ≡ 3 (mod 5),⌊√ 2xn ⌋ + 1 if ⌊√ 2xn ⌋ ≡ 4 (mod 5),⌊√ 2xn ⌋ + 5 if ⌊√ 2xn ⌋ ≡ 0 (mod 5), (9b) z(m,xn) = z(0,xn) + 5m, (9c) where n ∈ N0 and m ∈ N. The five-tuple (p, k, x, y, z) = (2, k(m,xn), xn, 1, z(m,xn)) is a solution of (5). As an example, if n = 0 in x = xn = 3 + 4n, we have x = 3, z(0,3) = 5, and z(1,3) = 10 from (9a), (9b), and (9c), respectively. From (6a) and (6b), we have that M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 7 of 24 k(0,3) = 3 and k(1,3) = 18. Thus, the first two solutions when x = 3 and z ≡ 0 (mod 5) are (p, k, x, y, z) = (2, 3, 3, 1, 5) and (2, 18, 3, 1, 10). For these values of xn, we provide an easier way to solve for k(m,xn) by using the next theorem. Theorem 1. Suppose the exponential Diophantine equation (5), where x = xn = 1 + 4n, x = xn = 2 + 4n, or x = xn = 3 + 4n, for n ∈ N0, has a solution. Then, k(m,xn) = k(0,xn) + m−1∑ i=0 (αxn + 10i), (10) where m ∈ N. Proof. Consider the Diophantine equation 2x + (2+ 5k) = z2, where x = xn = 1+ 4n, x = xn = 2 + 4n, or x = xn = 3 + 4n, for n ∈ N0. We prove using the principle of mathematical induction on m that k(m,xn) = k(0,xn) + m−1∑ i=0 (αxn + 10i), where m ∈ N. Firstly, for m = 1, and by using the definition of αxn in (6c), we have: k(1,xn) = k(0,xn) + αxn . Thus, equation (10) is satisfied when m = 1. We now assume that equation (10) is true for m = a, that is, k(a,xn) = k(0,xn) + a−1∑ i=0 (αxn + 10i). We need to show that it is also true for m = a+ 1. By using (6b), we have the following simplification: k(a,xn) + (αxn + 10a) = ( k(0,xn) + a−1∑ i=0 (αxn + 10i) ) + (αxn + 10a), z2(a,xn) − 2xn − 2 5 + αxn + 10a = k(0,xn) + a∑ i=0 (αxn + 10i). Thus, we obtain z2(a,xn) − 2xn − 2 + 5αxn + 50a 5 = k(0,xn) + a∑ i=0 (αxn + 10i). (11) M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 8 of 24 By definition, αxn = k(1,xn) − k(0,xn). Using (6b) and (6a), we get αxn = z2(1,xn) − 2xn − 2 5 − z2(0,xn) − 2xn − 2 5 = z2(1,xn) − z2(0,xn) 5 . Also, recalling that z(m,xn) = z(0,xn) + 5m from (7d), (8d), and (9c), we have z(1,xn) = z(0,xn) + 5 and z(a,xn) = z(0,xn) + 5a. Hence, we can simplify the expression as follows: z2(a,xn) − 2xn − 2 + 5αxn + 50a 5 = z2(a,xn) − 2xn − 2 + z2(1,xn) − z2(0,xn) + 50a 5 = z2(a,xn) + z2(1,xn) − z2(0,xn) + 50a− 2xn − 2 5 = z2(a,xn) + (z(0,xn) + 5)2 − z2(0,xn) + 50a− 2xn − 2 5 = z2(a,xn) + z2(0,xn) + 10z(0,xn) + 25− z2(0,xn) + 50a− 2xn − 2 5 = z2(a,xn) + 10z(0,xn) + 50a+ 25− 2xn − 2 5 = z2(a,xn) + 10(z(0,xn) + 5a) + 25− 2xn − 2 5 = z2(a,xn) + 10z(a,xn) + 25− 2xn − 2 5 = (z(a,xn) + 5)2 − 2xn − 2 5 = z2(a+1,xn) − 2xn − 2 5 = k(a+1,xn). In the manipulations, we have used the fact that z(a+1,xn) = z(0,xn) + 5(a + 1) = z(0,xn) + 5a + 5 = z(a,xn) + 5, and the definition of k(m,xn). Thus, (11) can be written as k(a+1,xn) = k(0,xn) + a∑ i=0 (αxn + 10i), showing that equation (10) is also true for m = a + 1. Therefore, (10) is true for any x = xn = 1 + 4n, x = xn = 2 + 4n, or x = xn = 3 + 4n, where n ∈ N0. The next remark follows directly from Theorem 1. Remark 1. Consider the exponential Diophantine equation 2x + (2 + 5k) = z2, where x = xn = 1 + 4n, x = xn = 2 + 4n, or x = xn = 3 + 4n, for n ∈ N0. Then, k(m,xn) = k(m−1,xn) + αxn + 10(m− 1). M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 9 of 24 Given below are tables containing the first five solutions of equation (5), where x ≡ 1, 2 or 3 (mod 4). The expression for k(m,xn) is obtained by applying either (6b), Theorem 1, or Remark 1. If n = 0 in x = xn = 1 + 4n, then we have x = 1, and Tables 2 and 3 present some solutions where z ≡ 2 (mod 5) or z ≡ 3 (mod 5). For z ≡ 2 (mod 5), we have the following: z(0,1) = 7, z(1,1) = 12, k(0,1) = 9, k(1,1) = 28, α1 = 19. For z ≡ 3 (mod 5), we have z(0,1) = 3, z(1,1) = 8, k(0,1) = 1, k(1,1) = 12, α1 = 11. m k z Equation Solution (p, k, x, y, z) 9 7 2 + (2 + 5(9)) = 72 (2, 9, 1, 1, 7) 1 28 12 2 + (2 + 5(28)) = 122 (2, 28, 1, 1, 12) 2 57 17 2 + (2 + 5(57)) = 172 (2, 57, 1, 1, 17) 3 96 22 2 + (2 + 5(96)) = 222 (2, 96, 1, 1, 22) 4 145 27 2 + (2 + 5(145)) = 272 (2, 145, 1, 1, 27) Table 2: Some Solutions of (5) with x = 1, y = 1 and z ≡ 2 (mod 5) m k z Equation Solution (p, k, x, y, z) 1 3 2 + (2 + 5(1)) = 32 (2, 1, 1, 1, 3) 1 12 8 2 + (2 + 5(12)) = 82 (2, 12, 1, 1, 8) 2 33 13 2 + (2 + 5(33)) = 132 (2, 33, 1, 1, 13) 3 64 18 2 + (2 + 5(64)) = 182 (2, 64, 1, 1, 18) 4 105 23 2 + (2 + 5(105)) = 232 (2, 105, 1, 1, 23) Table 3: Some Solutions of (5) with x = 1, y = 1 and z ≡ 3 (mod 5) If n = 1 in x = xn = 1 + 4n, we get x = 5, and Tables 4 and 5 present some solutions where z ≡ 2 (mod 5) or z ≡ 3 (mod 5). For z ≡ 2 (mod 5), we have z(0,5) = 7, z(1,5) = 12, k(0,5) = 3, k(1,5) = 22, and α5 = 19. For z ≡ 3 (mod 5), we have z(0,5) = 8, z(1,5) = 13, k(0,5) = 6, k(1,5) = 27, and α5 = 21. M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 10 of 24 m k z Equation Solution (p, k, x, y, z) 3 7 25 + (2 + 5(3)) = 72 (2, 3, 5, 1, 7) 1 22 12 25 + (2 + 5(22)) = 122 (2, 22, 5, 1, 12) 2 51 17 25 + (2 + 5(51)) = 172 (2, 51, 5, 1, 17) 3 90 22 25 + (2 + 5(90)) = 222 (2, 90, 5, 1, 22) 4 139 27 25 + (2 + 5(139)) = 272 (2, 139, 5, 1, 27) Table 4: Some Solutions of (5) with x = 5, y = 1 and z ≡ 2 (mod 5) m k z Equation Solution (p, k, x, y, z) 6 8 25 + (2 + 5(6)) = 82 (2, 6, 5, 1, 8) 1 27 13 25 + (2 + 5(27)) = 132 (2, 27, 5, 1, 13) 2 58 18 25 + (2 + 5(58)) = 182 (2, 58, 5, 1, 18) 3 99 23 25 + (2 + 5(99)) = 232 (2, 99, 5, 1, 23) 4 150 28 25 + (2 + 5(150)) = 282 (2, 150, 5, 1, 28) Table 5: Some Solutions of (5) with x = 5, y = 1 and z ≡ 3 (mod 5) If n = 0 in x = xn = 2+4n, we get x = 2, and we have Tables 6 - 7 below that present some solutions where z ≡ 1 (mod 5) or z ≡ 4 (mod 5). m k z Equation Solution (p, k, x, y, z) 6 6 22 + (2 + 5(6)) = 62 (2, 6, 2, 1, 6) 1 23 11 22 + (2 + 5(23)) = 112 (2, 23, 2, 1, 11) 2 50 16 22 + (2 + 5(50)) = 162 (2, 50, 2, 1, 16) 3 87 21 22 + (2 + 5(87)) = 212 (2, 87, 2, 1, 21) 4 134 26 22 + (2 + 5(134)) = 262 (2, 134, 2, 1, 26) Table 6: Some Solutions of (5) with x = 2, y = 1 and z ≡ 1 (mod 5) m k z Equation Solution (p, k, x, y, z) 2 4 22 + (2 + 5(2)) = 42 (2, 2, 2, 1, 4) 1 15 9 22 + (2 + 5(15)) = 92 (2, 15, 2, 1, 9) 2 38 14 22 + (2 + 5(38)) = 142 (2, 38, 2, 1, 14) 3 71 19 22 + (2 + 5(71)) = 192 (2, 71, 2, 1, 19) 4 114 24 22 + (2 + 5(114)) = 242 (2, 114, 2, 1, 24) Table 7: Some Solutions of (5) with x = 2, y = 1 and z ≡ 4 (mod 5) If n = 1 in x = xn = 2+ 4n, we get x = 6 and Tables 8 - 9 show some solutions where z ≡ 1 (mod 5) or z ≡ 4 (mod 5). m k z Equation Solution (p, k, x, y, z) 3 9 26 + (2 + 5(3)) = 92 (2, 3, 6, 1, 9) 1 26 14 26 + (2 + 5(26)) = 142 (2, 26, 6, 1, 14) 2 59 19 26 + (2 + 5(59)) = 192 (2, 59, 6, 1, 19) 3 102 24 26 + (2 + 5(102)) = 242 (2, 102, 6, 1, 24) 4 155 29 26 + (2 + 5(155)) = 292 (2, 155, 6, 1, 29) Table 9: Some Solutions of (5) with x = 6, y = 1 and z ≡ 4 (mod 5) M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 11 of 24 m k z Equation Solution (p, k, x, y, z) 11 11 26 + (2 + 5(11)) = 112 (2, 11, 6, 1, 11) 1 38 16 26 + (2 + 5(38)) = 162 (2, 38, 6, 1, 16) 2 75 21 26 + (2 + 5(75)) = 212 (2, 75, 6, 1, 21) 3 122 26 26 + (2 + 5(122)) = 262 (2, 122, 6, 1, 26) 4 179 31 26 + (2 + 5(179)) = 312 (2, 179, 6, 1, 31) Table 8: Some Solutions of (5) with x = 6, y = 1 and z ≡ 1 (mod 5) If n = 0, 1 in x = xn = 3 + 4n, we get x = 3, 7, respectively, and Tables 10 - 11 show some solutions wherein z ≡ 0 (mod 5). m k z Equation Solution (p, k, x, y, z) 3 5 23 + (2 + 5(3)) = 52 (2, 3, 3, 1, 5) 1 18 10 23 + (2 + 5(18)) = 102 (2, 18, 3, 1, 10) 2 43 15 23 + (2 + 5(43)) = 152 (2, 43, 3, 1, 15) 3 78 20 23 + (2 + 5(78)) = 202 (2, 78, 3, 1, 20) 4 123 25 23 + (2 + 5(123)) = 252 (2, 123, 3, 1, 25) Table 10: Some Solutions of (5) with x = 3, y = 1 and z ≡ 0 (mod 5) m k z Equation Solution (p, k, x, y, z) 19 15 27 + (2 + 5(19)) = 152 (2, 19, 7, 1, 15) 1 54 20 27 + (2 + 5(54)) = 202 (2, 54, 7, 1, 20) 2 99 25 27 + (2 + 5(99)) = 252 (2, 99, 7, 1, 25) 3 154 30 27 + (2 + 5(154)) = 302 (2, 154, 7, 1, 30) 4 219 35 27 + (2 + 5(219)) = 352 (2, 219, 7, 1, 35) Table 11: Some Solutions of (5) with x = 7, y = 1 and z ≡ 0 (mod 5) The last result for this subsection discusses the case when x ≡ 0 (mod 4). Theorem 2. The exponential Diophantine equation 2x + (2+ 5k) = z2 has no solution if x ≡ 0 (mod 4). Proof. Consider the Diophantine equation 2x + (2 + 5k) = z2, where x ≡ 0 (mod 4). Then, 2x ≡ 1 (mod 5). Hence, 2x + (2 + 5k) ≡ 1 + 2 ≡ 3 (mod 5). On the other hand, z2 ≡ 0, 1, or 4 (mod 5). Thus, the equation can never be true whenever x ≡ 0 (mod 4). 