<s>[INST] If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$. If $(a,b)=1$, prove that $(a^2+b^2,a+b)=1$ or $2$.
So far, I let
$d=(a^2+b^2,a+b)$
$\implies d|(a^2+b^2-(a+b)^2)$
$\implies d|(a^2+b^2-(a^2+2ab+b^2))$
$\implies d|(-2ab)$
I have heard from other people that this somehow leads to a conclusion, but I am not seeing it. Alternatively, is there some other way of showing it?
EDIT(10:41PM):
Just noticed that
$(a^2+b^2\pm (a+b)(a-b),a+b)$
results in
$d|2a^2$ and $d|2b^2$.
$\implies d|(2a^2,2b^2)$.
$\implies d|2(a^2,b^2)$.
Now I just have to fully understand why
$(a,b)=1\implies (a^2,b^2)=1$.
I sort of intuitively understand this.
 [/INST] Many gcd questions can be answered by finding a way to use the Euclidean algorithm. By the Euclidean algorithm, I mean $(a, aq+r) = (a,r)$.
Here, $a$ is played by $a+b$, $q$ is played by $a-b$, and $r$ is played by $2b^2$. The Euclidean algorithm then implies that
 $$(a+b, (a+b)(a-b) + 2b^2) = (a+b, 2b^2).$$
This reads as $(a^2+b^2, a+b) = (a+b, 2b^2)$. The conclusion should now be clear.
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