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

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    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!
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    Quote Originally Posted by Ragnarok View Post
    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
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