2.1.2. Part II: 2x + (2 + 5k)y = z2, where y ̸= 1. For this part, we study 2x + (2 + 5k)y = z2, (12) where y ̸= 1. The discussion begins with the case where x = 0 or y = 0, followed by a claim when min (x, y) > 1. M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 12 of 24 Theorem 3. Consider the Diophantine equation (12). Let y = 0. Then, (12) has a solution only when x = 3. Consequently, (p, k, x, y, z) = (2, k, 3, 0, 3) is a solution for all k ∈ N. On the other hand, (12) has no solution when x = 0. Proof. Case 1. Let y = 0 in (12). Then, for all k ∈ N, (2 + 5k)y = 1. We can express (12) as 2x+1 = z2, or equivalently, z2− 2x = 1. If z = 0 or z = 1, it is obvious that it has no solution. However, if z > 1, then this makes it a Catalan equation, so it has a unique solution. Hence, (p, k, x, y, z) = (2, k, 3, 0, 3) is a solution of (12). Therefore, if y = 0, (12) has a solution only if x = z = 3 for all k ∈ N. Case 2. Let x = 0 in (12). Then, equation (12) will be equivalent to z2−(2+5k)y = 1. By Mihailescu’s theorem, this will only have a solution when z = 3, y = 3 and k = 0. However, we choose k ∈ N and so it is impossible for k = 0 to happen. Therefore, (12) has no solution if x = 0. Solutions wherein min (x, y) > 1 remains an open problem. Below is a claim that has been verified numerically but has not been proven rigorously. Conjecture 1. Let k ≤ 5000 and x ≤ 50. If y = 2, then equation (12) has 21 solutions, and if y = 3 then (12) has 9 solutions. Tables 12 and 13 are given below to illustrate the list of solutions when y = 2 and y = 3, respectively. M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 13 of 24 k z Equation Solution (p, k, x, y, z) 1 9 25 + (2 + 5(1))2 = 92 (2, 1, 5, 2, 9) 2 20 28 + (2 + 5(2))2 = 202 (2, 2, 8, 2, 20) 12 66 29 + (2 + 5(12))2 = 662 (2, 12, 9, 2, 66) 25 129 29 + (2 + 5(25))2 = 1292 (2, 25, 9, 2, 129) 50 260 212 + (2 + 5(50))2 = 2602 (2, 50, 12, 2, 260) 6 96 213 + (2 + 5(6))2 = 962 (2, 6, 13, 2, 96) 22 144 213 + (2 + 5(22))2 = 1442 (2, 22, 13, 2, 144) 204 1026 213 + (2 + 5(204))2 = 10262 (2, 204, 13, 2, 1026) 409 2049 213 + (2 + 5(409))2 = 20492 (2, 409, 13, 2, 2049) 38 320 216 + (2 + 5(38))2 = 3202 (2, 38, 16, 2, 320) 818 4100 216 + (2 + 5(818))2 = 41002 (2, 818, 16, 2, 4100) 198 1056 217 + (2 + 5(198))2 = 10562 (2, 198, 17, 2, 1056) 406 2064 217 + (2 + 5(406))2 = 20642 (2, 406, 17, 2, 2064) 3276 16386 217 + (2 + 5(3276))2 = 163862 (2, 3276, 17, 2, 16386) 806 4160 220 + (2 + 5(806))2 = 41602 (2, 806, 20, 2, 4160) 102 1536 221 + (2 + 5(102))2 = 15362 (2, 102, 21, 2, 1536) 358 2304 221 + (2 + 5(358))2 = 23042 (2, 358, 21, 2, 2304) 3270 16416 221 + (2 + 5(3270))2 = 164162 (2, 3270, 21, 2, 16416) 614 5120 224 + (2 + 5(614))2 = 51202 (2, 614, 24, 2, 5120) 3174 16896 225 + (2 + 5(3174))2 = 168962 (2, 3174, 25, 2, 16896) 1638 24576 229 + (2 + 5(1638))2 = 245762 (2, 1638, 29, 2, 24576) Table 12: Solutions of 2x + (2 + 5k)y = z2 when y = 2 k z Equation Solution (p, k, x, y, z) 3 71 27 + (2 + 5(3))3 = 712 (2, 3, 7, 3, 71) 6 192 212 + (2 + 5(6))3 = 1922 (2, 6, 12, 3, 192) 6 256 215 + (2 + 5(6))3 = 2562 (2, 6, 15, 3, 256) 54 4544 219 + (2 + 5(54))3 = 45442 (2, 54, 19, 3, 4544) 102 12288 224 + (2 + 5(102))3 = 122882 (2, 102, 24, 3, 12288) 102 16384 227 + (2 + 5(102))3 = 163842 (2, 102, 27, 3, 16384) 870 290816 231 + (2 + 5(870))3 = 2908162 (2, 870, 31, 3, 290816) 1638 786432 236 + (2 + 5(1638))3 = 7864322 (2, 1638, 36, 3, 786432) 1638 1048576 239 + (2 + 5(1638))3 = 10485762 (2, 1638, 39, 3, 1048576) Table 13: Solutions of 2x + (2 + 5k)y = z2 when y = 3 2.2. On the Diophantine Equation px + (p+ 5k)y = z2 for Prime Pairs p and p+ 5k This subsection discusses some findings regarding solutions of (3), where p and p+ 5k are prime pairs. This is further divided into two sub-cases based on the parity of k; that is, when k is odd, and when k is even. M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 14 of 24 2.2.1. Sub-case