# Suppose that$\ gcd(b, a) = 1$. Prove that $\gcd(b + a, b − a) \leq 2$

Suppose that $\gcd(b, a) = 1$. Prove that $\gcd(b + a, b − a) \leq 2$

I've been given a hint I should use divisor rules, so I have if $d \mid b+a$ and $d \mid b-a$, then $d \mid 2a$ and $d \mid 2b$, but then I'm stumped on what to do after

• You're pretty much done once you've proved that $d|2a$ and $d|2b$. What if $d$ is odd? If $d$ is even, what can you say about $d/2$? – anomaly Oct 23 '14 at 14:21

## 2 Answers

Since $d|2a$ and $d|2b$, we have $d|\text{gcd}(2a,2b) = 2$. Hence $\text{gcd}(b+a,b-a) \le 2$.

• gcd(2a,2b) = 2 because gcd(a,b) = 1 right? – user2980566 Oct 23 '14 at 14:23
• @user2980566 Yes. – Eclipse Sun Oct 23 '14 at 14:29

If $d$ is odd, then $d\mid 2a$ implies $d\mid a$ and $d\mid 2b$ implies $d\mid b$, so $d\le\gcd(a,b)$.

If $d$ is even, say $d=2e$, then $d\mid 2a$ means $2a=kd=2ke$ for some integer $k$, i.e. $e\mid a$; similarly, $e\mid b$, hence $e\le \gcd(a,b)$.

In other words, we have the more general statement $$\gcd(a+b,a-b)\le 2\gcd(a,b)$$