# 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 \$\displaystyle x^2\$ is divisible by \$\displaystyle a\$ (where \$\displaystyle a\$ is prime and \$\displaystyle x\$ is an integer), then \$\displaystyle x\$ is also divisible by \$\displaystyle a\$?

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

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

then \$\displaystyle x\$ is also divisible by \$\displaystyle 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 \$\displaystyle a\$ being prime), but somehow i dont know how to demonstrate it. I guess I would start by saying that

\$\displaystyle x^2=am\$

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

\$\displaystyle x=an\$

is true also (\$\displaystyle 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 \$\displaystyle x^2\$ is divisible by \$\displaystyle a\$ (where \$\displaystyle a\$ is prime and \$\displaystyle x\$ is an integer), then \$\displaystyle x\$ is also divisible by \$\displaystyle a\$?

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

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

then \$\displaystyle x\$ is also divisible by \$\displaystyle 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 \$\displaystyle a\$ being prime), but somehow i dont know how to demonstrate it. I guess I would start by saying that

\$\displaystyle x^2=am\$

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

\$\displaystyle x=an\$

is true also (\$\displaystyle 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 \$\displaystyle a\$ is prime , then \$\displaystyle 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