Prove that $\pm 5, \pm 5^2 ,...,\pm5^{2^{a-2}}$ forms a reduced residue system mod $2^a$

Question

Let $a \ge 3$ Prove that $\pm 5, \pm 5^2 ,...,\pm5^{2^{a-2}}$ forms a reduced residue system mod $2^a$

My attempt: We are given $\phi(2^a) = 2^{a-1} $number of elements, hence this is a possible candidate for a reduced residue system( $\phi$ denotes the Euler totient function). I approached this problem by Induction. For $a=3$, $\,\, \pm 5, \pm 5^2 \equiv 1,3,5,7 \bmod(8)$, which is clearly a reduced residue system $\bmod (2^3)$.

But after assuming it for $a=k$, for some $k>3$, I am not able to see how this holds for $a=k+1$. Am I on the right track? Is there any other way to solve this problem?

Answer 1

Hint:

Prove by induction that $\; 5^{2^k}\equiv 1+2^{k+2}\mod 2^{k+3}$, and deduce that

1. The congruence class has order $2^{a-2}\mod 2^a$,
2. The subgroups $\bigl\langle\,\overline 5\,\bigr\rangle$ and $\bigl\langle\,\overline{\mkern-4mu -1}\,\bigr\rangle$ have a trivial intersection, whence $\;\mathbf Z/2^a\mathbf Z$ is the product of these two subgroups.