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.

Lemma 1: If $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$.
Proof: We will prove the contrapositive. Suppose $\gcd(a^2,b^2)>1$. Then $a^2$ and $b^2$ must have a prime common divisor, say $p$. Since $p|a^2$, Euclid's lemma implies that $p|a$. Similarly, $p|b$. Thus $p$ is a common divisor of $a,b$ and so $\gcd(a,b)>1$.
Lemma 2: If $\gcd(u,v)=1$, then $\gcd(u+v,u-v) \leq 2$.
Proof: Let $d$ be a common divisor of $u+v$ and $u-v$. Then $d|(u+v+u-v)$ and so $d|2u$. Moreover, $d|(u+v-(u-v))$ and so $d|2v$. Thus $d|\gcd(2u,2v)$. But $\gcd(2u,2v)=2\gcd(u,v)=2$. It follows that $\gcd(u+v,u-v) \leq 2$.
Given these two lemmas, we'll prove the problem statement. Since $\gcd(a,b)=1$, $\gcd(a^2,b^2)=1$ by Lemma 1. Now, by setting $u=a^2$, $v=b^2$ in Lemma 2, we can see that $\gcd(a^2+b^2,a^2-b^2) \leq 2$. 
Finally, suppose $d$ is a common divisor of $a^2+b^2$ and $a+b$. Because $d|(a+b)$, $d|(a+b)(a-b)=a^2-b^2$. So $d$ is a common divisor of $a^2+b^2$ and $a^2-b^2$. It follows that $d \leq 2$, completing the proof.
