EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 4, 2023, 2066-2081 ISSN 1307-5543 – ejpam.com Published by New York Business Global On the Diophantine Equations ax + by + cz = w2 Kittipong Laipaporn1, Saowapak Kaewchay1, Adisak Karnbanjong1,∗ 1 Department of Mathematics and Statistics, Center of Excellence for Ecoinformatics, School of Science, Walailak University, Nakhon Si Thammarat , 80160, Thailand Abstract. Over the past decade, exponential Diophantine equations of the form ax + by = wn have been studied as if they were a phenomenon. In particular, numerous articles have focused on the cases where n = 2 or n = 4 and 2 ≤ a, b ≤ 200. However, these articles are primarily concerned with determining whether the left-hand side of the equation needs to consist of more than two exponentials. Therefore, in this article, we investigate the exponential Diophantine equation in the form ax + by + cz = w2, using only elementary tools related to modulo concepts. We present three theorems in which the variables a, b and c vary under certain conditions, and three additional theorems where the variable c is fixed at 7. Furthermore, if we restrict our parameters a, b and c to 2 ≤ a ≤ b ≤ c ≤ 20, then 1,330 equations have been considered. Our results confirm that 135 of these equations have been fully clarified. 2020 Mathematics Subject Classifications: 11A07 Key Words and Phrases: Exponential Diophantine Equation, Modulo 1. Introduction Many mathematicians have proposed generalized forms of the Diophantine equation in various ways. One example is the exponential Diophantine equation 3x + 4y = 5z, which has no solutions for any natural numbers x, y, z except when x = y = z = 2. This was proven in 1956 by W. Sierpinski, [17]. In the same year, L. Jamanowicz published an article that seemed to follow in W. Sierpinski’s footsteps by selecting other equations such as 5x + 12y = 13z, 7x + 24y = 25z, 9x + 40y = 41z, and 11x + 60y = 61z[see 13]. Since then, many mathematicians have explored variations by changing the base values a, b, and c in the exponential Diophantine equation ax + by = cz. For example, in 2005, D. Acu [1] considered three cases: Case 1 where a = b = c = p, Case 2 where a = b = p and c = 2p, and the last case where a = p, b = q, and c = pq, with p and q being prime numbers. Furthermore, he proposed the alternative form 2x + 5y = z2 in 2007, and noted that Catalan’s conjecture was an important tool for finding solutions,[see 2]. It’s also worth noting that the exponential term cz was interchanged with the polynomial term z2. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4936 Email addresses: lkittipo@wu.ac.th (K. Laipaporn), saowapak.ka@mail.wu.ac.th (S. Kaewchay), kadisak@mail.wu.ac.th (A. Karnbanjong) https://www.ejpam.com 2066 © 2023 EJPAM All rights reserved. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2067 Since 2007, hundreds of articles inspired by the equation 2x + 5y = z2 have been published. Many were authored by B. Sroysang, as seen in references [18–20], among others listed in [5, 9, 16, 22]. In addition, there are exponential Diophantine equations similar to 2x + 5y = z2 but with more than three variables. Some of these are showcased in Table 1. Ever since D. Acu introduced the equation 2x + 5y = z2 in 2007, researchers have explored alternative forms by altering the base numbers 2 and 5. They have also tried to generalize the equation, creating