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 $.

If $\gcd(a,b)=1$, then there are $x,y\in\mathbb{Z}$ so that $ax+by=1$.
Then, since
$$
\begin{align}
2
&=\overbrace{[(a+b)+(a-b)]}^{\large2a}\,x+\overbrace{[(a+b)-(a-b)]}^{\large2b}\,y\\[4pt]
&=(a+b)(x+y)+(a-b)(x-y)
\end{align}
$$
we have that
$$
\gcd(a+b,a-b)\mid2
$$
Furthermore
$$
\begin{align}
1
&=(ax+by)^3\\
&=a^2(ax^3+3bx^2y)+b^2(3axy^2+by^3)
\end{align}
$$
implies that $\gcd(a^2,b^2)=1$. Apply the result above to get
$$
\gcd(a^2+b^2,a^2-b^2)\mid2
$$
