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Problem

Suppose $X$ and $Y$ are greater than $0$. Show that $\gcd(X,Y)$ is $1$ iff $\gcd(X^m,Y^m)= 1$.

Please help with the above. I have no idea what's going on. An explanation would be nice.

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Suppose that $\gcd(X,Y)=1$. We show that $\gcd(X^m,Y^m)=1$. Suppose to the contrary that $\gcd(X^m,Y^m)=d\gt 1$. Then there is a prime $p$ that divides $d$. So $p$ divides $X^m$ and $p$ divides $Y^m$.

Recall that if a prime $p$ divides a product, then $p$ divides at least one of the terms. So $p$ divides $X$ and $p$ divides $Y$. This contradicts the fact that $\gcd(X,Y)=1$.

Next we need to show that if $\gcd(X^m,Y^m)=1$ then $\gcd(X,Y)=1$. The argument is similar to the one above, but easier, and is left to you.

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  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Jan 27 '14 at 18:13
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I would like to give a simple explanation based on prime factorization: for any integer $N$, we will have $$N=p_1\times...\times{p_m}$$

So $\gcd(X,Y)=1$ iff $X,Y$ have no common prime factors.

Proof From the assumption, we have that $X^m,Y^m$ have no common prime factors. And same are the $X,Y$. Then easily $\gcd(X,Y)=1$

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Let $\,\cal P(n) = $ set of primes factors of $\,n.\,$ By unique factorization, $\,\cal \color{#c00}{P(n^k) = \cal P(n)}\,$ for $\,k\ge 1,\,$ so

$$\gcd(x,y)=1\iff \cal P(x)\cap \cal P(y) = \emptyset\color{#c00}\iff P(x^m)\cap \cal P(y^m) =\emptyset \iff \gcd(x^m,y^m) =\rm 1$$

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