new forms and investigating their solutions. Interestingly, there are few equations like 3x + 5y + 7z = w2 that consider three base numbers for the exponential terms. This particular equation was established by J. B. Bacani and J. F. T. Rabago, and serves as the inspiration for this article. Here, we focus on the form ax + by + cz = w2 under certain conditions for a, b, c, x, y, z, and w, using only elementary tools related to modulo concepts. Most equations in Table 1 have more than three variables, but they still involve only two base numbers. Even the equation px + (p+ 1)y + (p+ 2)z = M2, cited in [6], appears to have three exponential terms, but its parameters x, y, and z are restricted to the set {1, 2, 3}. Table 1: Example of the exponential Diophantine equations during the past ten years which were considered more than three variables. Ref. Author Equation [3] J.B. Bacani and J.F.T. Rabago 3x + 5y + 7z = w2 [4] J.B. Bacani and J.F.T. Rabago px + qy = z2 [6] Nechemia Burshtein px + (p+ 1)y + (p+ 2)z = M2 [7] Nechemia Burshtein p3 + qy = z3 [10] R. Dokchan and A. Pakapongpun px + (p+ 20)y = z2 [14] K. Laipaporn, S. Wananiyakul 3x + p5y = z2 and P. Khachorncharoenkul [21] S. Subburam lax +mby = ncz [23] A. Suvarnamani px + (p+ 1)y = z2 2. Main Theorem Our main results are divided into two groups. In the first group, we focus on the equation ax+by+cz = w2 with the variable c fixed at 7, while varying (a, b) in specific cases. These cases consider all elements in the set A, which includes (3, 4), (9, 4), (3, 16), (9, 16), (4, 6), (16, 6), (9, 10), and (6, 10). This is discussed in Theorems 1–4 and Corollaries 1–2. The second group consists of Theorems 5–7 and Corollary 3, where all variables a, b, and c are allowed to vary under certain conditions. Additionally, we introduce an auxiliary result, referred to as ’Result 1,’ which is utilized in sections 1 and 2. Lemma 1. For any non-negative integers w and z, the equation 5 + 7z = w2 has no solution. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2068 Proof. It is clear that w has to be even. So, we have that z is odd because of 0 ≡ w2 ≡ 5 + 7z ≡ 1 + (−1)z ≡ { 0 (mod 4) if z is odd, 2 (mod 4) if z is even. If z = 1 then w is not integer, so we can let z = 2k + 1 for some integer k ≥ 1. Then the eqaution 5 + 7z = w2 becomes (w − 2)(w + 2) = (7 + 1)(72k − 72k−1 + · · · − 7 + 1) So 8|(w − 2)(w + 2). Since 2|(w − 2), 2|(w + 2), and w + 2 = (w − 2) + 4, we conclude that 4|(w − 2) and 4|(w + 2). This implies that 0 ≡ w2 − 4 ≡ 72k+1 + 1 ≡ 8 (mod 16), so it is a contratiction. Hence the equation 5 + 7z = w2 has no solution. With the previous lemma, we are ready to solve the following equation. 2.1. The exponential Diophantine equation ax + by + 7z = w2 First, we note from [3], J. B. Bacani and J. F. T. Rabago that they focused on the equation 3x + 5y + 7z = w2. In case x = y = 0, the equation becomes to 2 + 7z = w2 that already considered that why we examined our theorems in this section over the set U = N4 0 − {(0, 0, z, w)|z, w ∈ N0} where N0 is the set of all non-negative integers. From now on, it causes us to investigate x and y are unequal to zero simultaneously. Then we have the results of the exponential Diophantine equation ax + by + 7z = w2 where a and b are positive integers and the variables (x, y, z, w) are elements in U as follows: Theorem 1. The equation 3x + 4y + 7z = w2 (1) has no solution for any (x, y, z, w) ∈ U . Proof. The proof of the theorem is separated into 3 cases. Case 1 z = 0. Then the equation (1) becomes 3x + 4y + 1 = w2. (2) We note that 3x + 4y + 1 ≡ { 2 (mod 3) for x ≥ 1 and y ≥ 0 2 (mod 4) for x = 0 and y ≥ 1 (3) Since w2 ≡ 0, 1 (mod 3) and w2 ≡ 0, 1 (mod 4), we conclude that 3x + 4y + 1 = w2 has no solution for all x, y, w ∈ N0. Case 2 z = 1. For any x ≥ 1 and y ≥ 0, we see that 3x + 4y + 7 ≡ 2 (mod 3). Again, w2 ≡ 0, 1 (mod 3) and this fact forces the equation 3x+4y+7 = w2 has no solution. Now, we remain to consider x = 0 and y ≥ 1 and then the equation (1) is in the form 8 = (w − 2y)(w + 2y). So, we can let w − 2y = 2u and w + 2y = 23−u where u = 0 or u = 1. Since 2w = { 9 if u = 0 6 if u = 1 and w has to be integer, we have w = 3 and y = 0. It contradics to y ≥ 1. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2069 Case 3 z ≥ 2. First, we can see that 3x + 4y + 7z ≡ 2 (mod 3) for any x ≥ 1 and y ≥ 0, but w2 ̸≡ 2 (mod 3), so the equation (1) has no solution. Next, we consider x = 0 and y = 1. By lemma 1, we know that the equation (1) has no solution. Finally, we assume that x = 0 and y ≥ 2. So w is even. It implies that 0 ≡ w2 ≡ 1 + 4y + 7z ≡ 1+ (−1)z (mod 4) and then z has to be odd. Again w2 ≡ 1+4y +7z ≡ 1+0+7 ≡ 8 (mod 16) but we have the fact that w2 ≡ 0, 1, 4, 9 (mod 16) which is a contradiction. Hence 3x + 4y + 7z = w2 has no solution for any (x, y, z, w) ∈ U . Corollary 1. Each of the following equations: 9x + 4y + 7z = w2, (4) 3x + 16y + 7z = w2 (5) and 9x + 16y + 7z = w2 (6) have no solution for any (x, y, z, w) ∈ U . Proof. Suppose that (x0, y0, z0, w0) ∈ U is a solution of the equation (4). Then w2 0 = 32x0 + 4y0 + 7z0, i.e. (2x0, y0, z0, w0) is a solution of the equation (1). It contradicts to 1. With the same trace of the proof of equation (4), we can conclude that equation (5) and equation (6) have no solution for any (x, y, z, w) ∈ U . Theorem 2. The equation 4x + 6y + 7z = w2 (7) has no solution for all (x, y, z, w) ∈ U . Proof. The proof of 2 follows from the footprint of 1 by using modulo 3, 4 and 16 but dividing the value z to two cases. Case 1 z = 0. Since w2 ̸≡ 2 (mod 3) but 4x + 6y + 1 ≡ 2 (mod 3) for all x ≥ 0 and y ≥ 1, the equation (7) has no solution. So, we remain to proof the case x ≥ 1 and y = 0, the equation (7) becomes to 4x + 2 = w2. By the fact that x ≥ 1, we have w is even and then w2 ≡ 0 (mod 4) that contradicts to w2 ≡ 4x + 2 ≡ 2 (mod 4). Case 2 z ≥ 1. Again w2 ̸≡ 2 (mod 3) and 4x + 6y + 7z ≡ 2 (mod 3) for any x ≥ 0 and y ≥ 1. So the equation (7) has no solution. Next, we have to consider only subcase x ≥ 1 and y = 0. If x = 1 then the equation (7) has also no solution by 1. Now, we focus on the equation 4x+1+7z = w2 for x ≥ 2. Since w2 ≡ 0, 1, 4, 9 (mod 16) and 4x + 1 + 7z ≡ { 2 (mod 16) if z is even, 8 (mod 16) if z is odd. we have 4x + 1 + 7z ̸≡ w2 (mod 16). Thus the equation (7) has no solution. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2070 Corollary 2. The equation 16x + 6y + 7z = w2 (8) has no solution for any (x, y, z, w) ∈ U . Proof. With the same trace of 1, we can conclude that equation (8) has no solution by using 2. Theorem 3. The equation 9x + 10y + 7z = w2 (9) has no solution for any (x, y, z, w) ∈ U − T , where T = {(0, 3, z, w)|z ≥ 2 and z is odd}. Proof. For proving this theorem, we still use modulo 3 and 16 that play the main role to verify the existence of its solution and also use modulo 