We have $p>2$ - prime number and we know that $2^n\equiv 1\mod p$ and $2^n$ is not congruent to $1 \mod p^2$ ($n$-natural number). Prove that $2^d$ is not congruent to $1 \mod p^2$ where order $2 = d \mod p$.
I will be grateful when someone tell me for what I have to start.
Assuming that $ d $ is the order of $ 2 $ in $ \mathbb{Z}_p $, let $ n $ be given such that $ 2^n = 1 $ in $ \mathbb{Z}_p $. We then have $ d|n $. Let $ n = dm $ for some $ m $, then in $ \mathbb{Z}_{p^2} $ we have that $ 2^n = 2^{dm} = (2^d)^m \neq 1 $ which implies $ 2^d \neq 1 $.
The statement does not hold generally if $ d $ is an arbitrary natural number. For instance, we have $ 2^2 \equiv 1 \pmod{3} $, $ 2^2 \neq 1 \pmod{9} $, but choosing $d = 6 $ we have that $ 2^6 \equiv 1 \pmod{9} $.