scholarly journals The Exponential Diophantine Equation4m2+1x+5m2-1y=(3m)z

2014 ◽  
Vol 2014 ◽  
pp. 1-5 ◽  
Author(s):  
Juanli Su ◽  
Xiaoxue Li
Keyword(s):  
Positive Integer ◽  
Diophantine Equations ◽  
Integer Solution ◽  
Lucas Numbers ◽  
Positive Integer Solution ◽  
Exponential Diophantine Equations ◽  
Primitive Divisors

Letmbe a positive integer. In this paper, using some properties of exponential diophantine equations and some results on the existence of primitive divisors of Lucas numbers, we prove that ifm>90and3|m, then the equation4m2+1x + 5m2-1y=(3m)zhas only the positive integer solution(x,y,z)=(1,1,2).

The exponential Diophantine equation x2 + (3n 2 + 1)y = (4n 2 + 1)z

2014 ◽  
Vol 64 (5) ◽  
Author(s):  
Jianping Wang ◽  
Tingting Wang ◽  
Wenpeng Zhang
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Diophantine Equations ◽  
Exponential Diophantine Equation ◽  
Lucas Numbers ◽  
Exponential Diophantine Equations ◽  
Primitive Divisors ◽  
Integer Solutions

AbstractLet n be a positive integer. In this paper, using the results on the existence of primitive divisors of Lucas numbers and some properties of quadratic and exponential diophantine equations, we prove that if n ≡ 3 (mod 6), then the equation x 2 + (3n 2 + 1)y = (4n 2 + 1)z has only the positive integer solutions (x, y, z) = (n, 1, 1) and (8n 3 + 3n, 1, 3).

scholarly journals On the Diophantine equations z2 = k(k2 + 3) and z2 = k(k2 + 12)

2021 ◽  
Vol 45 (1) ◽  
pp. 127-129
Author(s):  
Shah Mohammad Shahidul Islam ◽  
Abdullah Al Kafi Majumdar
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Analytical Method ◽  
Diophantine Equations ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Integral Solutions

This paper provides an analytical method of finding all the (positive, integral) solutions of the Diophantine equation z2 = k(k2+3). We also prove analytically that the Diophantine equation z2 = k(k2+12) has no positive, integer solution. J. Bangladesh Acad. Sci. 45(1); 127-129: June 2021

A note on the ternary purely exponential diophantine equation Ax + By = Cz with A + B = C2

2020 ◽  
Vol 57 (2) ◽  
pp. 200-206
Author(s):  
Elif kizildere ◽  
Maohua le ◽  
Gökhan Soydan
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Primitive Divisors ◽  
Lehmer Numbers ◽  
Positive Integers

AbstractLet l,m,r be fixed positive integers such that 2| l, 3lm, l > r and 3 | r. In this paper, using the BHV theorem on the existence of primitive divisors of Lehmer numbers, we prove that if min{rlm2 − 1,(l − r)lm2 + 1} > 30, then the equation (rlm2 − 1)x + ((l − r)lm2 + 1)y = (lm)z has only the positive integer solution (x,y,z) = (1,1,2).

Complete solutions of the simultaneous Pell’s equations x2 − (a2 − 1)y2 = 1 and y2 − pz2 = 1

2019 ◽  
Vol 15 (05) ◽  
pp. 1069-1074 ◽  
Author(s):  
Hai Yang ◽  
Ruiqin Fu
Keyword(s):  
Positive Integer ◽  
Diophantine Equations ◽  
Integer Solution ◽  
System Of Equations ◽  
Positive Integer Solution ◽  
Elementary Methods ◽  
Integer Solutions ◽  
Positive Integers ◽  
Complete Solutions

Let [Formula: see text] be a positive integer with [Formula: see text], and let [Formula: see text] be an odd prime. In this paper, by using certain properties of Pell’s equations and quartic diophantine equations with some elementary methods, we prove that the system of equations [Formula: see text] [Formula: see text] and [Formula: see text] has positive integer solutions [Formula: see text] if and only if [Formula: see text] and [Formula: see text] satisfy [Formula: see text] and [Formula: see text], where [Formula: see text], [Formula: see text] and [Formula: see text] are positive integers. Further, if the above condition is satisfied, then [Formula: see text] has only the positive integer solution [Formula: see text]. By the above result, we can obtain the following corollaries immediately. (i) If [Formula: see text] or [Formula: see text], then [Formula: see text] has no positive integer solutions [Formula: see text]. (ii) For [Formula: see text], [Formula: see text] has only the positive integer solutions [Formula: see text], [Formula: see text], [Formula: see text], [Formula: see text], [Formula: see text] and [Formula: see text].

