scholarly journals A Note on Integer Solutions of the Diophantine Equation x2-dy2=1

1956 ◽  
Vol 3 (1) ◽  
pp. 55-56
Author(s):  
John Hunter

In the equationdis any positive integer which is not a perfect square. For convenience we shall consider only those solutions of (1) for which x and yare both positive. All the others can be obtained from these. In fact, it is well known that if (x0, y0) is the minimum positive integer solution of (1), then all integer solutions (x, y) are given byand, in particular, all positive integer solutions are given by

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.


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 $(x,y)$.


2018 ◽  
Vol 14 (05) ◽  
pp. 1223-1228
Author(s):  
Hai Yang ◽  
Ruiqin Fu

Let [Formula: see text] be a positive integer which is not a square. Further, let [Formula: see text] be the least positive integer solution of the Pell equation [Formula: see text], and let [Formula: see text] denote the class number of binary quadratic primitive forms of discriminant [Formula: see text]. If [Formula: see text] satisfies [Formula: see text] and [Formula: see text], then [Formula: see text] is called an exceptional number. In this paper, under the assumption that there have no exceptional numbers, we prove that the equation [Formula: see text] has no positive integer solutions [Formula: see text] satisfy [Formula: see text] and [Formula: see text].


Author(s):  
Sukrawan Mavecha

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.


2010 ◽  
Vol 81 (2) ◽  
pp. 177-185 ◽  
Author(s):  
BO HE ◽  
ALAIN TOGBÉ

AbstractLet a, b, c, x and y be positive integers. In this paper we sharpen a result of Le by showing that the Diophantine equation has at most two positive integer solutions (m,n) satisfying min (m,n)>1.


2021 ◽  
Vol 5 (1) ◽  
pp. 115-127
Author(s):  
Van Thien Nguyen ◽  
◽  
Viet Kh. Nguyen ◽  
Pham Hung Quy ◽  
◽  
...  

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 \(u < 100\).


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 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.


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.


2009 ◽  
Vol 51 (3) ◽  
pp. 659-667 ◽  
Author(s):  
BO HE ◽  
ALAIN TOGBÉ

AbstractLet n be a positive integer. In this paper, we consider the diophantine equation We prove that this equation has only the positive integer solutions (n, x, y, z) = (1, t, 1, 1), (1, t, 3, 2), (3, 2, 2, 2). Therefore we extend the work done by Leszczyński (Wiadom. Mat., vol. 3, 1959, pp. 37–39) and Makowski (Wiadom. Mat., vol. 9, 1967, pp. 221–224).


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