# How to solve these two simultaneous “divisibilities” : n+1∣m2+1n+1\mid m^2+1 and m+1∣n2+1m+1\mid n^2+1

Is it possible to find all integers $m>0$ and $n>0$ such that $n+1\mid m^2+1$ and $m+1\,|\,n^2+1$ ?

I succeed to prove there is an infinite number of solutions, but I cannot progress anymore.

Thanks !

Some further results along the lines of thought of @individ:

Suppose $p$ and $s$ are solutions to the Pell’s equation:

Then,

are solutions if $(a,b,c,d)$ are: (these are the only sets that I found using the computer)