I: k is odd. The case where p = 2 falls in this sub-case. Some results have already been discussed in Section 2.1. Additionally, using similar arguments as in proving claims in that section, we obtain the results below. We note that in the Diophantine equation px +(p+5k)y = z2, where p and p+5k are prime pairs and k is odd, the prime gap 5k is also odd. It is clear that p + 5k must be odd to be prime and for it to have a prime gap of an odd number, p must be even, that is, p = 2. Hence, some of the solutions were already included in Section 2.1. Let us start with the discussion of solutions when x is odd. That is, x ≡ 1 (mod 4) and x ≡ 3 (mod 4). Note that unlike the previous section that uses modulo 5 for z, this subsection uses modulo 10. We start the discussion with the following lemmas. Lemma 4. Let k be an odd integer. Suppose the Diophantine equation (5) has a solution. Then, x ≡ 1 (mod 4) if and only if z ≡ 3 (mod 10) or z ≡ 7 (mod 10). Proof. Consider equation (5) and let x ≡ 1 (mod 4) and k be odd. This means that 2x + (2 + 5k) ≡ 2 + 7 ≡ 9 (mod 10) ≡ z2. It follows that z2 must be odd and that z is odd too. We claim that z ≡ 3 (mod 10) or z ≡ 7 (mod 10). Suppose on the contrary that z ≡ 1, 5 or 9 (mod 10). If z ≡ 1, 9 (mod 10), then z2 ≡ 1 (mod 10). If z ≡ 5 (mod 10), then z2 ≡ 5 (mod 10). Any of these is a contradiction since we established that z2 ≡ 9 (mod 10). Now, if z ≡ 3, 7 (mod 10), then z2 ≡ 9 (mod 10). Hence, when x ≡ 1 (mod 4), (5) only has a solution when z ≡ 3 (mod 10) or z ≡ 7 (mod 10). Now, let z ≡ 3 (mod 10) or z ≡ 7 (mod 10) in (5), where k is odd. Then, z2 ≡ 9 (mod 10). Suppose on the contrary that x ≡ 2, 3, 0 (mod 4), then 2x ≡ 4, 8, 6 (mod 10), respectively. Hence, we will have 2x + (2 + 5k) ≡ 1, 5, 3 (mod 10), respectively. This will be a contradiction from the assumption that z2 ≡ 9 (mod 10). Thus, if z ≡ 3 (mod 10) or z ≡ 7 (mod 10), (5) only has a solution when x ≡ 1 (mod 4). Therefore, the equation 2x + (2 + 5k) = z2 where 2 and 2 + 5k are prime pairs and k is odd, has a solution when x ≡ 1 (mod 4) if and only if z ≡ 3 (mod 10) or z ≡ 7 (mod 10). Using this result, the first two solutions when x = 1 and z ≡ 3 (mod 10) are (p, k, x, y, z) = (2, 1, 1, 1, 3) and (2, 33, 1, 1, 13). On the other hand, the first solution when x = 1 and z ≡ 7 (mod 10) is (p, k, x, y, z) = (2, 9, 1, 1, 7). Lemma 5. Let k be an odd integer. Suppose the Diophantine equation (5) has a solution. Then, x ≡ 3 (mod 4) if and only if z ≡ 5 (mod 10). Proof. Let x ≡ 3 (mod 4) in (5), where p = 2 and p+5k are prime pairs and k is odd. This means that 2x+(2+5k) ≡ 8+7 ≡ 5 (mod 10) ≡ z2. Then, z2 is odd and z must be odd too. We claim that z ≡ 5 (mod 10). Suppose on the contrary that z ̸≡ 5 (mod 10). If z ≡ 1, 9 (mod 10), then z2 ≡ 1 (mod 10). If z ≡ 3, 7 (mod 10), then z2 ≡ 1 (mod 10). Any of these is a contradic- tion since we established that z2 ≡ 5 (mod 10). Now, if z ≡ 5 (mod 10), then z2 ≡ 5 (mod 10). Hence, when x ≡ 3 (mod 4), (5) only has a solution when z ≡ 5 (mod 10). M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 15 of 24 Now, let z ≡ 5 (mod 10). Then, z2 ≡ 5 (mod 10). Suppose on the contrary that x ≡ 1, 2, 0 (mod 4), then 2x ≡ 2, 4, 6 (mod 10), respectively. Thus, we will have equation (5) as 2x + (2 + 5k)y ≡ 9, 1, 3 (mod 10) ≡ z2, respectively. This will be a contradiction from the assumption that z2 ≡ 5 (mod 10). Hence, if z ≡ 5 (mod 10), (5) only has a solution when x ≡ 3 (mod 4). Therefore, the equation 2x + (2 + 5k) = z2, where p = 2 and p + 5k are prime pairs and k is odd, only has a solution when x ≡ 3 (mod 4) if and only if z ≡ 5 (mod 10). One can verify that the first solution when x = 3 and z ≡ 5 (mod 10) is (p, k, x, y, z) = (2, 3, 3, 1, 5). For values of xn that are either in the form 1 + 4n or 3 + 4n, an easier way to solve for k(m,xn) is given by the next theorem, which can be proven by induction. Theorem 4. Suppose the exponential Diophantine equation (5), where x = xn = 1 + 4n or x = xn = 3 + 4n, n ∈ N0, has a solution. Then, k(m,xn) = k(0,xn) + m−1∑ i=0 (αxn + 40i), (13) where m ∈ N. Proof. Consider the Diophantine equation 