<s>[INST] If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions?
 [/INST] We will do this using only divisibility.
Implying that you have this knowledge.

$(a+b,a^2-ab+b^2)=\{1\;\text{and}\;3\}$ if $(a,b)=1$

Calling $d$ the $(a+b,a^2-ab+b^2)$ 
$$(a+b,a^2-ab+b^2)=d\Longrightarrow d\mid a+b\;\;\text{and}\;\;d\mid a^2-ab+b^2$$$$a^2-ab+b^2=a^2-ab+b^2\underbrace{+3ab-3ab}_{=0})=a^2+2ab+b^2-3ab=(a+b)^2-3ab$$If $d\mid a+b\Longrightarrow d\mid (a+b)^2$ as $d\mid (a+b)^2-3ab$ For these data we have $d\mid3ab$

There are three possibilities
I) $d\mid 3\Longrightarrow d=1$ or $d=3$
II) $d\mid a$ Above we see that $d\mid a+b$ this implies that $d\mid b\Longrightarrow d\mid(a,b)\Longrightarrow d\mid1\Longrightarrow d=1$
III) $d\mid b$ Above we see that $d\mid a+b$ this implies that $d\mid a\Longrightarrow d\mid(a,b)\Longrightarrow d\mid1\Longrightarrow d=1$

Thus, we see that the unique possibilities for $d$ is $1$ or $3$.
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