Proof that $\gcd(a,c)=\gcd(b,c)$ if $ab+c=a^2c-b+a$

Question

I have the following problem that I am struggling with. The problem is this:

Let $a,b,c$ be integers. Prove that if $ab+c = a^2c-b+a$ then $\gcd(a,c)=\gcd(b,c)$

I tried lots of rearranging of the above formula to get it into a form that resembles: $ax+cy = bi+cj$ which is $\gcd(a,c)=\gcd(b,c)$ by $\gcd$ characterization but have had little success.

Any help would be appreciated! Thanks!

Answer 1

Actually, $D(a,c)=D(b,c)$, where $D(u,v)$ is the set of common divisors of $u$ and $v$.

Indeed, if $d \in D(a,c)$ then $d$ divides $ab+c$ and so $d$ divides $b=a^2c+a-(ab+c)$. Therefore, $D(a,c) \subseteq D(b,c)$.

Conversely, if $d \in D(b,c)$ then $d$ divides $ab+c$ so $d$ divides $a=ab+c-(a^2c-b)$. Therefore, $D(b,c) \subseteq D(a,c)$.

In particular, $\gcd(a,c)=\max D(a,c)=\max D(b,c)=\gcd(b,c)$.