Find the polynomials with integer coefficients $P$ such that:

$$
\forall n, n \text { divides } P\left(2^{n}\right)
$$
(let solved by Brieux Madeline-Derou)

Let $k$ be a positive integer and $q$ an odd prime integer. Set $n=2^{k} q$, then $n$ divides $P\left(2^{n}\right)$. Now $2^{n}=2^{(} 2^{k} \times 2^{q}$. By Fermat's Little Theorem, we have $0 \equiv P\left(2^{n}\right) \equiv P\left(2^{2^{k}} 2^{q}\right) \equiv P\left(2^{2^{k}} \times 2\right) \equiv P\left(2^{2^{k}+1}\right)$ $(\bmod q)$. Thus, every odd prime integer $q$ divides $P\left(2^{2^{k}}+1\right)$, and since it has an infinite number of prime factors, $P\left(2^{2^{k}+1}\right)=0$ for all positive integers $k$. Therefore, $P$ has an infinite number of zeros, and by rigidity, $P=0$. Conversely, $P=0$ works because for any positive $n$, $n$ divides 0.