2x + (2+ 5k) = z2, where x := xn = 1+ 4n or x := xn = 3 + 4n, n ∈ N0. We prove by induction on m that k(m,xn) = k(0,xn) + m−1∑ i=0 (αxn + 40i), where m ∈ N. We first note that by using the definition of αxn , equation (13) is satisfied when m = 1, as seen below: k(1,xn) = k(0,xn) + αxn . We now assume that equation (13) is true for m = a, that is, k(a,xn) = k(0,xn) + a−1∑ i=0 (αxn + 40i). By recalling that k(a,xn) = z2 (a,xn) −2xn−2 5 , we get k(a,xn) + (αxn + 40a) = ( k(0,xn) + a−1∑ i=0 (αxn + 40i) ) + (αxn + 40a), z2(a,xn) − 2xn − 2 5 + αxn + 40a = k(0,xn) + a∑ i=0 (αxn + 40i). M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 16 of 24 Thus, we have z2(a,xn) − 2xn − 2 + 5αxn + 200a 5 = k(0,xn) + a∑ i=0 (αxn + 40i). (14) Note that αxn can also be written as follows: αxn = z2(1,xn) − 2xn − 2 5 − z2(0,xn) − 2xn − 2 5 = z2(1,xn) − z2(0,xn) 5 . This, together with the equations z(1,xn) = z(0,xn) + 10 and z(a,xn) = z(0,xn) + 10a, will lead us to the following simplification: z2(a,xn) − 2xn − 2 + 5αxn + 200a 5 = z2(a,xn) − 2xn − 2 + z2(1,xn) − z2(0,xn) + 200a 5 = z2(a,xn) + z2(1,xn) − z2(0,xn) + 200a− 2xn − 2 5 = z2(a,xn) + (z(0,xn) + 10)2 − z2(0,xn) + 200a− 2xn − 2 5 = z2(a,xn) + z2(0,xn) + 20z(0,xn) + 100− z2(0,xn) + 200a− 2xn − 2 5 = z2(a,xn) + 20z(0,xn) + 200a+ 100− 2xn − 2 5 = z2(a,xn) + 20(z(0,xn) + 10a) + 100− 2xn − 2 5 = z2(a,xn) + 20z(a,xn) + 100− 2xn − 2 5 = (z(a,xn) + 10)2 − 2xn − 2 5 . Since z(a+1,xn) = z(0,xn) + 10(a+ 1) = z(0,xn) + 10a+ 10 = z(a,xn) + 10, and by definition of k(m,xn), we have z2(a,xn) − 2xn − 2 + 5αxn + 200a 5 = z2(a+1,xn) − 2xn − 2 5 = k(a+1,xn). Hence, k(a+1,xn) = k(0,xn) + a∑ i=0 (αxn + 40i), M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 17 of 24 showing that (13) is also true for m = a+1. Therefore, (13) is true for any m ∈ N, where xn = 1 + 4n or xn = 3 + 4n, n ∈ N0. The next remark follows directly from the theorem above. Remark 2. Consider the exponential Diophantine equation 2x + (2 + 5k) = z2, where x := xn = 1 + 4n or x := xn = 3 + 4n, n ∈ N0. Then, k(m,xn) = k(m−1,xn) + αxn + 40(m− 1). Tables 14 - 17 each contain the first five solutions of equation (5) where x ≡ 1 or 3 (mod 4), and k(m,xn) is obtained by applying equation (6b), Theorem 4, or Remark 2. m k z Equation Solution (p, k, x, y, z) 1 3 2 + (2 + 5(1)) = 32 (2, 1, 1, 1, 3) 1 33 13 2 + (2 + 5(33)) = 132 (2, 33, 1, 1, 13) 3 217 33 2 + (2 + 5(217)) = 332 (2, 217, 1, 1, 33) 4 369 43 2 + (2 + 5(369)) = 432 (2, 369, 1, 1, 43) 6 793 63 2 + (2 + 5(793)) = 632 (2, 793, 1, 1, 63) Table 14: Some Solutions of (5) with x = y = 1 and z ≡ 3 (mod 10) m k z Equation Solution (p, k, x, y, z) 9 7 2 + (2 + 5(9)) = 72 (2, 9, 1, 1, 7) 2 145 27 2 + (2 + 5(145)) = 272 (2, 145, 1, 1, 27) 3 273 37 2 + (2 + 5(273)) = 372 (2, 273, 1, 1, 37) 4 441 47 2 + (2 + 5(441)) = 472 (2, 441, 1, 1, 47) 7 1185 77 2 + (2 + 5(1185)) = 772 (2, 1185, 1, 1, 77) Table 15: Some Solutions of (5) with x = y = 1 and z ≡ 7 (mod 10) m k z Equation Solution (p, k, x, y, z) 27 13 25 + (2 + 5(27)) = 132 (2, 27, 5, 1, 13) 4 555 53 25 + (2 + 5(555)) = 532 (2, 555, 5, 1, 53) 6 1059 73 25 + (2 + 5(1059)) = 732 (2, 1059, 5, 1, 73) 7 1371 83 25 + (2 + 5(1371)) = 832 (2, 1371, 5, 1, 83) 12 3531 133 25 + (2 + 5(3531)) = 1332 (2, 3531, 5, 1, 133) Table 16: Some Solutions of (5) with x = 5, y = 1 and z ≡ 3 (mod 10) m k z Equation Solution (p, k, x, y, z) 3 7 25 + (2 + 5(3)) = 72 (2, 3, 5, 1, 7) 1 51 17 25 + (2 + 5(51)) = 172 (2, 51, 5, 1, 17) 5 643 57 25 + (2 + 5(643)) = 572 (2, 643, 5, 1, 57) 6 891 67 25 + (2 + 5(891)) = 672 (2, 891, 5, 1, 67) 7 1179 77 25 + (2 + 5(1179)) = 772 (2, 1179, 5, 1, 77) Table 17: Some Solutions of (5) with x = 5, y = 1 and z ≡ 7 (mod 10) Tables 18 - 19 present some solutions of (5), where z ≡ 5 (mod 10). Now, we discuss the case when x is even, i.e., x ≡ 0 (mod 4) and x ≡ 2 (mod 4). We begin with a remark, followed by a proven claim. M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 18 of 24 m k z Equation Solution (p, k, x, y, z) 3 5 23 + (2 + 5(3)) = 52 (2, 3, 3, 1, 5) 2 123 25 23 + (2 + 5(123)) = 252 (2, 123, 3, 1, 25) 3 243 35 23 + (2 + 5(243)) = 352 (2, 243, 3, 1, 35) 4 