Prove congruence for powers of 2

Question

Let $n=2^k$ where $k\ge 3$. Let $a$ be any odd natural number.

Prove that $a^{n/4}\equiv 1\pmod{n}$.

My attempt:

$\phi(n)=n/2$

So, by Euler's formula, $a^{n/2}\equiv 1\pmod{n}$.

I don't know how to proceed.

Answer 1

Thank you, Daniel! I think I've got it.

For $k=3$, $n=8$ and $a^{8/4}=a^2\equiv 1\pmod{8}$ because $8|(a-1)(a+1)$ [one of $(a-1)$ and $(a+1)$ is divisible by 4 and the other is divisible by 2].

Let the statement be true for some $k\ge 3$.

$n=2^k$ and $a^{n/4}\equiv 1\pmod{n}$

$a^{n/2}-1=(a^{n/4})^2-1=(a^{n/4}-1)(a^{n/4}+1)$

$n|(a^{n/4}-1)$ (induction hypothesis) and $2|(a^{n/4}+1)$

So, $2n|(a^{n/4}-1)(a^{n/4}+1)=a^{n/2}-1$

i.e. $a^{n/2}\equiv 1\pmod{2n}$

Does this look ok?

Answer 2

For $n=2^k$ with $k\ge 3$, the group $\mathbb Z_n^*$ is not cyclic, so it has no element of order $\phi(n)=\frac{n}{2}$.

Therefore, $a\equiv -1\mod \frac{n}{4}$ is impossible.