Let $a, b>1$ be odd such that $a+b=2^{l}$. Find the $k \in \mathbb{N}^{*}$ such that $k^{2} \mid a^{k}+b^{k}$.
Let's start by eliminating the case $k=1$ which works. Already $k$ is odd, because if $k$ were even, we would have a contradiction since $a^{k}+b^{k} \equiv 2[4]$. Next, take $p$ to be the minimal prime divisor of $k$ (possible since we have eliminated the case $k=1$). Since it is necessary that $a$ and $b$ are coprime, we have

$$
\left(\frac{a}{b}\right)^{2 k} \equiv 1[p]
$$

so by setting $\omega$ the order of $\frac{a}{b}$ modulo $p, \omega \mid 2 k$ and $\omega \mid p-1$ hence $\omega=1$ or 2. We study each of these cases and end up with a contradiction in each one.

The only $k$ that works is therefore 1.