5 and 10 in some subcaes of the variable z. Case 1 z = 0. Note that 9x + 10y + 1 ≡ { 2 (mod 3) if x ≥ 1 and y ≥ 0, 2 (mod 10) if x = 0 and y ≥ 1, and we know that neither w2 ̸≡ 2 (mod 3) nor w2 ̸≡ 2 (mod 10). Then it is clear that the equation (9) has no solution on U if z = 0. Case 2 z = 1. With the same fashion in Case 1, we note that 9x + 10y + 7 ≡ { 2 (mod 3) if x ≥ 1 and y ≥ 0, 3 (mod 5) if x = 0 and y ≥ 1. Since w2 ≡ 2 (mod 3) and w2 ̸≡ 3 (mod 5) the equation 9x + 10y + 7 = w2 has no solution on U if z = 1. Case 3 z ≥ 2. If x ≥ 1 and y ≥ 0 then it is easy to see that 9x + 10y + 7z = w2 has no solution by examing with modulo 3. Now, we remain to investigate the last subcase x = 0 and y ≥ 1. Then the equation (9) becomes 1 + 10y + 7z = w2. (10) Note that 1 + 7z ≡ { 2 (mod 16) if z is even, 8 (mod 16) if z is odd, and 10y ≡  10 (mod 16) if y = 1, 4 (mod 16) if y = 2, 8 (mod 16) if y = 3, 0 (mod 16) if y ≥ 4. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2071 So 1 + 10y + 7z ≡ { 2, 6, 10, 12 (mod 16) if z is even and y ≥ 1, 2, 8, 12 (mod 16) if z is odd, y ≥ 1 and y ̸= 3, but w2 ≡ 0, 1, 4, 9 (mod 16), this leads us to complete the proof that the equation (9) has no solution on U−T if z ≥ 2. From here we obtain the equation 9x+10y+7z = w2 has no solution for any (x, y, z, w) ∈ U − T . Theorem 4. The equation 6x + 10y + 7z = w2 (11) has no solution for any (x, y, z, w) ∈ U − T where T = {(0, 3, z, w)|z ≥ 2 and z is odd}. Proof. Note that, if x ≥ 1 and y ≥ 0 then 6x + 10y + 7z ≡ 2 (mod 3) for all z ≥ 0. From this fact and the fact that w2 ̸≡ 2 (mod 3), we conclude that the equation (11) has no solution for x ̸= 0. Now, it leads us to consider the equation (11) in the form 1 + 10y + 7z = w2 for any y ≥ 1, z and w are non-negative. For the case z = 0 or z = 1, we get 1 + 10y + 7z ≡ 2 or 3 (mod 5). but w2 ̸≡ 2, 3 (mod 5). So, we remain to examine the equation 1 + 10y + 7z = w2 in the case of y ≥ 1, z ≥ 2 and w ≥ 0.That is the same equation (10) in 3 and we immediately conclude that the equation (11) has no solution for all (x, y, z, w) ∈ U − T . 2.2. The exponential Diophantine equation ax + by + cz = w2 In this section, we present our results and discussion for ax + by + cz = w2, where all bases of the exponential are in terms of variables except 7 and 3. Theorem 5. The Diophantine equation ax + by + cz = w2 has no solution when a, b and c satisfy one of the following conditions: (i) a, b, c ≡ 1 (mod 4). (ii) a, b, c ≡ 1 (mod 5). (iii) a ≡ 0 (mod 4) and b, c ≡ 1 (mod 4). (iv) a ≡ 0 (mod 5) and b, c ≡ 1 (mod 5). (v) a ≡ 3 (mod 8) and b, c ≡ 1 (mod 8). (vi) a ≡ 1 (mod 8) and b, c ≡ 3 (mod 8). Proof. (i) Suppose that a, b, c ≡ 1 (mod 4). For any nonnegative integers x, y, z and w, it is obvious that ax + by + cz ≡ 3 (mod 4), but w2 ≡ 0, 1 (mod 4). This means that ax + by + cz ̸≡ w2 (mod 4). Hence, the Diophantine equation has no solution. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2072 (ii) Suppose that a, b, c ≡ 1 (mod 5). Given that w2 ≡ 0, 1, 4 (mod 5) and by the same trace of (i), we can conclude that the Diophantine equation has no solution. (iii) Suppose that a ≡ 0 (mod 4) and b, c ≡ 1 (mod 4). For any nonnegative integers x, y and z, we know that ax ≡ { 1 (mod 4) if x = 0 0 (mod 4) if x ≥ 1, and by, cz ≡ 1 (mod 4). Thus, ax + by + cz ≡ { 3 (mod 4) if x = 0 2 (mod 4) if x ≥ 1, for any nonnegative integers x, y and z. Since w2 ≡ 0, 1 (mod 4), the Diophantine