Three Integers Roots

The simplest way to find the positive integer roots of $u^2 + 3v^2 = 8068$ is trial and error! Since $3v^2 < 8068$ , we have $1 \le v \le 51$ . Just find which of $8068 - 3v^2$ are perfect squares.

A more elegant method is to consider the Euclidean domain $\mathbb{Z}[\omega]$ , where $\omega = e^{\frac{2\pi i}{3}}$ is a primitive cube root of unity. Then the distance function on this Euclidean domain is $N(a + b\omega) \; = \; |a + b\omega|^2 \; = \; a^2 - ab + b^2$ Since $2017 = |48 + 7\omega|^2$ , then the only prime factors of $2017$ are $\zeta(48 \pm 7\omega)$ , where $\zeta$ is one of the units $\pm1,\pm\omega,\pm\omega^2$ . This gives us the possible roots.