Prove that if x's square is divisible by a, then x is also divisible by a
How would you prove that, if is divisible by (where is prime and is an integer), then is also divisible by ?
For example, if , then how do you prove that
if is divisible by ,
then is also divisible by ?
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 being prime), but somehow i dont know how to demonstrate it. I guess I would start by saying that
with being an integer. So then I would want to prove that, if this is true, then
is true also ( 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.