As the title indicates, I do not know how to proceed. There is a hint to prove it. The hint says that:
show that if $p>2$ then a is of order $2^{n+1} \pmod p$.
But I do not see any connection between the hint and the question in demand.
Please help!
Thanks in advance,,
EDIT:
The original question in the title was to prove:
If $p|(a^{2n}+1)$, then $p = 2$ or $p \equiv 1 \pmod {2^{n+1}}$.
By the useful comment of @Zvi below, the statement appears to be false. The statement is originally from "An Introduction to The Theory of Numbers" by Ivan Niven, Herbert S.Zuckerman and Hugh L.Montgomery. Most probably, it is a typo, since the correct statement is:
If $p|(a^{2^{n}}+1)$, then $p = 2$ or $p \equiv 1 \pmod {2^{n+1}}$.
Please consider the latter statement. A useful clue or hint is very helpful.
Let $p>2$ be a prime number such that $p \mid a^{2^n}+1$ for some integers $a$ and $n\geq 0$. We claim that $p\equiv 1\pmod{2^{n+1}}$. Clearly, $p\nmid a$, so $a^{p-1}\equiv 1\pmod{p}$ by Fermat's Little Theorem.
Let $k$ be the order of $a$ modulo $p$. Then, we have $k\mid p-1$. Since $a^{2^n}\equiv -1\pmod{p}$, we get $$a^{2^{n+1}} =\left(a^{2^n}\right)^2\equiv (-1)^2=1\pmod{p},$$ we conclude that $k\mid 2^{n+1}$. This shows that $k=2^m$ for some integer $m$ such that $0\leq m\leq n+1$.
Note that $m$ must equal $n+1$. If $m\leq n$, then $a^{2^m}\equiv 1\pmod{p}$, so that $$a^{2^n}=\left(a^{2^{m}}\right)^{2^{m-n}}\equiv 1^{2^{m-n}}=1\pmod{p}.$$ This means $1\equiv a^{2^n}\equiv-1\pmod{p}$, so $p\mid 2$, but this contradicts the hypothesis that $p>2$. Thus, $m=n+1$ and $k=2^{n+1}$. Recall that $k\mid p-1$, so $2^{n+1}\mid p-1$ and the proof is finished.