Let (A) $B$ be positive integers such that $min{A,B}>1$, $gcd(A,B) = 1$ and $2|B.$ In this paper, using an upper bound for solutions of ternary purely exponential Diophantine equations due to R. Scott and R. Styer, we prove that, for any positive integer $n$, if $A >B^3/8$, then the equation $(A^2 n)^x + (B^2 n)^y = ((A^2 + B^2)n)^z$ has no positive integer solutions $(x,y,z)$ with $x > z > y$; if $B>A^3/6$, then it has no solutions $(x,y,z)$ with $y>z>x$. Thus, combining the above conclusion with some existing results, we can deduce that, for any positive integer $n$, if $Bequiv 2 pmod{4}$ and $A >B^3/8$, then this equation has only the positive integer solution $(x,y,z)=(1,1,1)$

Gökhan Soydan

Maohua Le

English

Hrčak - Portal of scientific journals of Croatia

GLASNIK MATEMATIČKIVol. 55(75)(2020), 195 – 201A NOTE ON THE EXPONENTIAL DIOPHANTINEEQUATION (A2n)x + (B2n)y = ((A2 +B2)n)zMaohua Le and Gökhan SoydanLingnan Normal College, P. R. China and Bursa Uludağ University, TurkeyAbstract. Let A, B be positive integers such that min{A,B} > 1,gcd(A,B) = 1 and 2|B. In this paper, using an upper bound for solutionsof ternary purely exponential Diophantine equations due to R. Scott andR. Styer, we prove that, for any positive integer n, if A > B3/8, thenthe equation (A2n)x + (B2n)y = ((A2 + B2)n)z has no positive integersolutions (x, y, z) with x > z > y; if B > A3/6, then it has no solutions(x, y, z) with y > z > x. Thus, combining the above conclusion with someexisting results, we can deduce that, for any positive integer n, if B ≡ 2(mod 4) and A > B3/8, then this equation has only the positive integersolution (x, y, z) = (1, 1, 1).1. IntroductionLet N be the set of all positive integers. Let n be a positive integer, and leta, b be positive integers such that min{a, b} > 1 and gcd(a, b) = 1. Recently,P.-Z. Yuan and Q. Han ([9]) proposed the following conjecture:Conjecture 1.1. For any n, if min{a, b} ≥ 4, then the equation(1.1) (an)x + (bn)y = ((a+ b)n)z, x, y, z ∈ Nhas only the solution (x, y, z) = (1, 1, 1).Since Conjecture 1.1 is much broader than Jeśmanowicz’ conjecture con-cerning Pythagorean triples (see [2] and the survey paper on the conjecturesof Jeśmanowicz and Terai which was published by G. Soydan, M. Demirci, I.N. Cangül and A. Togbé, ([5])), it is unlikely to be solved in the short term.There are only a few scattered results on Conjecture 1.1 at present (see [6]).2020 Mathematics Subject Classification. 11D61.Key words and phrases. Ternary purely exponential Diophantine equation.195196 M.-H. LE AND G. SOYDANLet A,B be positive integers such that min{A,B} > 1, gcd(A,B) = 1and 2|B. In the same paper, P.-Z. Yuan and Q. Han ([9]) deal with thesolutions (x, y, z) of (1.1) for the case that (a, b) = (A2, B2). Then (1.1) canbe rewritten as(1.2) (A2n)x + (B2n)y = ((A2 +B2)n)z , x, y, z ∈ N.For this equation, they proved that, for any n > 1, if B ≡ 2 (mod 4), then(1.2) has no solutions (x, y, z) with y > z > x; in particular, if B = 2, thenConjecture 1.1 is true for any n.In this paper, using an upper bound for solutions of ternary purely expo-nential Diophantine equations due to R. Scott and R. Styer ([4]), we prove ageneral result as follows:Theorem 1.2. For any n, if A > B3/8, then (1.2) has no solutions(x, y, z) with x > z > y; if B > A3/6, then (1.2) has no solutions (x, y, z)with y > z > x.Thus, combining Theorem 1.2 with the above mentioned results of [9], wecan deduce the following corollary:Corollary 1.3. For any n, if B ≡ 2 (mod 4) and A > B3/8, then (1.2)has only the solution (x, y, z) = (1, 1, 1).This implies that, for any fixed B with B ≡ 2 (mod 4), then Conjecture1.1 is true for (a, b) = (A2, B2) except for finitely many values of A.2. LemmasFor any positive integer m, let rad(m) denote the product of all distinctprime divisors of m, and let rad(1) = 1. Obviously, rad(m) is