<s>[INST] Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
 [/INST] First show
$\gcd(a, b^2) = \gcd(a^2, b) = 1$
$\gcd(a,b) = 1$ means that there are $x$, $y$ ($x$, $y$ are integers) such that  $ax + by =1$
$ax + by(ax+by) =1$
$a(x+bxy) + b^2y^2 =1$
That means $\gcd(a,b^2)=1$
By a similar derivation, $\gcd(a^2,b)=1$.
$ar_1(ar_2+b^2s_2) +b^2s_1 = 1$
$a^2r_1^2+b^2(ar_1s_1+s_1)=1$
Therefore,
$\gcd(a^2, b^2) = 1$
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