Related papers: A solution to a problem and the Diophantine equati…

As an application of the method of Thue-Siegel, we will resolve a conjecture of Walsh to the effect that the Diophantine equation $aX^{4} - bY^2=1$, for fixed positive integers $a$ and $b$, possesses at most two solutions in positive…

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…

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 give a specific way of describing positive integer solutions of a Diophantine equation $(x+y)^2+(y+z)^2+(z+x)^2=12xyz$ and introduce a generalized cluster pattern behind it.

In this short note we study the existence and number of solutions in the set of integers ($Z$) and in the set of natural numbers ($N$) of Diopahntine Equations of second degree with two variables of the general form $ax^2-by^2=c$.

In this paper, we deal with the quartic Diophantine equation $X^4-Y^4=R^2-S^2$ to present its infinitely many integer solutions.

This paper investigates the exponential Diophantine equation of the form $a^x+b=c^y$, where $a, b, c$ are given positive integers with $a,c \ge 2$, and $x,y$ are positive integer unknowns. We define this form as a "Type-I transcendental…

First, we consider the equation $ax^2 - by^2 + c = 0$, with $a,b \in N*$ and $c \in Z*$, which is a generalization of Pell's equation. Here, we show that: if this equation has an integer solution and $ab$ is not a perfect square, then it…

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 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.

The study of finiteness or infiniteness of integer solutions of a Diophantine equation has been considered as a standard problem in the literature. In this paper, for f(x) in Z[x] monic and q1 ,...., qm in Z, we study the conditions for…

In this article the Diophantine equations of the form $x^{2}-8C_{n}xy+ 16y^{2}=\pm2^{r}$, $x^{4}-8C_{n}x^{2}y+ 16y^{2}=\pm2^{r}$ and $x^{2}-8C_{n}xy^{2}+ 16y^{4}=\pm2^{r}$ are taken into consideration for the investigations of existence of…

In 2016 Izadi and Nabardi (b) showed (4-2-4) has infinitely many integer solutions. They used a specific congruent number elliptic curve.In 2019 Janfada and Nabardi,item C, showed that a necessary condition for n to have an integral…

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…

We consider the number of solutions in positive integers $(x,y,z)$ for the purely exponential Diophantine equation $a^x+b^y =c^z$ (with $\gcd(a,b)=1$). Apart from a list of known exceptions, a conjecture published in 2016 claims that this…

We give the complete solution in integers $(n,a,b,x,y)$ of the title equation when $\gcd(x,y)=1$, except for the case when $xab$ is odd.

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…

In this paper we consider the Diophantine equation \begin{align*}b^k +\left(a+b\right)^k &+ \cdots + \left(a\left(x-1\right) + b\right)^k=\\ &=d^l + \left(c+d\right)^l + \cdots + \left(c\left(y-1\right) + d\right)^l, \end{align*} where…

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…

Stiller proved that the Diophantine equation $x^2+119=15\cdot 2^{n}$ has exactly six solutions in positive integers. Motivated by this result we are interested in constructions of Diophantine equations of Ramanujan-Nagell type…