How to prove that $2^{2^{m}}-1 \equiv 0$ (mod $2^{2^{n}}+1$)

Question

I'd like to solve this problem but I can't

$\exists \ m,n \ \in \mathbb{Z}$ & $ m\gt n\ge 0$

$2^{2^{m}}-1 \equiv 0$ (mod $2^{2^{n}}+1$)

Any ideas? Thanks in advance.

Answer 1

$2^{2^m}-1=(2^{2^{m-1}}-1)(2^{2^{m-1}}+1)$.

Then, not only there exists a pair $m,n$, but there are infinitely many, namely $m$ is any positive integer and $n=m-1$.

In fact, since $2^{2^{m-1}}-1$ can be factored again the same way (provided that $m>1$) the congruence is always true, given the conditions of the problem.

Answer 2

Hint: $2^{2^m}-1$ is a difference of two squares if $m>0$.