Let $a$, $b$, $c$, $r$ be positive integers such that $a^{2}+b^{2}=c^{r}$, $\min (a,b,c,r)>1$, $\gcd (a,b)=1, a$ is even and $r$ is odd. In this paper we prove that if $b\equiv 3\hspace{4.44443pt}(\@mod \; 4)$ and either $b$ or $c$ is an odd prime power, then the equation $x^{2}+b^{y}=c^{z}$ has only the positive integer solution $(x,y,z)=(a,2,r)$ with $\min (y,z)>1$.
For any positive integer $D$ which is not a square, let $(u_1,v_1)$ be the least positive integer solution of the Pell equation $u^2-Dv^2=1,$ and let $h(4D)$ denote the class number of binary quadratic primitive forms of discriminant $4D$. If $D$ satisfies $2\nmid D$ and $v_1h(4D)\equiv0 \pmod D$, then $D$ is called a singular number. In this paper, we prove that if $(x,y,z)$ is a positive integer solution of the equation $x^y+y^x=z^z$ with $2\mid z$, then maximum $\max\{x,y,z\}<480000$ and both $x$, $y$ are singular numbers. Thus, one can possibly prove that the equation has no positive integer solutions $(x,y,z)$., Xiaoying Du., and Obsahuje bibliografické odkazy
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^ap^b=y^n$, $x, y, n\in \mathbb {N}$, $\gcd (x, y)=1$, $n\geq 3$, $a, b\in \mathbb {Z}$, $a\geq 0$, $b\geq 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 $x^2+2^a\cdot 17^b=y^n$, $x, y, n\in \mathbb {N}$, $\gcd (x, y)=1$, $n\geq 3$, $a, b\in \mathbb {Z}$, $ a\geq 0$, $ b\geq 0$, are determined.