GCD proof: $\gcd(a+b,a-b)\geq\gcd(a,b)$

Question

How to prove that $\gcd(a+b,a-b)\geq\gcd(a,b)$, where $a$ and $b$ are integers?

I'm not sure how to frame the equations and derive the result.

Answer 1

This is deceptively simple. Let $d=gcd(a,b)$. That is to say $d$ is the greatest integer that divides $a$ and $b$. All we need though is that it divides both. Next, we use the fact that if $d|a$ and $d|b$ then $d|a+b$ and $d|a-b$. With this, we have that $d$ is a common divisor of both $a+b$ and $a-b$, and thus must be less than or equal to the greatest common divisor of $a+b$ and $a-b$.