So assume that $a^n|b^n$ then $b^n=a^n(x)$ for some integer $x$ taking the nth root of both sides you get $b=a(x^{\frac{1}{n}})$. Now I'm not sure if this is ok since $x^{\frac{1}{n}}$ is not always an integer. But it seems like it should always work so I"m not sure if this is the correct approach.
Prove if $a^n|b^n$ then $a|b$
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divisibility
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0It *could* be made to work *iff* you first proved that $\sqrt[n]{|x|}$ is either an integer or irrational. – 2017-01-24
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0Why either? I get showing its an integer but I don't think I can do that. Also my problem with the other proofs is that we have only covered divisibility not prime factorization yet. So I can't use that technique. – 2017-01-24
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0You only know $\sqrt[n]{|x|}$ is an integer *after* you proved $a|b\,$. If you have some other proof in mind, please post it. – 2017-01-24