I have an example I am trying to prove but I can't quite see the trick and would appreciate some help:
Suppose p is a prime factor of $10^{2^k} +1$. Show that $p \equiv 1$ (mod $2^{k+1}$).
(Also: once I have this result, does it generalise to other bases as well?)
Thank you!
We have $$10^{2^k}\equiv -1\pmod p\implies 10^{2^{k+1}}\equiv 1 \pmod p$$ From which we deduce that the order of $10\pmod p$ is exactly $2^{2k+1}$. It follows immediately that $$2^{2k+1}\,|\,p-1$$ (since the order of any non-zero element $\pmod p$ must be a divisor of $p-1$) and this is the desired claim.