Prove that for any $n\in\mathbb{N}$, $2^{2^{4n+1}}+7$ is always composite.
I compute the expression for n=1, and see that it is divisible by 11. So my natural guess is that $2^{2^{4n+1}}+7\equiv 0 \text{ mod } 11$. But I don't have any idea to proceed further. Thanks in advance for any help or suggestion.
If your conjecture is correct, you should have $$ 2^{2^{4n+1}}\equiv4\pmod{11} $$ that's the same as $$ 2^{2^{4n+1}-2}\equiv1\pmod{11} $$ which is granted if you show that $$ 2^{4n+1}-2\equiv0\pmod{10} $$ that is, $$ 16^n\equiv1\pmod{5} $$ Fix a proof for every statement and you're done.
