Finding the possible values of $\gcd(a^2,b)$

Question

If $\gcd (a,b)=p\qquad p\text{ is a prime.}$

What are the possible values of $\gcd(a^2,b)$

I saw this solution:

$a:=\alpha p,\qquad b:=\beta p,\qquad \gcd(\alpha,\beta)=1$

$(a^2,p)=(\alpha^2 p^2,\beta p)=p(\alpha ^2 p,\beta)$

$\color{red}{=p(p,\beta)}\in \{p,p^2\}$

I don't understand the red, how did they got this?

Answer 1

Write $a=\alpha p$ and $b=\beta p$; we know that $\gcd(\alpha,\beta)=1$. There are two cases

1. $p\mid \beta$, or
2. $p\nmid\beta$

In case 1, $\beta=\gamma p$, so $$ a^2=p^2\alpha^2,\quad b=p^2\gamma $$ so $\gcd(a^2,b)=p^2$, because $\gcd(\alpha^2,\gamma)=1$ (prove it).

In case 2 and 3, $\gcd(a^2,b)=p\gcd(p\alpha^2,\beta)=p$ (prove it).

In general, if $d$ is a common divisor of $m$ and $n$, then $$ \gcd(m,n)=d\gcd(m/d,n/d) $$