TITLE: prove that if $\gcd(a,b)=1$ then $\gcd(a^2, b^2)=1$
QUESTION [6 upvotes]: Using Bezout's identity I have shown
$$(a,b^2)=(b,a^2)=1$$
but what should be the next step?
Thanks.

REPLY [2 votes]: For any two integers $x,y$, $gcd(x,y)=1$ iff there exist $r,s\in \mathbb Z$ such that $rx+sy=1$. 
So now, suppose $gcd(a,b)=1$ and write $ra+sb=1$. Square both sides and rearrange terms appropriately, and this proves that $gcd(a^2,b)=1$, and also that $gcd(a,b^2)=1$. This is presumably what you've done. So, for all $a,b\in \mathbb Z$, $gcd(a,b)=1$ implies $gcd(a^2,b)=gcd(a,b^2)=1$. Apply again and you get $gcd(a^2,b^2)=1$.