Prove that gcd(a, b) divides a−b.

Question

Having some trouble with this. It makes snese, but need to prove it in an articulate way.

Have: Let d = gcd(a,b). There exists m,n in Z s.t. d = ma + nb. d >= 1, d|a and d|b.

Want: d|a-b

Am I missing something?

Thanks so much!

Answer 1

Let $d=\gcd(a,b)$. By definition, $a=da'$ and $b=db'$ for some $a'$ and $b'$. Now show that $a-b$ is of the form $dm$ for some $m$.

Answer 2

There exist $m,n \in \mathbb{Z}$ s.t $ma + nb = d$. So we have $m(a-b) + (n+m)b = d$. So from this we get that $d \mid a-b$

Answer 3

You don't need Bezout's identity. It's much simpler than that. Actually, if $d$ is any common divisor of $a$ and $b$, then $a=dm$, $b=dn$ and $a-b=d(m-n)$, so $d$ divides $a-b$.

A thumb's rule: you need Bezout's identity to prove a theorem about gcd if that theorem does not hold for other common divisors.