Croatian Mathematical Society Competitions

Note that $n^5+n^4+1=(n^3-n+1)(n^2+n+1)$, and that $$\begin{array}{rl}d& =\gcd (n^3-n+1,n^2+n+1)\\ & =\gcd (n^2+n+1,-n^2-2n+1)\\ & =\gcd (n^2+n+1,n-2)\\ & =\gcd (n-2,7),\end{array}$$ hence we have two cases:

1) $d=7$

This implies $7\mid m$ and $7\nmid n$, from which we get $7\mid m-7n$ and $7\nmid m-4n$. Therefore, there is no solution in this case.

2) $d=1$

This implies that $n^3-n+1$ and $n^2+n+1$ are both squares of integers. That is true only for $n=0$ and $n=-1$, since $n^2<n^2+n+1<(n+1{)}^2$ holds for $n\ge 1$, and $(n+1{)}^2<n^2+n+1<n^2$ holds for $n<-1$. Both $n=0$ and $n=-1$ yield $m^2=1$, i.e. $m=\pm 1$, and among four possibilities only two satisfy the given conditions: $(m,n)=(-1,0)$ and $(m,n)=(1,0)$.