scholarly journals The Exponential Diophantine Equation2x+by=cz

2014 ◽  
Vol 2014 ◽  
pp. 1-3 ◽  
Author(s):  
Yahui Yu ◽  
Xiaoxue Li

Letbandcbe fixed coprime odd positive integers withmin{b,c}>1. In this paper, a classification of all positive integer solutions(x,y,z)of the equation2x+by=czis given. Further, by an elementary approach, we prove that ifc=b+2, then the equation has only the positive integer solution(x,y,z)=(1,1,1), except for(b,x,y,z)=(89,13,1,2)and(2r-1,r+2,2,2), whereris a positive integer withr≥2.

2010 ◽  
Vol 107 (2) ◽  
pp. 161
Author(s):  
Bo He ◽  
Alain Togbé ◽  
Shichun Yang

Let $a,b,$ and $c$ be positive integers. We show that if $(a,b) =(N^k-1,N)$, where $N,k\geq 2$, then there is at most one positive integer solution $(x,y)$ to the exponential Diophantine equation $|a^x-b^y|=c$, unless $(N,k)=(2,2)$. Combining this with results of Bennett [3] and the first author [6], we stated all cases for which the equation $|(N^k \pm 1)^x - N^y|=c$ has more than one positive integer solutions $(x,y)$.


2019 ◽  
Vol 15 (05) ◽  
pp. 1069-1074 ◽  
Author(s):  
Hai Yang ◽  
Ruiqin Fu

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].


Author(s):  
YASUTSUGU FUJITA ◽  
MAOHUA LE

Abstract Jeśmanowicz conjectured that $(x,y,z)=(2,2,2)$ is the only positive integer solution of the equation $(*)\; ((\kern1.5pt f^2-g^2)n)^x+(2fgn)^y=((\kern1.5pt f^2+g^2)n)^x$ , where n is a positive integer and f, g are positive integers such that $f>g$ , $\gcd (\kern1.5pt f,g)=1$ and $f \not \equiv g\pmod 2$ . Using Baker’s method, we prove that: (i) if $n>1$ , $f \ge 98$ and $g=1$ , then $(*)$ has no positive integer solutions $(x,y,z)$ with $x>z>y$ ; and (ii) if $n>1$ , $f=2^rs^2$ and $g=1$ , where r, s are positive integers satisfying $(**)\; 2 \nmid s$ and $s<2^{r/2}$ , then $(*)$ has no positive integer solutions $(x,y,z)$ with $y>z>x$ . Thus, Jeśmanowicz’ conjecture is true if $f=2^rs^2$ and $g=1$ , where r, s are positive integers satisfying $(**)$ .


2021 ◽  
Vol 6 (10) ◽  
pp. 10596-10601
Author(s):  
Yahui Yu ◽  
◽  
Jiayuan Hu ◽  

<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>


2019 ◽  
Vol 103 (556) ◽  
pp. 101-110
Author(s):  
Ken Surendran ◽  
Desarazu Krishna Babu

There are recursive expressions (see [1]) for sequentially generating the integer solutions to Pell's equation:p2 −Dq2 = 1, whereDis any positive non-square integer. With known positive integer solutionp1 andq1 we can compute, using these recursive expressions,pnandqnfor alln> 1. See Table in [2] for a list of smallest integer, orfundamental, solutionsp1 andq1 forD≤ 128. These (pn,qn) pairs also formrational approximationstothat, as noted in [3, Chapter 3], match with convergents (Cn=pn/qn) of the Regular Continued Fractions (RCF, continued fractions with the numerator of all fractions equal to 1) for.


