If $d \mid a,b$ and $\exists x,y \in Z : d = ax + by$, then $d = \gcd(a,b)$

Question

If $d \mid a,b$ and $\exists x,y \in Z : d = ax + by$, then $d = \gcd(a,b)$

Well, I've tried by saying that $\gcd(a,b) \iff d = a s \cdot b t$, and now $s=x,t=y$, and I've proved it.

I've also tried to find counter examples and failed. Although my proof seems to be odd, I'm looking forward to see what you can say about this. Thanks

This proof of the gcd universal property shows that linear common divisors are necessarily greatest. — Bill Dubuque, Apr 29 '23 at 18:38

score 1 · Accepted Answer · answered Nov 18 '16 at 08:09

1

To show the statement "$d = \gcd(a,b)$", we should show (i) that $d$ divides $a$ and $b$, and (ii) that $d$ is the greatest such divisor, i.e. for any common divisor $k$ of $a$ and $b$, $k$ divides $d$.

Conveniently, (i) is given to us as a hypothesis.

For (ii), assume that $k$ divides $a$ and $b$. Now use the fact that $d = ax + by$. Can you conclude that $k$ divides $d$?

answered Nov 18 '16 at 08:09

Caleb Stanford

45,895

Well if I divide by $d$ the equation. I get $1 = a/dx + b/dy$, It means that $k < d$, which means $d$ is the greatest common divisor, what do you think? – Ilan Aizelman WS Nov 18 '16 at 08:23
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@IlanAizelmanWS how did you conclude $k < d$ from that? – Caleb Stanford Nov 18 '16 at 08:31
The gcd can't be lower than 1, so if we divide by $ k > d $ than we'll get $ p = ax + by $ and $ p < 1 $ which can't be true. – Ilan Aizelman WS Nov 18 '16 at 08:34
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@IlanAizelmanWS I see, yes, that is correct. – Caleb Stanford Nov 18 '16 at 08:35

If $d \mid a,b$ and $\exists x,y \in Z : d = ax + by$, then $d = \gcd(a,b)$

1 Answers1

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