Prove that if $d$ is a common divisor of two integers $a$ and $b$, then $d=\gcd(a,b)$ if and only if $\gcd(a/d,b/d)=1$.
So far I used what was given so I have $a=dk$, $b=ld$ and $\gcd(a,b)=d$ can be written as a linear combination of $ax+by=d$ but I am unsure how to use the information.
Where do I go from here? Can someone show me how to solve this using Bezout's Identity if possible?
Hint $\ $ By the $ $ GCD Distributive Law, $ $ and $\ d\mid a,b\iff d\mid (a,b)\ \,$ [gcd Universal Property]
$$\begin{eqnarray} d\,\left(\dfrac{a}d,\dfrac{b}d\right) &\!\!=& (a,b)\\[.4em] \Rightarrow\ \left(\dfrac{a}d,\dfrac{b}d\right) &\!\!=& (a,b)\,/\,d\\[.4em] \!{\rm Thus}\ \ \ 1 = \left(\dfrac{a}d,\dfrac{b}d\right) &\!\!\iff\!& (a,b)\!=\!d\end{eqnarray}$$
Comment: Dividing by d will give the equality $1$, but there is much more to be said. Yes, by definition the smallest positive integer that can be expressed by a linear combination is $1$ which would imply that $1$ is the gcd,but a better way I think is to proceed by contradiction.
So, you have: $\frac{a}{d}+\frac{b}{d}=1$, where $d=\gcd(a,b)$.
Suppose $\gcd(\frac{a}{d},\frac{b}{d})$=e. We will show that $e=1$.
Here: $e|\frac{a}{d},e|\frac{b}{d}\Rightarrow \frac{a}{d}=ex, \frac{b}{d}=ey$ for $x,y \in \mathbb{Z}$
Thus: $a=dex,b=dey \Rightarrow$ $de$ is a common divisor of $a,b$, but $de>d$ which is a contradiction by assumption of $d=\gcd(a,b)$. Hence, $e=1$.
($\Leftarrow$) $$\begin{aligned}(a/d)x+(b/d)y=1&\implies ax+by=d\implies(\forall c,;c\mid a\land c\mid b\implies c\mid ax+by=d)\implies c\le d\&\implies \gcd(a,b)=d\end{aligned}$$
– rosshjb Feb 14 '25 at 05:31