2014 ◽  
Vol 90 (1) ◽  
pp. 9-19 ◽  
Author(s):  
TAKAFUMI MIYAZAKI ◽  
NOBUHIRO TERAI

AbstractLet $m$, $a$, $c$ be positive integers with $a\equiv 3, 5~({\rm mod} \hspace{0.334em} 8)$. We show that when $1+ c= {a}^{2} $, the exponential Diophantine equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(c{m}^{2} - 1)}\nolimits ^{y} = \mathop{(am)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$ under the condition $m\equiv \pm 1~({\rm mod} \hspace{0.334em} a)$, except for the case $(m, a, c)= (1, 3, 8)$, where there are only two solutions: $(x, y, z)= (1, 1, 2), ~(5, 2, 4). $ In particular, when $a= 3$, the equation $\mathop{({m}^{2} + 1)}\nolimits ^{x} + \mathop{(8{m}^{2} - 1)}\nolimits ^{y} = \mathop{(3m)}\nolimits ^{z} $ has only the positive integer solution $(x, y, z)= (1, 1, 2)$, except if $m= 1$. The proof is based on elementary methods and Baker’s method.


Author(s):  
K. L. Yakuto

The problem of the positive integer solution of the equation Xn = A for different-order matrices is important to solve a large range of problems related to the modeling of economic and social processes. The need to solve similar problems also arises in areas such as management theory, dynamic programming technique for solving some differential equations. In this connection, it is interesting to question the existence of positive and positive integer solutions of the nonlinear equations of the form Xn = A for different-order matrices in the case of the positive integer n. The purpose of this work is to explore the possibility of using analytical methods to obtain positive integer solutions of nonlinear matrix equations of the form Xn = A where A, X are the third-order matrices, n is the positive integer. Elements of the original matrix A are integer and positive numbers. The present study found that when the root of the nth degree of the third-order matrix will have zero diagonal elements and nonzero and positive off-diagonal elements, the root of the nth degree of the third-order matrix will have two zero diagonal elements and nonzero positive off-diagonal elements. It was shown that to solve the problem of finding positive integer solutions of the matrix equation for third-order matrices in the case of the positive integer n, the analytical techniques can be used. The article presents the formulas that allow one to find the roots of positive integer matrices for n = 3,…,5. However, the methodology described in the article can be adopted to find the natural roots of the third-order matrices for large n. 


2016 ◽  
Vol 95 (1) ◽  
pp. 5-13 ◽  
Author(s):  
MOU-JIE DENG ◽  
DONG-MING HUANG

Let $a,b,c$ be a primitive Pythagorean triple and set $a=m^{2}-n^{2},b=2mn,c=m^{2}+n^{2}$, where $m$ and $n$ are positive integers with $m>n$, $\text{gcd}(m,n)=1$ and $m\not \equiv n~(\text{mod}~2)$. In 1956, Jeśmanowicz conjectured that the only positive integer solution to the Diophantine equation $(m^{2}-n^{2})^{x}+(2mn)^{y}=(m^{2}+n^{2})^{z}$ is $(x,y,z)=(2,2,2)$. We use biquadratic character theory to investigate the case with $(m,n)\equiv (2,3)~(\text{mod}~4)$. We show that Jeśmanowicz’ conjecture is true in this case if $m+n\not \equiv 1~(\text{mod}~16)$ or $y>1$. Finally, using these results together with Laurent’s refinement of Baker’s theorem, we show that Jeśmanowicz’ conjecture is true if $(m,n)\equiv (2,3)~(\text{mod}~4)$ and $n<100$.


2015 ◽  
Vol 11 (04) ◽  
pp. 1107-1114 ◽  
Author(s):  
Hai Yang ◽  
Ruiqin Fu

Let D1, D2, D, k, λ be fixed integers such that D1 ≥ 1, D2 ≥ 1, gcd (D1, D2) = 1, D = D1D2 is not a square, ∣k∣ > 1, gcd (D, k) = 1 and λ = 1 or 4 according as 2 ∤ k or not. In this paper, we prove that every solution class S(l) of the equation D1x2-D2y2 = λkz, gcd (x, y) = 1, z > 0, has a unique positive integer solution [Formula: see text] satisfying [Formula: see text] and [Formula: see text], where z runs over all integer solutions (x,y,z) of S(l),(u1,v1) is the fundamental solution of Pell's equation u2 - Dv2 = 1. This result corrects and improves some previous results given by M. H. Le.


2020 ◽  
Vol 57 (2) ◽  
pp. 200-206
Author(s):  
Elif kizildere ◽  
Maohua le ◽  
Gökhan Soydan

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).


Sign in / Sign up

Export Citation Format

Share Document