Prove a statement about the GCD of $a$ and $b$ Prove that if $\gcd(a, b) = 1$, then $\gcd(a - b, a + b)$ must be $1$ or $2$.
I have it if $a$ and $b$ are odd, but what if they aren't? 
 A: If $a$ and $b$ are both even, then $\gcd(a, b) \geq 2$. Therefore only $a$ or only $b$ can be even in order for $\gcd(a, b) = 1$ to be true. Without loss of generality, let's say $a$ is even and $b$ is odd. Then $a - b$ is odd and so is $a + b$.
But if $a$ and $b$ are both odd, then $a - b$ is even and so is $a + b$.
Note that $a - b = (a + b) - 2b$ and likewise $a + b = (a - b) + 2b$. In other words, the difference between $a - b$ and $a + b$ is $2b$.
Well, I think you can take it from here...
A: How did you get it for $a$ and $b$ odd?  
But note $a$ and $b$ can't both be even else $\gcd(a,b)$ is even.  But if one of them is even, say $a$, and the other isn't and so $2$ is not a factor in common so $\gcd(a,b) = \gcd(\frac a2, b)$.
Which will reduce to a case where they are both odd.
But... you should have been able to get it without assuming $a,b$ were odd.
Note:  $\gcd(j,k) = \gcd(j, j \pm k)$. 
So $\gcd(a+b, a-b) = \gcd(a+b, a+b + a-b)=\gcd(a+b, 2a)$ and $\gcd(a+b,a-b)=\gcd(a+b, a+b - (a-b)) = \gcd(a+b, 2b)$.
So if $d = \gcd(a+b, a-b)$ let $d|2a$ and $d|2b$ but $a$ and $b$ have no factors in common.  
So $d|2$.
