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.
 A: It is known that the Eisenstein integers $\mathbb{Z}[\omega]$ is an unique factorization domain and it has six units
$$\pm 1, \pm \omega, \pm \omega^2$$
Over $\mathbb{Z}[\omega]$, the numbers $7, 13, 19$ factorize into its prime factors as
$$\begin{cases}
7  &= (3 + \omega)(3 + \omega^2)\\
13 &= (4 + \omega)(4 + \omega^2)\\
19 &= (5 + 2\omega)(5 + 2\omega^2)
\end{cases}$$
This mean if we want to factorize $1729 = 7 \times 13 \times 19$ over $\mathbb{Z}[\omega]$ as
$$1729 = ( x + y\omega )(x + y\omega^2) = x^2 - xy + y^2
\quad x, y \in \mathbb{Z}
$$
the corresponding factor $x + y\omega$ must have the form
$$x + y\omega = u A B C\quad\text{ with }\quad
\begin{cases}
A &=  3 + \omega &\text{or}& 3 + \omega^2\\
B &=  4 + \omega &\text{or}& 4 + \omega^2\\
C &=  5 + 2\omega &\text{or}& 5 + 2\omega^2
\end{cases}
$$
and $u$ is one of above six units. 
There are 8 possible choices of $A,B,C$. For each choice of $A,B,C$, 
multiply by one of the six units allow one to obtain an pair of $x,y$
that satisfies $x \ge y \ge 0$:


*

*$ABC = (3+\omega)(4+\omega)(5+2\omega) = 43+40\omega$.

*$ABC = (3+\omega)(4+\omega)(5+2\omega^2) = 45+8\omega$.

*$ABC = (3+\omega)(4+\omega^2)(5+2\omega) = 48+23\omega$.

*$ABC = (3+\omega)(4+\omega^2)(5+2\omega^2) = 32-15\omega \implies -\omega^2 ABC = (47+32\omega)$

*$ABC = (3+\omega^2)(4+\omega)(5+2\omega) = 47+15\omega$.

*$ABC = (3+\omega^2)(4+\omega)(5+2\omega^2) = 25-23\omega \implies -\omega^2 ABC = 48+25\omega$

*$ABC = (3+\omega^2)(4+\omega^2)(5+2\omega) = 37-8\omega \implies -\omega^2 ABC = 45+37\omega$

*$ABC = (3+\omega^2)(4+\omega^2)(5+2\omega^2) = 3-40\omega \implies -\omega^2 ABC =
43+3\omega$


As a result, there are $8$ pairs of $(a,b)$ that solves the original problem:
$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$
A: 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
$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$
A: Plotting $7 \times 13 \times 19 = a^2 − ab + b^2$, You will get an ellipse like this:    

.
But if we apply the condition $a$ & $b > 0$, both will be positive in only first Quadrant. And again $a > b$, we will end up with half of the ellipse in Quadrant 1 i.e. 1/8th of the total ellipse.  

.
Set of all the points on this curve is our solution. Infinite number of Real solutions.
Below are the integer solutions:     
(43,3), (43,40), (45,8), (45,37), (47,15),(47,32),(48,23),(48,25)