equation has no solution. (iv) By the same trace of (i) and (ii), we can conclude that the Diophantine equation has no solution. (v) Suppose that a ≡ 3 (mod 8) and b, c ≡ 1 (mod 8). For any nonnegative integers x, y and z, we can see that ax ≡ { 1 (mod 8) if x is even 3 (mod 8) if x is odd, and by, cz ≡ 1 (mod 8). Thus, ax + by + cz ≡ { 3 (mod 8) if x is even 5 (mod 8) if x is odd, for any nonnegative integers x, y and z. Since w2 ≡ 0, 1, 4 (mod 8), the Diophantine equation has no solution. (vi) Suppose that a ≡ 1 (mod 8) and b, c ≡ 3 (mod 8). First, we note that by ≡ { 1 (mod 8) if x is even 3 (mod 8) if x is odd, cz ≡ { 1 (mod 8) if x is even 3 (mod 8) if x is odd, ax ≡ 1 (mod 8) and w2 ≡ 0, 1, 4 (mod 8) for any nonnegative integers w, x, y and z. Next, we separate the proof into four cases. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2073 Case 1. Here, y = 0 and z = 0. Since ax + by + cz ≡ 3 (mod 8) and w2 ̸≡ 3 (mod 8), the Diophantine equation has no solution. Case 2. Here, y = 0 and z > 0. Since ax + 1 + cz ≡ { 3 (mod 8) if x is even 5 (mod 8) if x is odd, and w2 ̸≡ 3, 5 (mod 8), the Diophantine equation has no solution. Case 3. Here, y > 0 and z = 0. The proof of this case is the same as in Case 2. Case 4. Here, y > 0 and z > 0. Then, by + cz ≡  6 (mod 8) if y and z are odd, 2 (mod 8) if y and z are even, 4 (mod 8) if otherwise. However, w2 ≡ 0, 1, 4 (mod 8), so, the Diophantine equation has no solution. Theorem 6. If a + 2 is a perfect square number such that a ≡ 2 (mod 36) and b, c ≡ 9 (mod 36), then the solutions of Diophantine equation ax + by + cz = w2 are (x, y, z, w) = (1, 0, 0, √ a+ 2). Proof. Note that a ≡ 2 (mod 4), b, c ≡ 1 (mod 4), a ≡ 2 (mod 9) and b, c ≡ 0 (mod 9). Since 1+by+cz ≡ 3 (mod 4) and w2 ≡ 0, 1 (mod 4), equation ax+by+cz = w2 has no solution for the case x = 0. If x ≥ 2, then ax + by + cz ≡ 2 (mod 4), which contradicts w2 ≡ 0, 1 (mod 4). Thus, in this case, x ≥ 2, and there is no solution. We now consider that for x = 1, y and z are nonzero at the same time. Because a+ by + cz ≡ { 2 (mod 9) if y, z ≥ 1, 3 (mod 9) if either y or z is zero, and w2 ≡ 0, 1, 4, 7 (mod 9), a solution still does not exist. For the last case, that is, x = 1, y = z = 0, we see that ax + by + cz = a+ 2 is a perfect square number. Hence, the solution of equation ax + by + cz = w2 is (1, 0, 0, √ a+ 2). Theorem 7. If b, c ≡ 5 (mod 20), then the solution of the Diophantine equation 2x+by+ cz = w2 is (x, y, z, w) = (1, 0, 0, 2). Proof. First, we note that b, c ≡ 1 (mod 4) and b, c ≡ 0 (mod 5) since b, c ≡ 5 (mod 20). If x = 0, then we odd value, which leads to w2 ≡ 1 (mod 4). However, 1 + by + cz ≡ 3 (mod 4), contradicts the equation 2x + by + cz = w2, which has a solution in this case. Moreover, for the case x ≥ 2, there is no solution for nonnegative integer (x, y, z, w) because of 2x + by + cz ≡ 2 (mod 4) and w2 ≡ 0, 1 (mod 4). Now, it remains to consider only the case x = 1. The result of 2 + by + cz ≡ { 3 (mod 5) if y or z are zero, 2 (mod 5) if y or z are positive, K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2074 and w2 ≡ 0, 1, 4 (mod 5), we conclude that 2+by+cz = w2 has no solution for y, z, w ∈ N0, and y and z are not all zero simultaneously. Hence, 2x+by+cz = w2 has only one solution (x, y, z, w) = (1, 0, 0, 2). Corollary 3. For any nonnegative integers a, b, c, x, y, z and w, if a + 2 is a perfect square number, such that a ≡ 2 (mod 20) and b, c ≡ 5 (mod 20), then the solutions of the Diophantine equation ax + by + cz = w2 are (x, y, z, w) = (1, 0, 0, √ a+ 2). Proof. The proof of corollary 3 follows from the same manner as 7. Code Listing 1: Python source code. import math import collections def is_perfact_square(n): i = 1 while i<=math.floor(n** 0.5): if i*i==n: return True i = i+1 return False def check_theorem(a,b,c): theorem_dict = { (3,4,7):"Theoremm 1", (9,4,7):"Corollary 1", (3,16 ,7):"Corollary 1", (9,16 ,7):"Corollary 1", (4,6,7):"Theoremm 2", (16 ,6,7):"Corollary 2", (9,10 ,7):"Theoremm 3", (6,10 ,7):"Theoremm 4" } list_of_thm = [] if (a,b,c) in theorem_dict.keys(): list_of_thm.append(theorem_dict[(a,b,c )]) if b%4==1 and c%4==1: if a%4==1: list_of_thm.append("Theoremm 5(i)") elif a%4==0: list_of_thm.append("Theoremm 5(iii)") if b%5==1 and c%5==1: if a%5==1: list_of_thm.append("Theoremm 5(ii)") elif a%5==0: list_of_thm.append("Theoremm 5(iv)") if a%8==3 and b%8==1 and c%8==1: list_of_thm.append("Theoremm 5(v)") if a%8==1 and b%8==3 and c%8==3: list_of_thm.append("Theoremm 5(vi)") if a==2 and b%20==5 and c%20==5: list_of_thm.append("Theoremm 7") if is_perfact_square(a+2): if a%36==2 and b%36==9 and c%36==9: list_of_thm.append("TTheoremm 6") if a%20==2 and b%20==5 and c%20==5: list_of_thm.append("Corollary 3") return list_of_thm if __name__ == "__main__": my_dict = dict() base_min , base_max = 2, 20 K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2075 for a in range(base_min , base_max + 1): for b in range(base_min , base_max + 1): for c in range(base_min , base_max + 1): if len(check_theorem(a,b,c))>0: L = [a,b,c] L.sort() my_dict[tuple(L)] = check_theorem(a,b,c) ordered_my_dict = collections.OrderedDict(sorted(my_dict.items())) print(f"No.\ tCase\tReference of Theorem") k = 1 for abc in ordered_my_dict: print(f"{k}\t{abc}\t {’,’.join(ordered_my_dict[abc])}") k=k+1 2.3. Conclusion In summary, this article began by extending the number of exponential bases in Diophantine equations from two—represented as ax + by = z2—to three, represented as ax + by + cz = w2. Most of our results focus on situations where the equation ax + by + cz = w2 has no solution. Some of these cases relate to conditions where (a, b) is in the set containing pairs like (3, 4), (9, 4), (3, 16), (9, 16), (4, 6), (16, 6), (9, 10), and (6, 10), along with c being equal to 7. Other results are categorized as follows: (i) a, b, c ≡ 1 (mod 4). (ii) a, b, c ≡ 1 (mod 5). (iii) a ≡ 0 (mod 4) and b, c ≡ 1 (mod 4). (iv) a ≡ 0 (mod 5) and b, c ≡ 1 (mod 5). (v) a ≡ 3 (mod 8) and b, c ≡ 1 (mod 8). (vi) a ≡ 1 (mod 8) and b, c ≡ 3 (mod 8). Some equations of the form ax + by + cz = w2 do have solutions, as discussed in Theorem 6, 7, and Corollary 3. Our research yielded 135 equations that have been clarified from a total of 1,330 equations, assuming we limit all variables a, b, and c to range from 2 to 20. We used Python code, as shown in Code Listing 1, to count the number of equations that satisfy our theorems and corollaries. Specifically, we offer code that counts all equations of the form ax + by + cz = w2 that are related to our results in the theorems and corollaries. The accompanying table lists the exponential Diophantine equations with variables a, b, and c limited to the range from 2 to 20. For future work, we aim to combine this research with a new form of Diophantine equation, as seen in reference [8], or we will expand our research using