<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] Suppose $gcd(a,b)=1$ then you have $ax+by=1$, cubing this, we get $(ax+by)^3=1$
i.e., $a^3x^3+b^3y^3+3a^2x^2by+3axb^2y^2=1$
i.e., $a^2(ax^3+3x^2by)+b^2(by^3+3axy^2)=1$
does this imply $gcd(a^2,b^2)=1$???
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