Related papers: On the Diophantine equation cy^l=(x^p-1)/(x-1)

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.

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…

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…

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…

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…

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 $p$…

We show that if {1, b, c, d} is a D(-1) diophantine quadruple with b<c<d and c=1+s^2, then the cases s=p^k, s=2p^k, c=p and c=2p^k do not occur, where p is an odd prime and k is a positive integer. For the integer d=1+x^2, we show that it…

Let $p$ be a prime integer, $\mathbb{Z}_p$ the finite field of order $p$ and $\mathbb{Z}^{*}_{p}$ is its multiplicative cyclic group. We consider the Diophantine equation $x^n + y^n = z^n$ with $1 \leq n \leq \frac{p - 1}{2}$. Our main aim…

We shall show that, for any given primes $\ell\geq 17$ and $p, q\equiv 1\pmod{\ell}$, the diophantine equation $(x^\ell-1)/(x-1)=p^m q$ has at most four positive integral solutions $(x, m)$ and give its application to odd perfect number…

The subject matter of this work is the diophantine equation x^n+y^m=c(x^k)(y^l), where n,m,k,l,c are natural numbers.We investigate this equation from the point of view of positive integer solutions.A preliminary examination of sources such…

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 prove that the Diophantine equation (a^2 c x^k - 1)(b^2 c y^k - 1) = (abc z^k - 1)^2 has no solutions in positive integers with x, y, z > 1, k \geq 7 and a^2x^k \neq b^2y^k.

Let $f(x)=x^{2}(x^{2}-1)(x^{2}-2)(x^{2}-3).$ We prove that the Diophantine equation $ f(x)=2f(y)$ has no solutions in positive integers $x$ and $y$, except $(x, y)=(1, 1)$.

Let $a$ and $b$ be two distinct fixed positive integers such that $\min \{a,b\}>1.$ First, we correct an oversight from \cite{X-Z}. Then, we show that the equation in the title with $b \equiv 3 \pmod 8$, $b$ prime and $a$ even has no…

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…

In this article, we study the solutions of certain type over $K$ of the Diophantine equation $x^2= By^p+Cz^p$ with prime exponent $p$, where $B$ is an odd integer and $C$ is either an odd integer or $C=2^r$ for $r \in \mathbb{N}$. Further,…

In this article, I study and solve the exponential Diophantine equation $M_p^{x} + (M_q + 1)^{y}= (lz)^2$ where $M_p$ and $M_q$ are Mersenne primes, $l$ is a prime number, and $x,y$, and $z$ are non-negative integers. Several illustrations…

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.

In this paper, we investigate the Diophantine equation \[ (2^k - 1)(3^k - 1) = x^n \] and prove that it has no solutions in positive integers $k, x, n > 2$.

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…