Prove or disprove: If $a>b$ and $\gcd(a,b)=1$ then $\gcd(a,a-b)=1$

Question

Let $a,b\in\Bbb{N}$. Prove or disprove that if $a>b$ and $\gcd(a,b)=1$ then $\gcd(a,a-b)=1$.

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I think it is true.

Proof If $\gcd(a,b)=1$ then $as+bt=1$ for some $s,t\in\Bbb{Z}$. Then $as+(b+a-a)t=as+(b-a)t+at=a(s+t)+(b-a)t=am+(a-b)k=1$, where $m=s+t\in\Bbb{Z}$ and $k=-t\in\Bbb{Z}$. Hence again by Bézout's identity, we conclude $\gcd(a,a-b)=1$. $\square$

Is it correct?

Answer 1

That works.

An alternative is that

• if $k$ divides $a$ and $a-b$
• then $k$ divides $a$ and $a-(a-b)=b$
• and so $k$ divides $1$, and the greatest such $k$ is $1$