Another proof that if $(a,b)=1$, then$(a+b,a-b) = 1$ or $2$

Question

I have seen this question a couple of times before on this website with very different, yet correct, proofs of this proposition. I came up with the following proof, and would like to know if it is correct.

Using that if $(a,b)=d$ then $d=ma+nb$, we have

if $(a+b,a-b) =d$, then

$$d=a+b+a-b=2a$$ $$d=a+b-(a-b)=2b$$

Thus,

$d | 2a$ and $d| 2b$

And,

$d=1$ or $d=2$

Answer 1

By Bezout $\ d := (a,b) = ma + nb\, $ for some $\,m,n,\, $ not any $\,m,n.\, $ So you cannot simply assume $\,m,n = 1,\pm1.\ $ However, since $\,(a,b)\mid a,b\,$ it is true that $\, (a,b)\mid ma+nb\,$ for all $\,m,n.\,$ Thus if you change $\, d =\ldots\, $ to $\,d\mid\ldots\,$ in your displayed equations then they are correct.

Finally you need to justify your claim that $\,d\mid 2a,2b\,\Rightarrow\, d\mid 2.\ $

Answer 2

$$d=gcd(a+b,a-b)\implies \begin{cases}d|(a+b)\\ d|(a-b)\end{cases}$$ So $$d|(a+b+a-b)\iff d|2a$$ and $$d|(a+b-(a-b))\iff d|2b$$ That is $$d|gcd(2a,2b)=2gcd(a,b)=2\implies d\in\{1,2\}.$$