Eisenstein integers and applications to Diophantine equations Solve the equation $7\times 13\times 19=a^2-ab+b^2$ for integers $a>b>0$. How many are there such solutions $(a,b)$?
I know that $a^2-ab+b^2$ is the norm of the Eisentein integer $z=a+b\omega$, but how can I make use of this? Thank you so much.

Note that $N(a+b\omega)=a^2-ab+b^2$ is the sum of squares, because
$$
a^2-ab+b^2=\frac{1}{4}((2a-b)^2+3b^2).
$$
Hence we have to solve the equation $(2a-b)^2+3b^2=4\cdot 7\cdot 13\cdot 19=6916$, which is
straightforward, since we only have to test a few integers $a,b \in \mathbb{N}$. In particular, $3b^2\le 6916$, so that $b<49$. Similarly, $(2a-b)^2\le 6916$ then gives $a< 66$.
We find, that the integer solutions with $b>a>0$ are given by
$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$
