In my textbook in the proof for the solving of Diophantine equation of the form $ax + by = c$ they causally used and mention this theorem (I don't know what to call it, sorry if it's not appropriate) without prior talk about it or proof.
"Note that $r$ and $s$ are relatively prime because $d$ is the greatest common divisor of $a$ and $b$."
Is it something that you should be able to intuitively see is the case? There is a post about it here (Prove divisibility with gcd: If $ar+bs=d=\gcd(a,b)$, then $r$ and $s$ are relatively prime) but the answer there doesn't answer for me how to prove that integer $k$ that divides both $r$ and $s$ must be less than or equal to $1$ because $kd\mid d$.
Is it something along the lines of $\,kd\mid d$
so $\,d=kd\!\cdot\!g\,$ for some integer $g$
and $g$ is either $1$ or $d$, so $k$ is either $1$ or $0$,
but I don't know if that is valid and how to formally state that.
Proof :
First of all, we have to suppose $\,d\neq0\,,\,$ otherwise the previous property would be false.
Since for hypothesis $\,kd\mid d\,,\,$ there exists an integer $\,g\,$ such that
$d=kd\!\cdot\!g$
and, by dividing by $\,d\neq0\,$ both sides of the previous equality, we get that
$1=k\!\cdot\!g$
hence
$\big(k=1\,\land\,g=1\big)\;$ or $\;\big(k=-1\,\land\,g=-1\big)$
consequently, in any case, it follows that
$k\leqslant1\,.$