<s>[INST] If $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$. and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $ If  $a,b$ are positive integers such that $\gcd(a,b)=1$, then show that $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
Progress
We have   $\gcd(a,b)=1\implies \exists u,v\in \mathbb Z$ such that $au+bv=1\implies a^2u'+bv'=1, u',v'\in \mathbb Z$.
Let $\gcd(a+b, a-b)=d$. Then $\mid (a+b)x+(a-b)y, \forall x,y\in \mathbb Z$.
How to show $\gcd(a+b, a-b)=1$ or $2$ and $\gcd(a^2+b^2, a^2-b^2)=1$ or $2 $.
 [/INST] Let $d=\text{gcd}(a+b,a-b)$ then $d$ divides $2a$ and $2b$ so that $d|\text{gcd}(2a,2b)=2$. The latter equality is because $\text{gcd}(a,b)=1$ so, in particular, $a$ and $b$ cannot be both even. In any case, $d|2$ means $d=1$ or $2$.
Edit: this is an edit to respond to your own edit:
Let $e=\text{gcd}(a^2+b^2,a^2-b^2)$ then $e$ divides $2a^2$ and $2b^2$ so that $e|\text{gcd}(2a^2,2b^2)$. Again, the latter equals to $2$: $a$ and $b$ cannot be both even and by the Prime Factorization Theorem, $\text{gcd}(a,b)=1$ means $a^2$ and $b^2$ share no common prime factor. Again, the conclusion follows.
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