Find the number of integral ordered pairs $(a, b)$ such that $a^2 + b^2 + ab = 1$

Question

I simplified it to $(|a+b|+1)(|a+b|-1) = ab$ because I thought that i could find the no. of ordered pairs by finding the number of divisors of RHS, but the RHS isn't a constant so now I'm stuck. Please help.

Answer 1

Hint (for an elementary answer) :

$$a^2+b^2+ab=1 \Longleftrightarrow \left( 2a + b\right)^2 + 3b^2=4$$

Answer 2

Perhaps first try to find all complex numbers $z=a+\frac{1+i\sqrt 3}2b$ with $a,b\in\Bbb Z$ and $|z|=1$.

Answer 3

$$a^2 + b^2 + ab = 1\implies b = \frac{\pm\sqrt{4 - 3 a^2} - a}{2} \quad\text{for}\quad -10\le a \le 10$$

$$\implies (a,b)= (-1,1)\quad (-1,0)\quad (0,1)\quad (0,-1)\quad (1,0)\quad (1,-1)\quad $$