Solve quadratic diophantine equation in two variables.

Question

Determine all $m,n\in \mathbb Z^+$ such that $m^2+1$ is a prime number and $10(m^2+1)=n^2+1$. Please provide complete explanation with solution.

Answer 1

Fermat numbers $F=2^2+1=5$ and $F=2^{2^2}+1= 17$ resulted from $k=1$ and $k=2$ in $F_n=2^{2^k}+1$ give $7$ and $13$ for n in equation $10(m^2+1)=n^2+1$ where it results in $m=2$ and $m=4$ respectively. So one set of solutions could be a combination of solutions $(m=± 2.. and.. m= ± 4)$ with $(n=± 17.. and.. n= ± 13)$.

It is not known all Fermat primes give integers for n.It can be checked by brute force.On RHS the last digit of $n^2$ must be 9 so that $10 | n^2+1$ , so the last digit of n can only be 3 or 7.This can also be checked by brute force to see whether there are more solutins.