405 45 23 + (2 + 5(405)) = 452 (2, 405, 3, 1, 45) 6 843 65 23 + (2 + 5(843)) = 652 (2, 843, 3, 1, 65) Table 18: Some Solutions of (5) with x = 3, y = 1 and z ≡ 5 (mod 10) m k z Equation Solution (p, k, x, y, z) 19 15 27 + (2 + 5(19)) = 152 (2, 19, 7, 1, 15) 2 219 35 27 + (2 + 5(219)) = 352 (2, 219, 7, 1, 35) 4 579 55 27 + (2 + 5(579)) = 552 (2, 579, 7, 1, 55) 11 3099 125 27 + (2 + 5(3099)) = 1252 (2, 3099, 7, 1, 125) 12 3619 135 27 + (2 + 5(3619)) = 1352 (2, 3619, 7, 1, 135) Table 19: Some Solutions of (5) with x = 7, y = 1 and z ≡ 5 (mod 10) Remark 3. From Theorem 2, the equation 2x +(2+ 5k) = z2, where 2+ 5k is prime and k is odd, has no solution when x ≡ 0 (mod 4). Theorem 5. Let 2+5k be prime, and x = 2+4n with odd number n. Then the exponential Diophantine equation (5), has solutions where k = 22+2n − 1 5 and z = 5k + 3 2 . Proof. Let x = 2 + 4n, n be odd, and q = 2 + 5k be prime. Then, we can express 2x + (2 + 5k) = z2 as 22+4n + q = z2. Finding for q we have, q = z2 − 22+4n = (z − 21+2n)(z + 21+2n). Since q is prime, we know that gcd (z − 21+2n, z + 21+2n) = 1. It follows that z − 21+2n = 1 and z + 21+2n = q. Having this system of equations, we solve for z. Thus, we get 2z = q + 1 =⇒ z = q + 1 2 =⇒ z = 5k + 3 2 . Substituting to find k, we obtain q = z + 21+2n 2 + 5k = 5k + 3 2 + 21+2n 2(2 + 5k) = 5k + 3 + 2 · 21+2n 4 + 10k = 5k + 3 + 22+2n 10k − 5k = 22+2n − 1 5k = 22+2n − 1 k = 22+2n − 1 5 . M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 19 of 24 We are sure that k is an odd integer because n is odd. Thus, 2x + (2 + 5k) = z2 has a solution if y = 1, x = 2 + 4n, k = 22+2n − 1 5 , and z = 5k + 3 2 , where n is an odd number. Using Theorem 5, one can verify that (p, k, x, y, z) = (2, 3, 6, 1, 9), (2, 51, 14, 1, 129), and (2, 13107, 30, 1, 32769) are indeed solutions of (3). Theorem 6. Consider the exponential Diophantine equation (3), where 2+5k is prime. If y = 0, the equation has only solution when x = 3. Consequently, (p, k, x, y, z) = (2, k, 3, 0, 3) is a solution for some odd k ∈ N. On the other hand, the equation has no solution when x = 0. The proof of Theorem 6 follows from Theorem 3. The next conjecture also follows from Conjecture 1. Conjecture 2. Let 2 + 5k be prime, k ≤ 5000, and x ≤ 50. If y = 2, then equation 2x+(2+5k)y = z2 has 2 solutions, namely, (p, k, x, y, z) = (2, 1, 5, 2, 9) and (2, 25, 9, 2, 129). If y = 3, then the equation has only the solution (p, k, x, y, z) = (2, 3, 7, 3, 71). 2.2.2. Sub-case II: k is even. This sub-case will discuss the solutions of (3) when k is even, that is, k ≡ 0 (mod 4) or k ≡ 2 (mod 4), and p and p+ 5k are prime pairs. We note that since k is even, then the prime gap, 5k, is also even. Thus, for p and p + 5k to be prime pairs, both of them must be odd integers. We begin the discussion with k ≡ 2 (mod 4). Lemma 6. Let p ≥ 3 and p+ 5k be prime pairs and k ≡ 2 (mod 4). Then, the equation (3) has no solution if x and y are both even integers. Proof. Consider the Diophantine equation px+(p+5k)y = z2, where p and q = p+5k are odd prime pairs and k ≡ 2 (mod 4). If p ≡ 1, 3 (mod 4), then q ≡ 3, 1 (mod 4), respectively. Let x and y be both even integers. In either case, we have px ≡ 1 (mod 4) and qy ≡ 1 (mod 4) so that px + qy ≡ 1 + 1 ≡ 2 (mod 4). Also, since px and qy are both odd, then their sum is even. Thus, z2 ≡ 0 (mod 4) and we’ll have a contradiction since px + qy ≡ 2 ̸≡ 0 ≡ z2 (mod 4). Conclusion follows. Notice that when x = y = 1, we have the equation p+ (p+ 5k) = z2. (15) We use the following variables to study (15). M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 20 of 24 Variable Meaning kn the value of k at a specific value of n z(0,kn) the first value of z at a specific value of kn z(m,kn) the (m+ 1)st value of z at a specific value of kn p(0,kn) the first value of p at a specific value of kn p(1,kn) the second value of p at a specific value of kn ωkn the difference between p(0,kn) and p(1,kn) at a specific value of kn p(m,kn) the (m+ 1)st value of p at a specific value of kn Table 20: Variables Considered for p+ (p+ 5k) = z2, where p ≥ 3 and p+ 5k are Prime Pairs We derive the