Show that $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $ Show that if  $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $  
$\gcd(a,b)=d\Rightarrow\ d\mid a,b\Rightarrow\ \ d^2\mid a^2,b^2\Rightarrow\ d^2\mid\gcd(a^2,b^2)$.
But to complete  the proof we must show that: $\gcd(a^2,b^2)\mid d^2$
How can I achieve this?

By dividing $d$ on both sides and using $\gcd(ac,bc)=c\gcd(a,b)$, we may assume that $\gcd(a,b)=d=1$, so you only need to show that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$. Suppose that $p$ is a prime that divides $a^2$ and $b^2$, then $p$ divides $a$ and $b$, this contradict to the assumption $\gcd(a,b)=1$, so there is no prime divides $a^2$ and $b^2$, it follows that $\gcd(a^2,b^2)=1$.
