How do I prove this statement about $n^\text{th}$-power residues?

Question

I am studying A Classical Introduction to Modern Number Theory by Ireland and Rosen, and the authors leave the proof of the following proposition (4.2.2) as "an exercise" ...

Suppose that $a$ is odd, $e \geq 3$, and consider the congruence $$x^n \equiv a \pmod{2^e}$$ Show that

• If $n$ is odd, a solution for $x$ in the above congruence always exists and is unique.

• If $n$ is even, a solution exists if and only if $a \equiv 1 \pmod{4}$ and $$a^{2^{e-2}/d} \equiv 1 \pmod{2^e}$$ where $d= \gcd\left(n,2^{e-2}\right)$.

I figured I would try the $e=3$ case first. We are working with $$x^n \equiv a \pmod{8}$$

If $a$ is odd, then $a$ must be $1$, $3$, $5$, or $7$ modulo $8$. Further, if $n$ is odd, then the solution $x$ is simply equal to $a$, since (as we can easily check) any odd number raised to any odd power is equal to itself, modulo $8$ - every odd number is a square root of $1$. As for the other values of $e$ in the $n$ odd case, I think it should be possible to use Hensel's lemma in some way.

I have no idea how to start the $n$ even case. Thank you in advance for help.

Answer 1

1. Suppose $n$ is even and write $d=\gcd(n,2^{e-2})$, assume that there exists a solution to the equation: $$x^n\equiv a\mod 2^e $$ because $a$ is odd then $x$ is odd and we have $a=1 \mod 4$ because $x^n\equiv (x^{\frac{n}{2}})^2\equiv 1\mod 4$, and we have also: $$x^{2}\equiv 1 \mod 8 $$ using Lifting The Exponent Lemma we obtain: $$ x^{2^{e-2}}\equiv 1 \mod 2^e$$ and this implies $a^{\frac{2^{e-2}}{d}}\equiv 1\mod 2^e$
2. Can you do the second implication?