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?

Simply, write $a=da'$ and $b=db'$ such that $$\gcd(a', b') =1$$ Hence, $a^2= d^2(a')^2$ and $b^2=d^2(b')^2$. 
Since,  $\gcd(a', b') =1$,  $\gcd((a')^2, (b')^2) =1$ and thus $$gcd(a^2, b^2) = d^2\gcd((a')^2, (b')^2) = d^2$$
