What are the $n$ such that $4^{n}+5^{n}$ is a perfect square?
Let $k, n$ be such that $4^{n}+5^{n}=k^{2}$.

Then $5^{n}=k^{2}-2^{2 n}=\left(k-2^{n}\right)\left(k+2^{n}\right)$. If 5 divides each of the two factors, 5 also divides their difference, namely $2^{n+1}$, which is a contradiction. Thus $k+2^{n}=5^{n}$, and $k-2^{n}=1$. By taking the difference again, we find $2^{n+1}=5^{n}-1$. This equation has a solution when $n=1$, and if $n \geq 2$ we have $5^{n}>2^{n+1}+1$ (by induction, for example). Therefore, $n=1$ is the only potential solution, and it is indeed a solution.