Prove that $a^2 \equiv 4 \mod 3^m$ if and only if $a \equiv 2 \mod 3^m$ or $a \equiv −2 \mod 3^m$

Question

Let $m ≥ 1$, and let $a$ be an integer. Prove that $a^ 2 \equiv 4\mod 3^m$ if and only if $a \equiv 2\mod 3^m$ or $a \equiv −2\mod 3^m$.

I know that i'm supposed to find $m$ factors $3$ in $a^ 2 − 4 = (a − 2)(a + 2)$, but I don't even know how to get started.

Anyone with tips to prove this?

Answer 1

I guess that the "if direction" in your statement should not be a problem, if else let me know. For the converse, you know that $3^m$ divides $a^2-4$ as $a^2 \equiv 4 \, \text{mod} \, 3^m$. As you already said, you can factor $a^2-4 = (a-2)(a+2)$ so you know that $3^m \mid (a-2)(a+2)$. Now think about what it would mean, if $3$ would divide both $a-2$ and $a+2$ and show that this is not possible. Then conclude that $3^m$ divides either $a-2$ or $a+2$ and your claim follows.