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Prove the following: $ \gcd(a^n,b^n) = 1 \Leftrightarrow \gcd(a,b)=1,\;$ where $ a,b,n \in \mathbb{N}$.

I can show one side by using Fermat's Theorem and breaking them down into unique factorizations and then putting them to the power of $n$ but how do go the other way?

amWhy
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Kirito
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    Easier, perhaps, to show the contrapositive: $\gcd(a,b)\neq 1\implies \gcd(a^n,b^n)\neq 1$. – lulu Feb 17 '20 at 23:23

1 Answers1

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In general if $\exists$ prime $p$, such that $p|a$ and $p|b \implies p^n|a^n$ and $p^n|b^n \implies (a^n, b^n) =(a,b)^n$. Therefore $gcd(a,b)=1 \iff gcd(a^n, b^n)=1$