How do I show $\gcd(a,b)=\gcd(b^2-a,b)$?

Question

How do I show $\gcd(a,b)=\gcd(b^2-a,b)$?

I know that $\gcd(a,b)=\gcd(a-b,b)$. I think the statement $\gcd(a,b)=\gcd(b^2-a,b)$ is false but im not finding any counterexample.

Answer 1

We have $d\mid a$ and $d\mid b$, if and only if $a=dr$ and $b=ds$ for some $r,s\in\Bbb Z$, if and only if $ma=drm$ and $nb=dsn$ for some $r,s\in\Bbb Z$ and all $m,n\in\Bbb Z$, if and only if $d \mid (ma+nb)$ for all $m,n\in\Bbb Z$.

Answer 2

$gcd(a,b)|gcd(b^2-a,b)$ and by way of Bezouts's identity, we have that if $ma+nb=gcd(a,b)$ where $m,n \in \mathbb{Z}$, then $m(b^2-a)+(n-mb)b=gcd(a,b)$ which shows that $gcd(b^2-a,b)|gcd(a,b)$.

Also, $gcd(b^2-a,b)=gcd(b^2-b-a,b)=gcd(b^2-2b-a,b)=\cdots=gcd(b^2-b\times b-a,b)=gcd(-a,b)=gcd(a,b)$