This problem is taken from the movie X+Y (A Brilliant Young Mind in the US). Apparently, it is a real problem from a British IMO qualifying exam:
Are there infinitely pairs of positive integers $(m,n)$ such that $m$ divides $n^2 + 1$ and $n$ divides $m^2 + 1$?
The answer is yes, with infinitely many solutions coming from alternating Fibonacci numbers:
Let $F_0 = 1$. Then $(F_{2n}, F_{2n+2})$ form a solution pair.
Besides the trivial solution $(1,1)$, all other solutions appear to be a pair of Fibonacci numbers of the above form. So I have been trying to prove this:
The only pairs of positive integers $(m,n)$ that satisfy $m|n^2+1$ and $n|m^2+1$ are $(1,1)$ and $(F_{2n}, F_{2n+2})$ for nonnegative $n$ where $F_{n}$ denotes the $nth$ Fibonacci number beginning with $F_0=1$.
Little success so far. Can anyone point me in the right direction? All I have so far is in suspecting that Vieta jumping might come in handy.