According to this lemma:
$$ (x, y) = (x + ky, y) \quad \text{where } k \in \mathbb{Z} $$
Let \(x=a^2 + b^2\), \(y=a+b\) and \(k=b-a\):
$$\begin{align} (a^2 + b^2, a+b) &= (a^2 + b^2 + (b-a)(a+b), a+b) \\ &= (2b^2, a+b) \end{align} $$
According to the same lemma:
$$\begin{gather} (a, b) = (a + b, b) \\ \therefore (a + b, b) = 1 \end{gather}$$
According to the this lemma:
$$ (a + b, b) = 1 \implies (a + b, b^2) = 1 $$
If \(b^2\) and \(a+b\) don't share any common factors, then:
$$ (a + b, 2b^2) = (a + b, 2) $$
If \(a+b\) is divisible by 2, then \((a + b, 2)=2\), otherwise it's 1.