We consider the diophantine equation (*)    x^p - x = y^q - y in integers (x, p, y, q). We prove that for given p and q with 2 ≤ p < q, (*) has only finitely many solutions. Assuming the abc-conjecture we can prove that p and q are bounded. In the special case p = 2 and y a prime power we are able to solve (*) completely.

Consider the equation in the title in unknown integers $(x,y,k,l,n)$ with $x\ge 1$, $y>1$, $n\ge 3$, $k\ge 0$, $l\ge 0$ and $gcd(x,y)=1$. Under the above conditions we give all solutions of the title equation (see Theorem 1).

We make more accessible a neglected simple continued fraction based algorithm due to Lagrange, for deciding the solubility of $a{x}^2+bxy+c{y}^2=N$ in relatively prime integers $x,y$, where $N\ne 0$, gcd$(a,b,c)=\text{gcd}(a,N)=1\text{et}D=b^2-4ac\>0$ is not a perfect square. In the case of solubility, solutions with least positive y, from each equivalence class, are also constructed. Our paper is a generalisation of an earlier paper by the author on the equation $x^2-D{y}^2=N$. As in that paper, we use a lemma on unimodular matrices that gives a much simpler proof than Lagrange’s...

A Diophantine $m$-tuple is a set of $m$ positive integers such that the product of any two of them is one less than a perfect square. In this paper we study some properties of elliptic curves of the form $y^2=(ax+1)(bx+1)(cx+1)$, where $\{a,b,c\}$, is a Diophantine triple. In particular, we consider the elliptic curve $E_k$ defined by the equation $y^2=(F_{2k}x+1)(F_{2k+2}x+1)(F_{2k+4}x+1),$ where $k\ge 2$ and $F_n$, denotes the $n$-th Fibonacci number. We prove that if the rank of $E_k\left(𝐐\right)$ is equal to one, or $k\le 50$, then all integer points on $E_k$ are given by $$\begin{array}{c}(x,y)\in \{(0\pm 1),(4F_{2k+1}{F}_{2k+2}{F}_{2k+3}\pm \left(2F_{2k+1}{F}_{2k+2}-1\right)\hfill \\ \hfill \times \left(2F_{2k+2}^2+1\right)\left(2F_{2k+2}{F}_{2k+3}+1\right)\}.\end{array}$$