<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 $gcd(a^2, b^2) = d$ and $d \ne 1$
Then $a^2 = dm, b^2 = dn \implies a^2b^2 = d^2mn$, where $gcd(m,n) = 1$
This implies both $m$ and $n$ are perfect squares. $m = p^2, n = q^2$
We have $a^2 = dp^2, b^2 = dq^2 \implies d$ is a perfect square. Let $d = x^2$
Now we have $a^2 = x^2p^2, b^2 = x^2q^2 \implies a = xp, b = xq \implies gcd(a,b) = x = 1$
Hence $d = x^2 = 1$
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