Proof that there exists an integer $n \geqslant2$ such that $n^2$ divides $2^n + 3^n$ Proof that there exists an integer $n \geqslant 2$ such that $n^2$ divides $2^n + 3^n$. I came up with this problem and I don't have a clue how to start, or even if it is not trivial at all.
Suppose there is an $n\in\mathbb{N}$ such that $2^n+3^n$ is divisible by $n^2$. Then, $2^n+3^n\equiv 0 \bmod n^2$. In particular, $2^n+3^n\equiv 0 \bmod n$. Suppose for convenience that $n=p\neq 2,3$ is a prime number. Then, by Fermat's little theorem, $2^p\equiv 2 \bmod p$ and $3^p\equiv 3 \bmod p$, so
$$2^p+3^p\equiv 2+3\equiv 5 \bmod p.$$
Hence, if $2^p+3^p\equiv 0 \bmod p$, we must have $p=5$. And indeed, $p=5$ works, 
$$2^5+3^5 = 275 = 11\cdot 25.$$
Hence, we have shown that the only prime $p$ such that $p^2$ divides $2^p+3^p$ is $p=5$.
Just plug in numnbers. n = 5 works.