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.

As $1 = ax + by$, then $ab = a^2bx + ab^2y$.
Therefore,
$1 = (ax + by)^2 = a^2x + b^2y + 2abxy$
$= a^2x + b^2y + 2(a^2bx + ab^2y)xy$
$= a^2(x + 2bx^2y) + b^2(y + 2axy^2)$
