Number Theory Greatest Common Divisor proof

Question

Prove the following: $\gcd(a^2,b^2)=\gcd(a,b)^2$

Answer 1

Solve the problem in $2$ cases:
(1) If $a$ and $b$ have a common factor.
(2) If $a$ and $b$ don't have a common factor, i.e. $\gcd(a,b) = 1$.

I believe the first case is fairly straightforward.

For the second case, see Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$.

Answer 2

Consider $a=p_1^{k_1}*p_2^{k_2}*p_3^{k_3}...$ and $b=p_1^{l_1}*p_2^{l_2}*p_3^{l_3}...$ and $\gcd(a,b)=p_1^{\min(k_1,l_1)}*p_2^{\min(k_2,l_2)}*p_3^{\min(k_3,l_3)}...$ Note that it is entirely possible for either $k_m$ or $l_n$ to be zero. Now square them.

$a^2=p_1^{2k_1}*p_2^{2k_2}*p_3^{2k_3}...$ and $b^2=p_1^{2l_1}*p_2^{2l_2}*p_3^{2l_3}...$ and $\gcd(a^2,b^2)=p_1^{\min(2k_1,2l_1)}*p_2^{\min(2k_2,2l_2)}*p_3^{\min(2k_3,2l_3)}...$ Notice that $a<b\implies ca<cb\forall c>0$, which means $\min(ca,cb)=c\min(a,b)$. We can now rewrite $\gcd(a^2,b^2)=p_1^{2\min(k_1,l_1)}*p_2^{2\min(k_2,l_2)}*p_3^{2\min(k_3,l_3)}...=(p_1^{\min(k_1,l_1)}*p_2^{\min(k_2,l_2)}*p_3^{\min(k_3,l_3)}...)^2=\gcd(a,b)^2$