Integral solutions to $20^m-10m^2+1=19^n$. Find postive integral solutions of the equation $20^m-10m^2+1=19^n$.
My solution,

*

*Using modulo  $10$, I got the information that $n$ is even.

*Using modulo $19$, I got the information that $m$ should be $19k+2$ or $19k-2$.

*Using modulo $20$, I got the information that $m$ is even.

By observation $(m,n) =(2,2)$. But how to arrive at a solution and show that only $(2,2)$ is the solution and no other solution exists.

Let $m$ and $n$ be positive integers such that
$$20^m-10m^2+1=19^n.$$
Reducing mod $3$ and $5$ shows that $m$ and $n$ are even, say $m=2M$ and $n=2N$. Then
$$10m^2-1=20^m-19^n=(20^M+19^N)(20^M-19^N).$$
This shows that $10m^2-1$ is divisible by $20^M+19^N$. In particular
$$20^M<20^M+19^N<10m^2-1=40M^2-1.$$
This shows that $M\leq1$, and hence $m=2$ and $n=2$.
