# Any way to simplify $\gcd(a+b,a-c)$?

If I have the expression $\gcd(a+b,a-c)$ is there a way to further reduce this? Are there any other properties?

-
According to my own interpretation of simplification, I doubt. Note that for any integers $p,q$ there exists integers $a,b,c$ such that $p=a+b$, $q=a-c$ –  Amr Aug 18 '13 at 22:47

No, there is no way to "simplify" the expression, since without further specification, we must take $a, b, c$ to be arbitrary integers.

-
@Software I am doing well. And you? –  amWhy Aug 19 '13 at 12:01

Theorem: $\gcd(a,b)=\gcd(a,b-ma)$ for any integer $m$.

Try and prove that, and then use it to answer your question.

Note, though that there are no simpler ways to express this using fewer letters. But you can express it in different ways. For example, $\gcd(a+b,a-c)=\gcd(a+b,b+c)$ (Why?)

-
I rearranged (a+b,b+c) to (a+b,a-c) to begin with –  user2175923 Aug 18 '13 at 22:55