other techniques, such as transforming the equation into an elliptic curve[12], extending into the quadratic field [see details 11], or using Legendre symbols with related theorems,[see also 15, 24]. K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2076 Table 2: The list of the equations ax + by + cz = w2 where 2 ≤ a ≤ b ≤ c ≤ 20 and all variables a, b and c are satisfied our results. No. ax + by + cz = w2 Result Ref. of Theorem 1 2x + 5y + 5z = w2 (1, 0, 0, 2) 7, Corollary 3 2 2x + 9y + 9z = w2 (1, 0, 0, 2) 6 3 3x + 3y + 9z = w2 No solution 5(vi) 4 3x + 3y + 17z = w2 No solution 5(vi) 5 3x + 4y + 7z = w2 No solution 1 6 3x + 7y + 16z = w2 No solution Corollary 1 7 3x + 9y + 9z = w2 No solution 5(v) 8 3x + 9y + 11z = w2 No solution 5(vi) 9 3x + 9y + 17z = w2 No solution 5(v) 10 3x + 9y + 19z = w2 No solution 5(vi) 11 3x + 11y + 17z = w2 No solution 5(vi) 12 3x + 17y + 17z = w2 No solution 5(v) 13 3x + 17y + 19z = w2 No solution 5(vi) 14 4x + 5y + 5z = w2 No solution 5(iii) 15 4x + 5y + 9z = w2 No solution 5(iii) 16 4x + 5y + 13z = w2 No solution 5(iii) 17 4x + 5y + 17z = w2 No solution 5(iii) 18 4x + 6y + 7z = w2 No solution 2 19 4x + 7y + 9z = w2 No solution Corollary 1 20 4x + 9y + 9z = w2 No solution 5(iii) 21 4x + 9y + 13z = w2 No solution 5(iii) 22 4x + 9y + 17z = w2 No solution 5(iii) 23 4x + 13y + 13z = w2 No solution 5(iii) 24 4x + 13y + 17z = w2 No solution 5(iii) 25 4x + 17y + 17z = w2 No solution 5(iii) 26 5x + 5y + 5z = w2 No solution 5(i) 27 5x + 5y + 8z = w2 No solution 5(iii) 28 5x + 5y + 9z = w2 No solution 5(i) 29 5x + 5y + 12z = w2 No solution 5(iii) 30 5x + 5y + 13z = w2 No solution 5(i) 31 5x + 5y + 16z = w2 No solution 5(iii) 32 5x + 5y + 17z = w2 No solution 5(i) 33 5x + 5y + 20z = w2 No solution 5(iii) 34 5x + 6y + 6z = w2 No solution 5(iv) 35 5x + 6y + 11z = w2 No solution 5(iv) K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2077 Table 3: The list of the equations ax + by + cz = w2 where 2 ≤ a ≤ b ≤ c ≤ 20 and all variables a, b and c are satisfied our results. No. ax + by + cz = w2 Result Ref. of Theorem 36 5x + 6y + 16z = w2 No solution 5(iv) 37 5x + 8y + 9z = w2 No solution 5(iii) 38 5x + 8y + 13z = w2 No solution 5(iii) 39 5x + 8y + 17z = w2 No solution 5(iii) 40 5x + 9y + 9z = w2 No solution 5(i) 41 5x + 9y + 12z = w2 No solution 5(iii) 42 5x + 9y + 13z = w2 No solution 5(i) 43 5x + 9y + 16z = w2 No solution 5(iii) 44 5x + 9y + 17z = w2 No solution 5(i) 45 5x + 9y + 20z = w2 No solution 5(iii) 46 5x + 11y + 11z = w2 No solution 5(iv) 47 5x + 11y + 16z = w2 No solution 5(iv) 48 5x + 12y + 13z = w2 No solution 5(iii) 49 5x + 12y + 17z = w2 No solution 5(iii) 50 5x + 13y + 13z = w2 No solution 5(i) 51 5x + 13y + 16z = w2 No solution 5(iii) 52 5x + 13y + 17z = w2 No solution 5(i) 53 5x + 13y + 20z = w2 No solution 5(iii) 54 5x + 16y + 16z = w2 No solution 5(iv) 55 5x + 16y + 17z = w2 No solution 5(iii) 56 5x + 17y + 17z = w2 No solution 5(i) 57 5x + 17y + 20z = w2 No solution 5(iii) 58 6x + 6y + 6z = w2 No solution 5(ii) 59 6x + 6y + 10z = w2 No solution 5(iv) 60 6x + 6y + 11z = w2 No solution 5(ii) 61 6x + 6y + 15z = w2 No solution 5(iv) 62 6x + 6y + 16z = w2 No solution 5(ii) 63 6x + 6y + 20z = w2 No solution 5(iv) 64 6x + 7y + 10z = w2 No solution 4 65 6x + 7y + 16z = w2 No solution Corollary 2 66 6x + 10y + 11z = w2 No solution 5(iv) 67 6x + 10y + 16z = w2 No solution 5(iv) 68 6x + 11y + 11z = w2 No solution 5(ii) 69 6x + 11y + 15z = w2 No solution 5(iv) 70 6x + 11y + 16z = w2 No solution 5(ii) 71 6x + 11y + 20z = w2 No solution 5(iv) 72 6x + 15y + 16z = w2 No solution 5(iv) 73 6x + 16y + 16z = w2 No solution 5(ii) K. Laipaporn, S. Kaewchay, A. Karnbanjong / Eur. J. Pure Appl. Math, 16 (4) (2023), 2066-2081 2078 Table 4: The list of the equations ax + by + cz = w2 where 2 ≤ a ≤ b ≤ c ≤ 20 and all variables a, b and c are satisfied our results. No. ax + by + cz = w2 Result Ref. of Theorem 74 6x + 16y + 20z = w2 No solution 5(iv) 75 7x + 9y + 10z = w2 No solution 3 76 7x + 9y + 16z = w2 No solution Corollary 1 77 8x + 9y + 9z = w2 No solution 5(iii) 78 8x + 9y + 13z = w2 No solution 5(iii) 79 8x + 9y + 17z = w2 No solution 5(iii) 80 8x + 13y + 13z = w2 No solution 5(iii) 81 8x + 13y + 17z = w2 No solution 5(iii) 82 8x + 17y + 17z = w2 No solution 5(iii) 83 9x + 9y + 9z = w2 No solution 5(i) 84 9x + 9y + 11z = w2 No solution 5(v) 85 9x + 9y + 12z = w2 No solution 5(iii) 86 9x + 9y + 13z = w2 No solution 5(i) 87 9x + 9y + 16z = w2 No solution 5(iii) 88 9x + 9y + 17z = w2 No solution 5(i) 89 9x + 9y + 19z = w2 No solution 5(v) 90 9x + 9y + 20z = w2 No solution 5(iii) 91 9x + 11y + 11z = w2 No solution 5(vi) 92 9x + 11y + 17z = w2 No solution 5(v) 93 9x + 11y + 19z = w2 No solution 5(vi) 94 9x + 12y + 13z = w2 No solution 5(iii) 95 9x + 12y + 17z = w2 No solution 5(iii) 96 9x + 13y + 13z = w2 No solution 5(i) 97 9x + 13y + 16z = w2 No solution 5(iii) 98 9x + 13y + 17z = w2 No solution 5(i) 99 9x + 13y + 20z = w2 No solution 5(iii) 100 9x + 16y + 17z = w2 No solution 5(iii) 101 9x + 17y + 17z = w2 No solution 5(i) 102 9x + 17y + 19z = w2 No solution 5(v) 103 9x + 17y + 20z = w2 No solution 5(iii) 104 9x + 19y + 19z = w2 No solution 5(vi) 105 10x + 11y + 11z = w2 No solution 5(iv) 106 10x + 11y + 16z = w2 No solution 5(iv) 107 10x + 16y + 16z = w2 No solution 5(iv) 108 11x + 11y + 11z = w2 No solution 5(ii) 109 11x + 11y + 15z = w2 No solution 5(iv) 110 11x + 11y + 16z = w2 No solution 5(ii) REFERENCES 2079 Table 5: The list of the equations ax + by + cz = w2 where 2 ≤ a ≤ b ≤ c ≤ 20 and all variables a, b and c are satisfied our results. No. ax + by + cz = w2 Result Ref. of Theorem 111 11x + 11y + 17z = w2 No solution 5(vi) 112 11x + 11y + 20z = w2 No solution 5(iv) 113 11x + 15y + 16z = w2 No solution 5(iv) 114 11x + 16y + 16z = w2 No solution 5(ii) 115 11x + 16y + 20z = w2 No solution 5(iv) 116 11x + 17y + 17z = w2 No solution 5(v) 117 11x + 17y + 19z = w2 No solution 5(vi) 118 12x + 13y + 13z = w2 No solution 5(iii) 119 12x + 13y + 17z = w2 No solution 5(iii) 120 12x + 17y + 17z = w2 No solution 5(iii) 121 13x + 13y + 13z = w2 No solution 5(i) 122 13x + 13y + 16z = w2 No solution 5(iii) 123 13x + 13y + 17z = w2 No solution 5(i) 124 13x + 13y + 20z = w2 No solution 5(iii) 125 13x + 16y + 17z = w2 No solution 5(iii) 126 13x + 17y + 17z = w2 No solution 5(i) 127 13x + 17y + 20z = w2 No solution 5(iii) 128 15x + 16y + 16z = w2 No solution 5(iv) 129 16x + 16y + 16z = w2 No solution 5(ii) 130 16x + 16y + 20z = w2 No solution 5(iv) 131 16x + 17y + 17z = w2 No solution 5(iii) 132 17x + 17y + 17z = w2 No solution 5(i) 133 17x + 17y + 19z = w2 No solution 5(v) 134 17x + 17y + 20z = w2 No solution 5(iii) 135 17x + 19y + 19z = w2 No solution 5(vi) Acknowledgements The authors would like to thank Pimchanok Pimton and Jutarat Wiriyadamrikul for her useful suggestions. 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