Prove $d = \gcd(a,b) \iff 1= \gcd (k_1, k_2)$.

Question

This is the assumption they give me: Let $a, b$ be integers and $d$ a positive integer. Let $d|a$ and $d|b$ so there there exists $a=dk_1$ and $b=dk_2$.

I can go the backwards direction but I'm going in circles when I go forward. Any hints?

Answer 1

If $\gcd(k_1,k_2)=k\gt1$ then let $k_1=k\alpha, k_2=k\beta$, then $\gcd(a,b)=\gcd(kdk_1,kdk_2)=kd$.

If $gcd(a,b)=d$, then $gcd(k_1d,k_2d)=d\gcd(k_1,k_2)$ and so $\gcd(k_1,k_2)=1$.

Answer 2

$d=gcd(a,b)$ if and only if there are $u,v$ with $d=ua+vb$, if and only if $1=uk_1+vk_2$, if and only if $1=gcd(k_1,k_2)$.