Prove that if $gcd(a, b)=d$ then $gcd(a^2, b^2 )=d^2$

Question

how can i prove that? Prove that if $gcd(a, b)=d$, then $gcd(a^2, b^2 )=d^2$

Please tell us what you've tried and where you are stuck. That will help people give you help that is better suited to what you know and need. Otherwise your question is likely to be downvoted and/or closed. — rogerl, Jun 11 '20 at 18:11
This has been asked and answered before, e.g. here https://math.stackexchange.com/q/1286904/42969 – found with Approach0 — Martin R, Jun 11 '20 at 18:12

score 0 · Answer 1 · answered Jun 11 '20 at 19:34

Suppose $a=da_1$ and $b=db_1$ for coprime integers $a_1$ and $b_1$ and so $\gcd(a,b)=d$

We have $a^2=d^2a_1^2$ and $b^2=d^2b_1^2$, and since $\gcd(a_1,b_1)=1$ means (considering the fundamental theorem of arithmetic) $a_1$ and $b_1$ have no common prime divisor, then it also means that $a_1^2$ and $b_1^2$ have no common prime divisor and so $\gcd(a_1^2,b_1^2)=1$ and thus $\gcd(a^2,b^2)=d^2$

Prove that if $gcd(a, b)=d$ then $gcd(a^2, b^2 )=d^2$

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