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 for...

Recently, Miyazaki and Togbé proved that for any fixed odd integer b ≥ 5 with b ≠ 89, the Diophantine equation $b^x+2^y={(b+2)}^z$ has only the solution (x,y,z) = (1,1,1). We give an extension of this result.

It is shown that for a given squarefree positive integer D, the equation of the title has no solutions in integers x > 0, m > 0, n ≥ 3 and y odd, nor unless D ≡ 14 (mod 16) in integers x > 0, m = 0, n ≥ 3, y > 0, provided in each case that n does not divide the class number of the imaginary quadratic field containing √(-2D), except for a small number of (stated) exceptions.

Let $\mathbb{Z}$, $\mathbb{N}$ be the sets of all integers and positive integers, respectively. Let $p$ be a fixed odd prime. Recently, there have been many papers concerned with solutions $(x,y,n,a,b)$ of the equation $x^2+2^a{p}^b=y^n$, $x,y,n\in \mathbb{N}$, $gcd(x,y)=1$, $n\ge 3$, $a,b\in \mathbb{Z}$, $a\ge 0$, $b\ge 0.$ And all solutions of it have been determined for the cases $p=3$, $p=5$, $p=11$ and $p=13$. In this paper, we mainly concentrate on the case $p=3$, and using certain recent results on exponential diophantine equations including the famous Catalan equation, all solutions $(x,y,n,a,b)$ of the equation $x^2+2^a·{17}^b=y^n$, $x,y,n\in \mathbb{N}$, $gcd(x,y)=1$, $n\ge 3$, $a,b\in \mathbb{Z}$, $a\ge 0$, $b\ge 0$, are determined....