Related papers: An exponential diophantine equation related to odd…
In this paper, we consider the exponential Diophantine equation \( (2^k-1)(b^k-1)=y^q \) with $k\ge 2$, odd integer $b$ and an odd prime exponent $q$ and obtain effective upper bounds for $q$ in terms of $b$. In particular, we show that…
We give a necessary condition for the existence of solutions of the Diophantine equation $p=x^{q}+ry^{q},$ with $p$, $q$, $r$ distinct odd prime natural numbers.
We study the exponential Diophantine equation $x^2+p^mq^n=2y^p$ in positive integers $x,y,m,n$, and odd primes $p$ and $q$ using primitive divisors of Lehmer sequences in combination with elementary number theory. We discuss the solvability…
Let $n$ be a positive integer. We discuss pairs of distinct odd primes $p$ and $q$ not dividing $n$ for which the Diophantine equations $pq=x^2+ny^2$ have integer solutions in $x$ and $y$. As its examples we classify all such pairs of $p$…
Let p, c be distinct odd primes, and l \ge 2 an integer. We find sufficient conditions for the Diophantine equation cy^l=(x^p-1)/(x-1) not to have integer solutions
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…
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…
In this paper we study the Diophantine equation $x^{4}-q^{4}=py^{5},$ with the following conditions: $p$ and $q$ are different prime natural numbers, $y$ is not divisible with $p$, $p\equiv3$ (mod20), $q\equiv4$ (mod5), $\overline{p}$ is a…
The Diophantine equation (x^n-1)/(x-1)=y^q has four known solutions in integers x, y, q and n with |x|, |y|, q > 1 and n > 2. Whilst we expect that there are, in fact, no more solutions, such a result is well beyond current technology. In…
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…
Consider the equation $q_1\alpha^{x_1}+\dots+q_k\alpha^{x_k} = q$, with constants $\alpha \in \overline{\mathbb{Q}} \setminus \{0,1\}$, $q_1,\ldots,q_k,q\in\overline{\mathbb{Q}}$ and unknowns $x_1,\ldots,x_k$, referred to in this paper as…
In this paper, we gave solutions of the Diophantine equations 16^{x}+p^{y}=z^{2}, 64^{x}+p^{y}=z^{2} where p is an odd prime, n is a positive integer and x,y,z are non-negative integers. Finally we gave a generalization of the Diophantine…
Motivated by the recent result of Farhi we show that for each $n\equiv \pm 1\pmod{6}$ the title Diophantine equation has at least two solutions in integers. As a consequence, we get that each (even) perfect number is a sum of three cubes of…
Let k => 1, m => 1 be small fixed integers, gcd(k, m) = 1. This note develops some techniques for proving the existence of infinitely many primes solutions x = p, and y = q of the linear Diophantine equation y = mx + k.
For two relatively prime positive integers $a, b\in \mathbb{N}$, it is known that exactly one of the two Diophantine equations $$ax + by \ =\ \frac{(a-1)(b-1)}{2}\ \mbox{ and }\ 1 + ax + by \ =\ \frac{(a-1)(b-1)}{2}$$ has a nonnegative…
We propose a method to determine the solvability of the diophantine equation $x^2-Dy^2=n$ for the following two cases: $(1)$ $D=pq$, where $p,q\equiv 1 \mod 4$ are distinct primes with $(\frac{q}{p})=1$ and…
In this note we consider the title Diophantine equation from both theoretical as well as experimental point of view. In particular, we prove that for $k=4, 6$ and each choice of the signs our equation has infinitely many co-prime positive…
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…
This work determine the entire family of positive integer solutions of the diophantine equation. The solution is described in terms of $\frac{(m-1)(m+n-2)}{2} $ or $\frac{(m-1)(m+n-1)}{2}$ positive parameters depending on $n$ even or odd.…
Let $N$ be an odd perfect number. Then, Euler proved that there exist some integers $n, \alpha$ and a prime $q$ such that $N = n^{2}q^{\alpha}$, $q \nmid n$, and $q \equiv \alpha \equiv 1 \bmod 4$. In this note, we prove that the ratio…