Divisibility by Quadratics $b^2+ba+1\mid a^2+ab+1\Rightarrow\ a=b$ 
The natural numbers $a$ and $b$ are such $a^2+ab+1$ is divisible by $b^2+ba+1$. Prove that $a = b$.

I tried to algebraically manipulate it as follows:
$(b^2 + ba + 1)k = a^2 + ab + 1$
$[b(a + b) + 1]k = a(a + b) + 1$
$kb(a + b) + k = a(a + b) + 1$
$k - 1 = (a - kb)(a + b)$
I'm stuck here. What should I do next? A case-by-case analysis of possible congruencies would be too tedious and inelegant. 

If $a^2+ab+1$ is divisible by $b^2+ba+1$, then so is $(a^2+ab+1)-(b^2+ba+1)=a^2-b^2$.
Note that $a+b$ and $b^2+ba+1$ are relatively prime. So $b^2+ab+1$ divides $a-b$. Now you should be able to finish, using considerations of size.
