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Let $(a,b)$ be the greatest common denominator for two whole numbers $a$ and $b$. By the Lemma of Bezout we know that if $(a,b)≠(0,0)$ then there exist whole numbers $r$ and $s$ such that $(a,b)=ra+sb$. I want to show that $(ca,cb)=c(a,b)$ for a nonzero constant c. I thought about just multiplying the equation by $c$ but I am not sure if this can just be done like that, what I was thinking is: $(ca,cb)=car+cbs=c(ar+bs)= c(a,b)$. Im not sure if just 'getting out' the $c$ would affect the bezout factors negatively, because a classmate meant that it cannot be done this way. This is a homework problem and I want to understand why this is wrong if its wrong. My classmate couldn't really elaborate as well so I wanted to find clearness to my question now. Thanks a lot for any help.

Mathman
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  • The converse of Bezout is false. That is, knowing that there are integers $m,n$ such that $d=ma+nb$ does not imply that $d=\gcd(a,b)$. Easy to find counterexamples. – lulu Oct 14 '21 at 17:08
  • here is a duplicate which addresses your underlying question. – lulu Oct 14 '21 at 17:09
  • thanks a lot I scrolled through the section and didn't see the other question. – Mathman Oct 14 '21 at 17:15
  • @lulu I always quote Bezout's with an additional line about what makes the gcd special among all possible linear combinations. That way the converse becomes true. But this is not that version, it seems. – Arthur Oct 14 '21 at 17:15

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