I'm don't know from where to start in this problem. I have to prove that
$(a+b,a^2-ab+b^2)=1 \text{ or } 3$
knowing that $(a,b) = 1$.
I've tried using the method they taught us on class, so I've written that:
$d\ |\ a+b \Rightarrow d\ | a^2 + 2ab + b^2$ and $d\ |\ a^2-ab+b^2 $
$\Rightarrow d\ | \ 3ab$
But from here, I dont know where to go. Usually, I should be able to find that d divides something in terms of a, and something in terms of b, and therefore, using that $(a:b)=1$ I would get the solution. But this is not the case. Any operation I've tried between the two terms, gives me an other term, but always in terms of a and b, and no a or b, so I'm not able to prove what I'm asked.
Any ideas?