following formulas: k = kn = 2 + 4n, (16a) z(0,kn) =  √ 6 + 5k if √ 6 + 5k is an even integer,⌊√ 6 + 5k ⌋ + 1 if ⌊√ 6 + 5k ⌋ is odd,⌊√ 6 + 5k ⌋ + 2 if ⌊√ 6 + 5k ⌋ is even, (16b) z(m,kn) = z(0,kn) + 2m, (16c) p(0,kn) = z2(0,kn) − 5kn 2 , (16d) p(m,kn) = z2(m,kn) − 5kn 2 , (16e) ωkn = p(1,kn) − p(0,kn), (16f) where n ∈ N0 and for some m ∈ N. We use the floor function of √ 6 + 5k for z(0,kn) since we know that the smallest possible value of p is 3. Also, from equation (15), we have 3 + 3 + 5k = z2 will mean that z = √ 6 + 5k. Since k is even, √ 6 + 5k is even if it is an integer. Otherwise, we have another two conditions as seen below. The five-tuple (p, k, x, y, z) = (p(m,kn), kn, 1, 1, z(m,kn)) will become a solution of equation (3). Below is another result. The proof is similar to the theorems above that use mathe- matical induction, hence we omit it. Theorem 7. Suppose that the exponential Diophantine equation (15), where k := kn = 2 + 4n, n ∈ N0, has a solution. Then, p(m,kn) = p(0,kn) + m−1∑ i=0 (ωkn + 4i). This remark follows directly from the Theorem 7. Corollary 1. Consider the Diophantine equation (15), where k = kn = 2 + 4n, n ∈ N0. Then, p(m,kn) = p(m−1,kn) + ωkn + 4(m− 1). M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 21 of 24 Given below are tables containing the first five solutions of equation (15), where k := kn = 2 + 4n, and p(m,kn) is obtained by applying (16e), or Theorem 7, or Remark 1. Tables 21 and 22 present some solutions of (15) when n = 0 and n = 1 in k = kn = 2 + 4n, respectively. If n = 0, we have k = 2, z(0,2) = 4, z(1,2) = 6, p(0,2) = 3, p(1,1) = 13, and ω2 = 10. For n = 1, we have k = 6, z(0,6) = 6, z(1,6) = 8, p(0,6) = 3, p(1,6) = 17, and ω6 = 14. m p z Equation Solution (p, k, x, y, z) 3 4 3 + 13 = 42 (3, 2, 1, 1, 4) 1 13 6 13 + 23 = 62 (13, 2, 1, 1, 6) 7 157 18 157 + 167 = 182 (157, 2, 1, 1, 18) 10 283 24 283 + 293 = 242 (283, 2, 1, 1, 24) 16 643 36 643 + 653 = 362 (643, 2, 1, 1, 36) Table 21: Some Solutions of (15) with k = 2, x = 1, y = 1 and z is even m p z Equation Solution (p, k, x, y, z) 1 17 8 17 + (17 + 5(6)) = 82 (17, 6, 1, 1, 8) 4 83 14 83 + (83 + 5(6)) = 142 (83, 6, 1, 1, 14) 8 227 22 227 + (227 + 5(6)) = 222 (227, 6, 1, 1, 22) 14 563 34 563 + (563 + 5(6)) = 342 (563, 6, 1, 1, 34) 19 953 44 953 + (953 + 5(6)) = 442 (953, 6, 1, 1, 44) Table 22: Some Solutions of (15) with k = 6, x = 1, y = 1 and z is even For the next theorem, we discuss the case when k is even and exactly either x or y is zero. The case when both x and y are zero are impossible since 2 is not a perfect square. Theorem 8. Consider the Diophantine equation (3) where p ≥ 3 and p + 5k are prime pairs, and k is a positive even number. Let x = 0 or y = 0. Then, (p, k, x, y, z) = (3, k, 1, 0, 2) for some even k ∈ N is the only solution of the equation. Proof. Consider the equation (3). Let p ≥ 3 and q = p+ 5k be prime pairs, k is even, and x = 0 or y = 0. Case 1. Let x = 0. Then, we have 1 + (p+ 5k)y = z2. If y > 1, then by Mihailescu’s theorem, this equation will only have a solution when z = 3 and p + 5k = 2, which is absurd since p ≥ 3. If y = 1, then the equation reduces to 1 + p + 5k = z2. We first claim that the only solution to the equation 1 + q = z2, where q is a prime number, is (q, z) = (3, 2). Indeed, rewriting the equation gives q = z2−1 = (z−1)(z+1). Since z−1 and z+1 are consecutive even integers, their product is greater than 1 for all z > 2, making q composite in those cases. Hence, the only valid solution occurs when z = 2, which gives q = 3. Returning to the original equation, this implies that p + 5k = 3. However, since k is assumed to be a positive integer, the left-hand side p + 5k > 3, and thus no prime value of p satisfies the equation. Therefore, there is no solution with p prime and y = 1. Case 2. Let y = 0. Then, (p + 5k)y = 1 for all even k. From that, we have px = z2 − 1 = (z + 1)(z − 1). It follows that gcd (z + 1, z − 1) = 1, which further implies M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 22 of 24 that z−1 = 1 and z+1 = px. From z−1 = 1, we obtain z = 2. Therefore, z+1 = 3 = px, which implies that p = 