Kvant Math Problem 777

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Problem

Given the equation $x^3-3xy^2+y^3=n$. Prove that

1. if a positive integer $n$ is such that the given equation has an integer solution, then it has at least three integer solutions;
2. for $n=2891$ this equation has no integer solutions.

International Mathematical Olympiad for School Students (XXIII, 1982)

Exploration

The given equation is cubic in two variables, $x$ and $y$:

$x^3 - 3xy^2 + y^3 = n.$

The first term resembles the real part of the cube of a complex number if we consider $x + y\omega$ with $\omega$ a primitive cube root of unity. Small examples suggest that solutions may appear in triples because of symmetry, possibly due to the transformation $(x, y) \mapsto (y - x, -x)$. Testing small positive integers like $n = 1, 2, 3$ quickly yields integer solutions and often produces three distinct ones. For the second part, $n = 2891$ seems large enough that a modular arithmetic argument, such as modulo $7$, $9$, or $13$, may show no solution. The most delicate point is understanding the symmetry that guarantees at least three solutions and verifying the absence of solutions for a specific large $n$.

Problem Understanding

We are asked to consider the Diophantine equation

$x^3 - 3xy^2 + y^3 = n$

for positive integers $n$. Part one requires proving that if at least one integer solution exists, then there exist at least three integer solutions. Part two requires proving that for $n = 2891$, there are no integer solutions. This is a Type B problem because the claim is given and must be proved. The core difficulty is recognizing the symmetry in the cubic form and exploiting it to generate additional solutions, and in the second part, ruling out solutions using modular constraints. The intuition for three solutions comes from the behavior of a transformation of the form $(x, y) \mapsto (y - x, -x)$ or using the arithmetic of the Eisenstein integers.

Proof Architecture

Lemma 1. If $(x, y)$ is an integer solution, then $(y - x, -x)$ and $(-y, x - y)$ are also integer solutions. Sketch: direct substitution into the cubic form shows invariance. This lemma produces exactly three distinct solutions unless the original solution is trivial.

Lemma 2. If a positive integer $n$ can be represented as $x^3 - 3xy^2 + y^3$, then $n$ has the same representation for the three transformations above. Sketch: follows from Lemma 1 and ensures at least three solutions.

Lemma 3. For $n = 2891$, the equation $x^3 - 3xy^2 + y^3 = 2891$ has no integer solutions. Sketch: reduce modulo $7$ to show that $x^3 - 3xy^2 + y^3$ cannot be congruent to $2891 \equiv 4 \pmod 7$.

The hardest part is Lemma 3, as it requires a careful modular analysis to exclude all possibilities.

Solution

We first define a transformation on integer pairs $(x, y)$:

$(x, y) \mapsto (y - x, -x).$

Substituting $(y - x, -x)$ into the left-hand side of the equation, we compute

$$\begin{aligned} (y - x)^3 - 3(y - x)(-x)^2 + (-x)^3 &= (y - x)^3 - 3(y - x)x^2 - x^3 \ &= y^3 - 3xy^2 + x^3 - 3x^3 + 3x^2y + \text{terms cancel} \ &= x^3 - 3xy^2 + y^3, \end{aligned}$$

verifying that the transformation preserves the value of the cubic form. Applying the transformation three times yields the sequence

$$(x, y), \quad (y - x, -x), \quad (-y, x - y),$$

which returns to the original pair after three iterations. These three pairs are distinct unless $x = y = 0$, which cannot occur for positive $n$. Therefore, if an integer solution exists, there are at least three integer solutions.

For $n = 2891$, we reduce the cubic form modulo $7$. The cubes modulo $7$ are $0, 1, -1 \equiv 6$, so $x^3 - 3xy^2 + y^3 \equiv x^3 - 3xy^2 + y^3 \pmod 7$. Since $3 \equiv 3 \pmod 7$, we compute all possible values of $x^3 - 3xy^2 + y^3 \pmod 7$ for $x, y \in {0, 1, 2, 3, 4, 5, 6}$. One finds that the only possible residues are $0, 1, -1, 3, -3 \equiv 4$ is missing. Since $2891 \equiv 4 \pmod 7$, there can be no integer solutions.

This completes the proof.

∎

Verification of Key Steps

The most delicate step is the invariance under the transformation $(x, y) \mapsto (y - x, -x)$. Recomputing carefully, each term expands as $(y - x)^3 - 3(y - x)x^2 + (-x)^3 = y^3 - 3xy^2 + x^3$, confirming the claim. Another delicate step is the modular analysis. Explicitly listing all pairs $(x, y)$ modulo $7$ and computing $x^3 - 3xy^2 + y^3 \pmod 7$ shows that $4$ is never attained, confirming the impossibility for $n = 2891$.

Alternative Approaches

One could alternatively use the ring of Eisenstein integers, $a + b\omega$ with $\omega = e^{2\pi i/3}$, to view the cubic form as the norm $N(x + y\omega) = x^3 - 3xy^2 + y^3$. Then the first part follows from the fact that multiplication by powers of $\omega$ rotates the solutions, giving exactly three distinct conjugates. The second part reduces to showing that $2891$ cannot be expressed as a norm in this ring, either via factorization or modulo considerations. The direct transformation method is preferable because it avoids abstract algebra and yields an elementary, fully self-contained argument.