Show that $\gcd(a,b)=d\Rightarrow\gcd(a^2,b^2)=d^2\ $

Question

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?

Answer 1

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 $$gcd(a^2, b^2) = d^2\gcd((a')^2, (b')^2) = d^2$$

Answer 2

Write $a=da'$ and $b=db'$ with $\gcd(a',b')=1$.

Then $$ \dfrac{a^2}{b^2}=\dfrac{d^2(a')^2}{d^2(b')^2}=\dfrac{(a')^2}{(b')^2} $$ which cannot be further reduced.

Answer 3

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 $\gcd(a^2,b^2)=1$.