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.