Question

If $p$is prime and $\left(a,b\right)=p$, then$\left(a^2,b^2\right)=?$

Short answer

If $\left(a,b\right)=p$, then $\left(a^2,b^2\right)=p^2$.

Step-by-step solution

Use the given part

It is given that, $p$is a prime and $\left(a,b\right)=p$.

It is known that if $p|a^n$, then $p^n|a^n$.So, for the given problem, $p|a^2$implies $p^2|a^2$, which also implies that $p^2|a^2$. Therefore, $p^2|\left(a^2,b^2\right)$.

Proof part

Since $p^2|\left(a^2,b^2\right)$, assume there exists some integer $d$such that $\left(a^2,b^2\right)=d{p}^2$. Now solve for $d$.

$$\begin{gathered} d=\frac{1}{p^2}\left(a^2,b^2\right) \\ =\left(\frac{a^2}{p^2},\frac{b^2}{p^2}\right) \end{gathered}$$

It is known that $\left(\frac{a}{p},\frac{b}{p}\right)=1$;if $d\ne 1$,then integer $d$has some prime factor $p_1$such that role="math" localid="1645867043093" $p_1|a$and $p_1|b$, which shows that $p_1$is a factor of $a$and $b$other than $p$, implying $\left(a,b\right)=p_{1}p$.

Thisis a contradiction to $\left(a,b\right)=p$.Hence,$d=1$ , and $\left(a^2,b^2\right)=p^2$.