<s>[INST] How do I show that $\gcd(a^2, b^2) = 1$ when $\gcd(a,b)=1$? How do I show that $\gcd(a^2, b^2) = 1$ when $\gcd(a,b)=1$?
I can show that $\gcd(a,b)=1$ implies $\gcd(a^2,b)=1$ and $\gcd(a,b^2)=1$.  But what do I do here?
 [/INST] If $\gcd(a, b) = 1$, this means exactly that $a$ and $b$ do not have any prime factors in common. Any common factor would constitute a common divisor. If $a$ and $b$ have some prime factors in common, then the product of those common factors (including any repeated ones of higher multiplicity) is the greatest common divisor.

For instance $24 = 2 \cdot 3\cdot 4$ and $60 = 3\cdot 4\cdot 5$. The common factors are $3$ and $4$ and so $\gcd(24, 60) = 3\cdot 4 = 12$. 

If we take an integer $a$ and square it, the resulting integer $a^2$ does not have any new prime factors which are not already present in $a$. It simply has the same factors, with double the multiplicity. 

For instance $12 = 3\cdot 2\cdot 2$, and $144 = 12^2 = (3\cdot 2\cdot 2)(3\cdot 2 \cdot 2) = 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 2$.

So if $a$ and $b$ have no prime factors in common, then $a^2$ and $b^2$ have no prime factors in common either, and so $\gcd(a^2, b^2) = 1$.
</s>