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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

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  • $\begingroup$ 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 $d/2$? $\endgroup$ – anomaly Oct 23 '14 at 14:21
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Since $d|2a$ and $d|2b$, we have $d|\text{gcd}(2a,2b) = 2$. Hence $\text{gcd}(b+a,b-a) \le 2$.

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  • $\begingroup$ gcd(2a,2b) = 2 because gcd(a,b) = 1 right? $\endgroup$ – user2980566 Oct 23 '14 at 14:23
  • $\begingroup$ @user2980566 Yes. $\endgroup$ – Eclipse Sun Oct 23 '14 at 14:29
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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 $$ \gcd(a+b,a-b)\le 2\gcd(a,b)$$

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