equal to thelargest squarefree divisor of m.Lemma 2.1 ([6, Theorem 1.1], [9, Proposition 3.1]). Assume n > 1in (1.1) and let (x, y, z) be a solution of (1.1) with (x, y, z) 6= (1, 1, 1). Ifmin{a, b} > 2, then eitherx > z > y, rad(n) | b, b = b1b2, by1 = nz−y, b1, b2 ∈ N, b1 > 1, gcd(b1, b2) = 1ory > z > x, rad(n) | a, a = a1a2, ax1 = nz−x,a1, a2 ∈ N, a1 > 1, gcd(a1, a2) = 1.Remark 2.2. Because when min{a, b} = 2, there might be a solution(x, y, z) to (1.1) with y > z = x (see [1, 3, 7, 8]), the condition min{a, b} > 2in Lemma 2.1 is necessary.Lemma 2.3. If B ≡ 2 (mod 4) and (x, y, z) 6= (1, 1, 1) is a solution to(1.2), then x > z > y.AN EXPONENTIAL DIOPHANTINE EQUATION 197Proof. When n > 1, Lemma 2.1 shows that Lemma 2.3 is equivalent to[9, Theorem 1.3].So we can assume n = 1. Suppose (1.2) has a solution (x, y, z) 6= (1, 1, 1),so that(2.1) A2x +B2y = (A2 +B2)z .Clearly (1, 1, 1) is the only possible solution to (1.2) with z = 1, so in (2.1)we have(2.2) z > 1.Since z ≥ 2, if max{x, y} ≤ z, then we have(A2 +B2)z = A2x +B2y ≤ A2z +B2z < (A2 +B2)z ,a contradiction from which we get(2.3) z < max{x, y}.Next, we show that y < z using a straightforward approach which works whenn = 1 (as well as when n > 1 as in [9]).It is a familiar elementary result (see, for example, [9, Lemma 3.2]) that,if (2.1) holds, there are positive integers u and v such that 2 | v, u2 + v2 =A2 +B2, (u, v) = 1, and±(u± v√−1)z = Ax +By√−1,with(2.4) ν2(v) + ν2(z) = ν2(By)where, for any positive integer m, 2ν2(m) || m. 2 || B, so A2+B2 ≡ 5 (mod 8),so 2 || v, so that (2.4) becomes1 + ν2(z) = y,so that(2.5) z ≥ 2y−1 ≥ yand z = y implies y ≤ 2. Since z > 1 and y = z = 2 impliesA2x = A2(A2 + 2B2)which contradicts (A, 2B) = 1, we must have(2.6) y < z.(2.3) and (2.6) combine to give y < z < x.Lemma 2.4 ([9, Theorem 1.4]). For any n, if B = 2, then (1.2) has onlythe solution (x, y, z) = (1, 1, 1).198 M.-H. LE AND G. SOYDANLemma 2.5 ([4, Theorem 3]). Let G,H,K be fixed positive integers withmin{G,H,K} > 1, gcd(G,H) = 1 and 2 ∤ K. Further, let PQ be the largestsquarefree divisor of GH, with P and Q chosen so that (GH/P )1/2 is aninteger. If there exists a positive integer Z such that G + H = KZ , then Zsatisfies(2.7) Z≤ 12Q, if P = 1,≤ 12(Q + 1), if P = 2,<12P 1/2Q logP, if P ≥ 3.Lemma 2.6. Under the assumptions of Lemma 2.5, we have(2.8) Z ≤ 12PQ.Proof. Obviously, by (2.7), (2.8) holds for P ≤ 2. Let(2.9) f(t) =log tt1/2, t ≥ 3.Then we have(2.10) f ′(t) =2− log t2t3/2, t ≥ 3,where f ′(t) is the derivative of f(t). We see from (2.9) and (2.10) that f(e2) =2/e is the maximum value of f(t). Therefore, if P ≥ 3, then from (2.7) and(2.9) we getZ <12P 1/2Q logP =(12PQ)(logPP 1/2)=(12PQ)(f(P )) ≤(12PQ)(2e)<12PQ.This implies that (2.8) holds for P ≥ 3. The lemma is proved.Lemma 2.7. For any n, the solutions (x, y, z) of (1.2) satisfy z ≤ AB/2.Proof. Since AB/2 ≥ 3, the lemma holds for (x, y, z) = (1, 1, 1). Wenow assume that (x, y, z) is a solution of (1.2) with (x, y, z) 6= (1, 1, 1). Then,by Lemma 2.1, we have either x > z > y or y > z > x.Since min{A2, B2} ≥ 4, by Lemma 2.1, if x > z > y, then we have(2.11) B = B1B2, B1, B2 ∈ N, gcd(B1, B2) = 1,(2.12) B2y1 = nz−yand(2.13) A2xnx−z +B2y2 = (A2 +B2)z .AN EXPONENTIAL DIOPHANTINE EQUATION 199Take G = A2xnx−z, H = B2y2 , K = A2 + B2 and Z = z. Let PQ be thelargest squarefree divisor of GH . Since gcd(A,B) = 1, by (2.11) and (2.12),we have(2.14)PQ = rad(GH) = rad(A2xnx−z) · rad(B2y2 )= rad(AB1) · rad(B2) = rad(AB) ≤ AB.Therefore, applying Lemma 2.6 to (2.13), we get from (2.14) that(2.15) z ≤ PQ2≤ AB2.Similarly, if y > z > x, then we have(2.16) A = A1A2, A1, A2 ∈ N, gcd(A1, A2) = 1,(2.17) A2x1 = nz−xand(2.18) A2x2 +B2yny−z = (A2 +B2)z .Take G = A2x2 , H = B2yny−z, K = A2+B2 and Z = z. By (2.16) and (2.17),we have(2.19)PQ = rad(GH) = rad(A2x2 ) · rad(B2yny−z)= rad(A2) · rad(BA1) = rad(AB) ≤ AB,where