# Prove that if x's square is divisible by a, then x is also divisible by a

• May 7th 2010, 11:29 PM
Ragnarok
Prove that if x's square is divisible by a, then x is also divisible by a
How would you prove that, if $x^2$ is divisible by $a$ (where $a$ is prime and $x$ is an integer), then $x$ is also divisible by $a$?

For example, if $a=3$, then how do you prove that

if $x^2$ is divisible by $3$,

then $x$ is also divisible by $3$?

I get the feeling this might be blindingly obvious, but then, those are the types of proofs I tend to have trouble on. I know the reason for it intuitively (it has to do with $a$ being prime), but somehow i dont know how to demonstrate it. I guess I would start by saying that

$x^2=am$

with $m$ being an integer. So then I would want to prove that, if this is true, then

$x=an$

is true also ( $n$ being an integer). Is that right? Where would I go from there? A proof by contradiction? I'm still not totally fluent in abstract language, so it takes me a long time to "translate" things.

Thanks!
• May 7th 2010, 11:49 PM
tonio
Quote:

Originally Posted by Ragnarok
How would you prove that, if $x^2$ is divisible by $a$ (where $a$ is prime and $x$ is an integer), then $x$ is also divisible by $a$?

For example, if $a=3$, then how do you prove that

if $x^2$ is divisible by $3$,

then $x$ is also divisible by $3$?

I get the feeling this might be blindingly obvious, but then, those are the types of proofs I tend to have trouble on. I know the reason for it intuitively (it has to do with $a$ being prime), but somehow i dont know how to demonstrate it. I guess I would start by saying that

$x^2=am$

with $m$ being an integer. So then I would want to prove that, if this is true, then

$x=an$

is true also ( $n$ being an integer). Is that right? Where would I go from there? A proof by contradiction? I'm still not totally fluent in abstract language, so it takes me a long time to "translate" things.

Thanks!

This follows at once from the very definition of "prime number": as $a$ is prime , then $a\mid bc\Longrightarrow a\mid b\,\,\,or\,\,\,a\mid c$

A simple way to prove it, without relying on the prime definition, is by checking the prime decomposition of x...

Tonio