Related papers: On the Diophantine Equation $x^{2}+5^{a}\cdot 11^{…
The title equation is completely solved in integers $(n,x,y,a,b)$, where $n\geq 3$, $\gcd(x,y)=1$ and $a,b\geq 0$. The most difficult stage of the resolution is the explicit resolution of a quintic Thue-Mahler equation. Since it is for the…
In this note, we find all the solutions of the Diophantine equation x^2 +2^a.3^b.11^c=y^n in nonnegative integers a, b, c, x, y, n>= 3 with x and y coprime.
The object of this paper is to give a new proof of all the solutions of the Diophantine equation x^2+11^m=y^n; in positive integers x, y with odd m>1 and n>=3.
We find all solutions of Diophantine equation x^{2}+11^{2k} = y^{n} where x>=1, y>=1, n>=3 and k is natural number. We give p-adic interpretation of this equation.
Consider the diophantine equation $(3^{x_1}-1)(3^{x_2}-1)=(5^{y_1}-1)(5^{y_2}-1)$ in positive integers $x_1\le x_2$, and $y_1\le y_2$. Each side of the equation is a product of two terms of a given binary recurrence, respectively. In this…
In this paper, we give all the solutions of the Diophantine equation x^2+7^{alpha}.11^{beta}=y^n, in nonnegative integers x, y, n>=3 with x and y coprime, except for the case when alpha.x is odd and beta is even.
We prove that for given integers b and c, the diophantine equation x^2+bx+c=y^2, has finitely many integer solutions(i.e. pairs in ZxZ),in fact an even number of such solutions(including the zero or no solutions case).We also offer an…
We completely solve the Diophantine equation $x^2+2^k11^\ell19^m=y^n$ in integers $x,y\geq 1;~ k,\ell, m\geq 0~$ and $n\geq 3$ with $\gcd(x,y)=1$, except the case $2\mid k, 2\nmid \ell m$ and $5\mid n$. We use this result to recover some…
The objective of the paper is to determine the complete solutions for the Diophantine equation $x^2 + 3^{\alpha}113^{\beta} = y^{\mathfrak{n}}$ in positive integers $x$ and $y$ (where $x, y \geq 1$), non-negative exponents $\alpha$ and…
It is a generalization of Pell's equation $x^2-Dy^2=0$. Here, we show that: if our Diophantine equation has a particular integer solution and $ab$ is not a perfect square, then the equation has an infinite number of solutions; in this case…
We give an infinite number of integer solutions to the Diophantine equation x^5 - (x+1)^5 -(x+2)^5 +(x+3)^5 = 5^m + 5^n, and some solutions to some similar equations.
The title equation, where $p>3$ is a prime number $\not\equiv 7 \pmod 8$, $q$ is an odd prime number and $x,y,n$ are positive integers with $x,y$ relatively prime, is studied. When $p\equiv 3\pmod 8$, we prove (Theorem 2.3) that there are…
We solve completely the Lebesgue-Nagell equation x^2+D=y^n, in integers x, y, n>2, for D in the range 1 =< D =< 100.
We investigate the solvability of the Diophantine equation in the title, where $d>1$ is a square-free integer, $p, q$ are distinct odd primes and $x,y,a,b$ are unknown positive integers with $\gcd(x,y)=1$. We describe all the integer…
In this article, we consider the equation x^2+19^{m}=y^n, n>2, m>0. We find the solutions of the title equation for not only 2 \mid m but also 2\notdividesm.
We study the solutions of a Diophantine equation of the form $a^x+b^y=c^z$, where $a\equiv 2 \pmod 4$, $b\equiv 3 \pmod 4$ and $\gcd (a,b,c)=1$. The main result is that if there exists a solution $(x,y,z)=(2,2,r)$ with $r>1$ odd then this…
Suppose that $x$ is odd, $n\geq7$ and $p\notin\{2,5\}$ are primes. In this paper, we prove that the Diophantine equations $x^{2}\pm5^{\alpha}p^{n}=y^{n}$ have no solutions in positive integers $\alpha,x,y$ with $gcd(x,y)=1$.
Using elementary number theory we study Diophantine equations over the rational integers of the following form, $y^2=(x+a)(x+a+k)(x+b)(x+b+k)$, $y^2=c^2x^4+ax^2+b$ and $y^2=(x^2-1)(x^2-\alpha^2)(x^2-(\alpha+1)^2).$ We express their integer…
We prove that for each odd prime p, positive integer alpha, and non-negative integers beta and gamma, the Diophantine equation X^{2N} + 2^{2 alpha} 5^{2 beta} p^{2 gamma} = Z^5 has no solution with X, Z, N in Z^+, N > 1, and gcd(X,Z) = 1.
Given odd, coprime integers $a$, $b$ ($a>0$), we consider the Diophantine equation $ax^2+b^{2l}=4y^n$, $x, y\in\Bbb Z$, $l \in \Bbb N$, $n$ odd prime, $\gcd(x,y)=1$. We completely solve the above Diophantine equation for…