PQ is the largest squarefree divisor of GH . Therefore, applying Lemma2.6 to (2.18), we see from (2.19) that z satisfies (2.15). Thus, the lemma isproved.3. ProofsProof of Theorem 1.2. By Lemma 2.4, the theorem holds for B = 2.We may therefore assume that B ≥ 4.We now prove the first part of the theorem. Since 2 ∤ A and A > B3/8,we have A ≥ 9. Let (x, y, z) be a solution of (1.2) with x > z > y. By (2.13),we have A2xnx−z < (A2+B2)z , whence we get (A2n)x−z < (1+B2/A2)z and(3.1) log(A2n) ≤ (x− z) log(A2n) < z log(1 +B2A2).Since log(1 + t) < t for any t > 0, by (3.1), we have(3.2)A2B2log(A2n) < z.On the other hand, by Lemma 2.7, we have z ≤ AB/2. Hence, by (3.2),we get(3.3)A2B2log(A2n) <AB2.200 M.-H. LE AND G. SOYDANFurther, since A > B3/8, we see from (3.3) that(3.4) log(A2n) < 4.But, since A ≥ 9 and n ≥ 1, (3.4) is false. Therefore, the first part of thetheorem is proved.Using the same method as before, we can easily prove the second partof the theorem. Since 2 ∤ A and B > A3/6, we have A ≥ 3 and B ≥ 6.Let (x, y, z) be a solution of (1.2) with y > z > x. By (2.18), we haveB2yny−z < (A2 +B2)z , whence we get(3.5)B2A2log(B2n) ≤ B2A2(y − z) log(B2n) < z.Further, by Lemma 2.7, we have z ≤ AB/2. Hence, by (3.5), we get(3.6)B2A2log(B2n) <AB2.Furthermore, since B > A3/6, we see from (3.6) that(3.7) log(B2n) < 3.But, since B ≥ 6 and n ≥ 1, (3.7) is false. Thus, the second part of thetheorem is proved. The proof is complete.Proof of Corollary 1.3. Combining Theorem 1.2 with Lemma 2.3,we obtain the corollary immediately.Acknowledgements.The authors are grateful for the referees for carefully reading our manu-script and for giving such constructive comments which substantially helpedimproving the presentation of the paper. We would like to thank ProfessorsReese Scott and Robert Styer for reading the previous version of this man-uscript carefully and giving valuable advice. Especially thanks to ProfessorRobert Styer for verifying some details of the previous version of this paperand providing other technical assistance. The second author was supportedby TÜBİTAK (the Scientific and Technological Research Council of Turkey)under Project No: 117F287.References[1] W.-J. Guan and S. Che, On the Diophantine equation 2yny−x = (b + 2)x − bx, J.Northwest Univ. Nat. Sci. 44 (2014), 534–536. (in Chinese)[2] L. Jeśmanowicz, Several remarks on Pythagorean numbers, Wiadom. Math. 1 (1955/56),196–202.[3] M.-H. Le, R. Scott and R. Styer, A survey on the ternary purely exponential Diophantineequation ax + by = cz , Surv. Math. Appl. 14 (2019), 109–140.[4] R. Scott and R. Styer, On px − qy = c and related three term exponential Diophantineequations with prime bases, J. Number Theory 105 (2004), 212–234.AN EXPONENTIAL DIOPHANTINE EQUATION 201[5] G. Soydan, M. Demirci, I. N. Cangül and A. Togbé, On the conjecture of Jesmanowicz,Int. J. Appl. Math. Stat. 56 (2017), 46–72.[6] C.-F. Sun and M. Tang, On the Diophantine equation (an)x + (bn)y = (cn)z , ChineseAnn. Math. Ser. A 39 (2018), 87–94. (in Chinese)[7] M. Tang and Q.-H. Yang, The Diophantine equation (an)x+(2n)y = ((b+2)n)z , Colloq.Math. 132 (2013), 95–100.[8] Y.-H. Yu and Z.-P. Li, The exceptional solutions of the exponential Diophantine equa-tion (bn)x + (2n)y = ((b+ 2)n)z , Math. Pract. Theo. 44 (2014), 290–293. (in Chinese)[9] P.-Z. Yuan and Q. Han, Jeśmanowicz’ conjecture and related equations, Acta Arith.184 (2018), 37–49.M.-H. LeInstitute of Mathematics, Lingnan Normal CollegeGuangdong, 524048 ZhangjiangChinaE-mail : lemaohua2008@163.comG. SoydanDepartment of MathematicsBursa Uludağ University16059 BursaTurkeyE-mail : gsoydan@uludag.edu.trReceived : 8.12.2019.Revised : 13.5.2020.

A note on the exponential Diophantine equation ((A^2n)^x+(B^2n)^y=((A^2+B^2)n)^z)

Abstract

Similar works

Full text

Available Versions

Hrčak - Portal of scientific journals of Croatia