Prove that $2^{2^5} +1 \equiv 0 \mod 641$ 
$2^{(2^5)} +1 \equiv 0 \mod 641$

From the totient funcion we have:

$2^{32^{20}} \equiv 1$

Thus:

Either $2^{32} \equiv 1$ or $2^{32} \equiv -1$

But how do I prove that $2^{32} \equiv -1$?

You only need $$641=2^4+5^4$$
and $$641=2^7\cdot 5+1$$
This gives you $$2^{32}=(2^7)^4\cdot 2^4\equiv -(2^7)^4\cdot 5^4=-(2^7\cdot 5)^4\equiv -1\mod 641$$
