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

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?

## marked as duplicate by Bill Dubuque modular-arithmetic StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 22 at 14:33

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## 1 Answer

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.

• So what you're saying is, find some example for $a$ that divides either $a-2$ or $a+2$, but not both (like $a=11$). But you can also find values for $a$ that do not work (like $a=12$) right? – Flint Feb 22 at 13:58
• No what I meant is the following: By the assumptions you get that $3^m \mid (a-2)(a+2)$, so in particular you get $3 \mid (a-2)$ or $3 \mid (a+2)$, since $3$ is a prime number. Lets assume w.l.o.g. that $3 \mid (a-2)$. Now you have to conclude that $3$ does not divide $a+2$. But now, as $3^m$ divides $(a-2)(a+2)$, you get that $3^m$ divides $a-2$ (in the prime factorization of the product, no power of $3$ appears in the prime factorization of $a+2$, so $3^m$ has to appear in the prime factorization of $a-2$). – Wos07573 Feb 22 at 16:31