A Note on the Exponential Diophantine Equation x2+q2m=2yp

2012 ◽  
Vol 241-244 ◽  
pp. 2650-2653
Author(s):  
Jian Ping Wang
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Primitive Divisors ◽  
Lehmer Numbers

ln this paper, using a deep result on the existence of primitive divisors of Lehmer numbers given by Y. Bilu, G. Hanrot and P. M. Voutier, we prove that the equation has no positive integer solution (x, y, m, p, q), where p and q are odd primes with p>3, gcd(x, y)=1 and y is not the sum of two consecutive squares.

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

2020 ◽  
Vol 55 (2) ◽  
pp. 195-201
Author(s):  
Maohua Le ◽  
◽  
Gökhan Soydan ◽  
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Upper Bound ◽  
Diophantine Equations ◽  
Exponential Diophantine Equation ◽  
Positive Integer Solution ◽  
Exponential Diophantine Equations ◽  
Integer Solutions ◽  
Positive Integers ◽  
Upper Bound For Solutions

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 >B3/8, then the equation (A2 n)x + (B2 n)y = ((A2 + B2)n)z has no positive integer solutions (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 some existing 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 integer solution (x,y,z)=(1,1,1).

scholarly journals On the Diophantine Equation x2 – kxy + ky2 + ly = 0, l = 2n

2017 ◽  
Vol 55 (1) ◽  
pp. 115-118
Author(s):  
Sukrawan Mavecha
Keyword(s):  
Positive Integer ◽  
Diophantine Equation ◽  
Integer Solution ◽  
Positive Integer Solution

AbstractWe consider the Diophantine equation x2-kxy+ky2+ ly = 0 for l = 2nand determine for which values of the odd integer k, it has a positive integer solution x and y.

scholarly journals On the generalized Ramanujan-Nagell equation $ x^2+(2k-1)^y = k^z $ with $ k\equiv 3 $ (mod 4)

2021 ◽  
Vol 6 (10) ◽  
pp. 10596-10601
Author(s):  
Yahui Yu ◽  
◽  
Jiayuan Hu ◽  
Keyword(s):  
Positive Integer ◽  
Unsolved Problem ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Elementary Methods ◽  
Fixed Positive Integer ◽  
Almost All ◽  
Terai’S Conjecture ◽  
Positive Integers

<abstract><p>Let $ k $ be a fixed positive integer with $ k &gt; 1 $. In 2014, N. Terai <sup>[<xref ref-type="bibr" rid="b6">6</xref>]</sup> conjectured that the equation $ x^2+(2k-1)^y = k^z $ has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. This is still an unsolved problem as yet. For any positive integer $ n $, let $ Q(n) $ denote the squarefree part of $ n $. In this paper, using some elementary methods, we prove that if $ k\equiv 3 $ (mod 4) and $ Q(k-1)\ge 2.11 $ log $ k $, then the equation has only the positive integer solution $ (x, y, z) = (k-1, 1, 2) $. It can thus be seen that Terai's conjecture is true for almost all positive integers $ k $ with $ k\equiv 3 $(mod 4).</p></abstract>

scholarly journals A note on Jeśmanowicz’ conjecture for non-primitive Pythagorean triples

2021 ◽  
Vol 5 (1) ◽  
pp. 115-127
Author(s):  
Van Thien Nguyen ◽  
◽  
Viet Kh. Nguyen ◽  
Pham Hung Quy ◽  
◽  
...  
Keyword(s):  
Positive Integer ◽  
Simple Proof ◽  
Diophantine Equation ◽  
Necessary Conditions ◽  
Integer Solution ◽  
Positive Integer Solution ◽  
Pythagorean Triples ◽  
Pythagorean Triple

Let \((a, b, c)\) be a primitive Pythagorean triple parameterized as \(a=u^2-v^2, b=2uv, c=u^2+v^2\), where \(u>v>0\) are co-prime and not of the same parity. In 1956, L. Jesmanowicz conjectured that for any positive integer \(n\), the Diophantine equation \((an)^x+(bn)^y=(cn)^z\) has only the positive integer solution \((x,y,z)=(2,2,2)\). In this connection we call a positive integer solution \((x,y,z)\ne (2,2,2)\) with \(n>1\) exceptional. In 1999 M.-H. Le gave necessary conditions for the existence of exceptional solutions which were refined recently by H. Yang and R.-Q. Fu. In this paper we give a unified simple proof of the theorem of Le-Yang-Fu. Next we give necessary conditions for the existence of exceptional solutions in the case \(v=2,\ u\) is an odd prime. As an application we show the truth of the Jesmanowicz conjecture for all prime values \(u < 100\).

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