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?
 A: gcd$(a,b)=1$ if and only if no prime divides a and b. A prime divides $a^2$ if and only if it divides a. Therefore a number divides $a^2$ and $b^2$ if and only if it divides $a$ and $b$.
A: The following proof is less informative than the posted proofs, but it is kinda cute. Recall the theorem of Bézout which says that $c$ and $d$ are relatively prime if and only if there exist integers $s$ and $t$ such that $cs+dt=1$.
Let $x$ and $y$ be integers such that $ax+by=1$. Now cube both sides. We get 
$$a^2(ax^3+3x^2by)+b^2(3axy^2+by^3)=1,$$
and therefore by the theorem of Bézout $a^2$ and $b^2$ are relatively prime. 
A: Hint: You've shown that $\gcd(y,b^2)=1$ when $\gcd(y,b)=1.$ What happens when $y=a^2$?
A: 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 $\gcd(a^2, b^2) = 1$.
