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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!

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    $\begingroup$ Notice that $ab + c - a = a^2c - b$. So now, if we let $g = \gcd(a, c)$ and $h = \gcd(b, c)$. Then we have that $g$ divides left hand side and $h$ divides ring hand side, so $h$ divides $ab + c - a$ and $g$ divides $a^2c - b$. Thus by the first part, $h \le g$ and by the second $g \le h$ so must be equal. $\endgroup$ – mdave16 Jun 6 '17 at 23:36
  • $\begingroup$ I am starting to doubt you did lots of rearranging... my solution is $x = b-1, y = 1, i = -1, j = a^2$. And many other solutions are available, including $x= 1, y = a^2, i = a +1, j=1$ $\endgroup$ – mdave16 Jun 6 '17 at 23:40
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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, $\gcd(a,c)=\max D(a,c)=\max D(b,c)=\gcd(b,c)$.

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