2

My attempt:

$ax+by=d$(as gcd(a,b)=d).

Dividing by c

$\frac{a}{c}x+\frac{b}{c}y=\frac{d}{c}$

Let $k$ be the gcd of$ (\frac{a}{c},\frac{b}{c})$ so $k|\frac{a}{c}$ and $ k|\frac{b}{c}$ then $k|\frac{a}{c} x + \frac{b}{c}y$ and so $k|\frac{d}{c}$.

As $d|a $ implies that $\frac{d}{c}|\frac{a}{c} $ similarly it divides $\frac{b}{c}$ . Hence $\frac{d}{c}|k$(as $k$ is the greatest common factor). Thus $k= \frac{d}{c}$

Antimony
  • 3,442
  • Presumably, you mean $gcd\left(\frac{a}{c},\frac{\color{red}b}{c}\right)=\frac{d}{c}.$ And also I assume that $c$ is a divisor of $a$ and $b?$ – Thomas Andrews Jan 09 '20 at 18:16
  • 2
    Yes . c is a divisor was not mentioned but then I thought about a counterexample and found that c has to be a divisor – Antimony Jan 09 '20 at 18:18
  • 2
    Your proof works, but you have $k \mid d$ on line 5 when it should be $k \mid \frac{d}{c}$. – Derek Luna Jan 09 '20 at 18:54
  • Yes, as in the first proof here, it follows simply by scaling the Bezout-based characterization of the gcd, i.e.

    $$\begin{align} &\gcd(ca,cb) = d\ \iff\ &d\mid ca,cb\ \ &\ \exists x,y!:\ cax+cby = d\ \iff\ &d/c\mid a,b\ \ &\ \ \exists x,y!:\ ax, +, by = d/c\ \iff\ &\gcd(a,b) = d/c\end{align}\qquad$$

     – Bill Dubuque Jan 09 '20 at 19:48
  • As a corollary you also get "If $\gcd(a,b) = d \implies \gcd( \frac{a}{d} ,\frac{b}{d}) = 1 $" – john Oct 20 '22 at 03:50

1 Answers1

0

Clearly $\frac {d}{c} \mid \frac {a}{c}$ and $\frac {b}{c}$ as $\gcd(a,b)=d$. Let $m$ be a common divisor of $\frac {a}{c}$ and $\frac{b}{c}$. Then $m$ divides the left side which implies that $m \mid \frac{d}{c}$ and $\gcd(\frac{a}{c},\frac{b}{c}) = \frac{d}{c}$.

Derek Luna
  • 2,732
  • 8
  • 19