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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.

Bill Dubuque
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    Your "and g is either 1 or d so k is either 1 or 0" is wrong. Replace it by: $a,b$ are supposed to be $\ne0$ (or at least one of them), so that $d\ne0.$ Hence $kd\mid d$ implies $k\mid1.$ The only integers dividing $1$ are $1$ and $-1.$ – Anne Bauval Mar 10 '23 at 06:56
  • $,kd\mid d\iff k\mid 1,$ by cancelling $,d\neq 0, $ as in the linked dupe. – Bill Dubuque Mar 10 '23 at 08:41

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Prove that if $\,kd\mid d\,$ then $\,k\leqslant1$.

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

$k\leqslant1\,.$

Angelo
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