Eisenstein integers and applications to Diophantine equations

Question

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.

Answer 1

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)

Answer 2

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 $$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$

Answer 3

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:

$$(a,b) = (43, 3), (43,40), (45, 8), (45, 37), (47, 15), (47,32), (48,23), (48,25)$$