elementary number theory - How can I prove that $gcd(a,b)=1implies gcd(a^2,b^2)=1$ without using prime decomposition?

How can I prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$, without using prime decomposition? I should only use definition of gcd, division algorithm, Euclidean algorithm and corollaries to those. Thanks for your help!

Answer

The golden rule for all problems about greatest common divisors, I claim (especially if you're specifically trying to avoid using prime decomposition), is the fact that $\gcd(a,b)$ is equal to the smallest positive integer $g$ that can be written in the form $g=ax+by$ with $x,y$ integers.

In particular, $\gcd(a,b)=1$ if and only if there exist integers $x$ and $y$ such that $ax+by=1$. (This is not true with 1 replaced by a larger integer! It's true because 1 is the smallest positive integer there is.)

That's my general hint; my specific hint is to cube both sides of the equation $ax+by=1$.