3 and x = 1. Therefore, (p, k, x, y, z) = (3, k, 1, 0, 2) is the only solution of px + (p + 5k)y = z2 for even k, where p ≥ 3 and p+ 5k are prime pairs, and x = 0 or y = 0. The following result is obtained when k ≡ 0 (mod 4). Theorem 9. The Diophantine equation (3), where p ≥ 3 and p+5k are prime pairs such that k ≡ 0 (mod 4) and x and y have the same parity, has no solution. Proof. Suppose x ≥ 1 and y ≥ 1 are of the same parity. Let x and y be odd integers, then px ≡ (p + 5k)y ≡ 1 (mod 4) or px ≡ (p + 5k)y ≡ 3 (mod 4). If x and y are even integers, then px ≡ 1 (mod 4) and (p+5k)y ≡ 1 (mod 4). In any case, px +(p+5k)y ≡ 2 (mod 4). On the other hand, since px and (p + 5k)y are both odd, their sum is even. Hence, z2 ≡ 0 (mod 4) and thus, px + (p+ 5k)y ̸≡ z2. 3. Conclusions In this paper, the authors tried to provide solutions of an exponential Diophantine equation of the form px + (p+ 5k)y = z2, where k ∈ N. The authors considered the cases (i) when p = 2; or (ii) when p and p+5k are prime pairs are given. In addition, the study is limited only to integer solutions (x, y, z), where x and y are not simultaneously greater than 1. For the first case, it is concluded that if y = 1, then the Diophantine equation has infinitely many solutions that can be further divided based on the value of x taking modulo 4. If x ≡ 1 (mod 4), z can be either z ≡ 2, 3 (mod 5). If x ≡ 2 (mod 4), then z ≡ 1, 4 (mod 5). If x ≡ 3 (mod 4), then z ≡ 0 (mod 5). In the case where z have two possible equivalence modulo 4, the union of all these solutions form all the solutions at a specific value of x. Moreover, when y = 0, (p, k, x, y, z) = (2, k, 3, 0, 3) is the only solution of the equation for all k ∈ N. When y = 2, 3 such that k ≤ 5000 and x ≤ 50, there are finitely many solutions mentioned. In particular, 21 solutions were mentioned when y = 2 and 9 solutions were mentioned when y = 3. The equation has no solution when y = 1 and x is a multiple of 4, i.e., x ≡ 0 (mod 4), and if x = 0. The discussion for the second case was divided into two, based on the parity of k. If k is odd, some solutions are already mentioned in the first case since p = 2. In this case, modulo 10 was used for the value of z since we are dealing with prime numbers and z should be odd for p + 5k to be an odd integer and prime number. Similar to that of the first case, when y = 1, the Diophantine equation also have infinitely many solutions for some value of x with the note that there are infinitely many prime numbers of the form p + 5k. If x ≡ 1 (mod 4), then the equation has a solution for some z ≡ 3, 7 (mod 10) and if x ≡ 3 (mod 4), then the equation has a solution for some z ≡ 5 (mod 10). M. C. Avenilla, J. B. Bacani / Eur. J. Pure Appl. Math, 18 (3) (2025), 6593 23 of 24 There are finitely many solutions when y = 1 and x = 2 + 4n, n ∈ N0, where the value of k and z are given by k = (22+2n − 1)/5 and z = (5k + 3)/2. Three of them were mentioned and they are the following: (p, k, x, y, z) = (2, 3, 6, 1, 9), (2, 51, 14, 1, 129), and (2, 13107, 30, 1, 129). When y = 2 and y = 3 with the same values of k and x in the first case, there are three solutions that were mentioned. In other words, there are three (2+ 5k) out of the 30 solutions mentioned in the first case that were prime, namely, (p, k, x, y, z) = (2, 1, 5, 2, 9), (2, 25, 9, 2, 129), and (2, 3, 7, 3, 71). Similar to the first case, the equation has no solution when y = 1 and x ≡ 0 (mod 4), and when x = 0. For even integers k, the discussion was also divided into two: when k ≡ 2 (mod 4); and when k ≡ 0 (mod 4). It has infinitely many solutions when k ≡ 2 (mod 4) and x = y = 1. Moreover, (p, k, x, y, z) = (3, k, 1, 0, 2) is a solution for some even integers k, that is, only when 3 + 5k is a prime. It has no solution for the most part. For the last case of this study, it is concluded that the equation has infinitely many solution when x = y = 1, and that z and k have the same parity. For future study, the authors recommend the examination of the same cases, but to explore solutions where min (x, y) > 1. 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