Problem 1335

1. Finiteness of Solutions:
We start by showing that the number of integral solutions $(x, y)$ of the equation $x^2 + xy + y^2 = n$ is finite.

Given $n = x^2 + xy + y^2$, we can use the inequality:
$$n = x^2 + xy + y^2 \leq |x^2| + |xy| + |y^2| \leq (|x| + |y|)^2$$
This implies:
$$|x|, |y| \leq \sqrt{n}$$
Therefore, the possible values for $x$ and $y$ are bounded by $\sqrt{n}$, which means there are finitely many pairs $(x, y)$ that satisfy the equation.

2. Bound on the Number of Solutions:
Since $|x|, |y| \leq \sqrt{n}$, the number of possible pairs $(x, y)$ is bounded by:
$$(2\sqrt{n} + 1)^2 \approx 4n + 4\sqrt{n} + 1$$
This gives an upper bound on the number of solutions.

3. Assignment to Complex Numbers:
To each pair $(x, y)$, we assign the complex number $x + y\zeta_6 \in \mathbb{C}$, where $\zeta_6 = \frac{1 + i\sqrt{3}}{2}$ is a primitive 6th root of unity. This assignment is unique and bijective if $x, y \in \mathbb{R}$.

4. Modulus Calculation:
We calculate the modulus:
$$|x + y\zeta_6|^2 = x^2 + xy + y^2 = n$$
This shows that if $x + y\zeta_6$ has modulus $\sqrt{n}$, then the equation $x^2 + xy + y^2 = n$ holds.

5. Generating Six Solutions:
Consider the six numbers $a_k + b_k\zeta_6 = \zeta_6^k (x + y\zeta_6)$ for $k = 0, 1, \ldots, 5$. Since $|\zeta_6| = 1$ and $\zeta_6^k = 1 \iff 6|k$, all these numbers are different and have the same modulus squared $n$.

6. Integer Solutions:
We need to show that all $a_i, b_i$ are integers. For $k = 1$:
$$a_1 = -y, \quad b_1 = x + y$$
Since $x, y \in \mathbb{Z}$, it follows that $a_1, b_1 \in \mathbb{Z}$. By similar calculations, all $a_i, b_i$ are integers.

7. Grouping Solutions:
We can group the solutions into sets of six, showing that the number of solutions is a multiple of 6.

8. Stronger Result Using Eisenstein Integers:
The number of integer solutions $n > 0$ can be written as $x^2 + xy + y^2$ (with $x, y \in \mathbb{Z}$) in exactly:
$$6 \cdot \left( \sum_{\substack{d|n \\ d \equiv 1 \mod 3}} 1 - \sum_{\substack{d|n \\ d \equiv 2 \mod 3}} 1 \right)$$
If there is one or more prime divisor $p \equiv 2 \mod 3$ dividing $n$ an odd number of times, then this count is 0. Otherwise, it equals:
$$6 \prod_{\substack{p \in \mathbb{P} \\ p \equiv 1 \mod 3}} \left( v_p(n) + 1 \right)$$

$\blacksquare$