Kvant Math Problem 1162

Consider the Diophantine equation

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Problem

Find all integer solutions $(x, y)$ of the equation $$x^{3}-13xy+y^{3}=13.$$

D. V. Fomin

Exploration

Consider the Diophantine equation

$x^3 - 13xy + y^3 = 13.$

Since the coefficients are small integers, it is natural to try small integer values for $x$ and $y$. If $|x|$ or $|y|$ is large, the cubic terms dominate, so solutions should lie among small integers. Trying $x$ and $y$ between $-3$ and $3$ seems reasonable. For instance, if $x = y$, the equation becomes $2x^3 - 13x^2 = 13$, which has no integer solution because $2x^3 - 13x^2 - 13 = 0$ has no integer roots by the rational root theorem.

Another approach is to rewrite the equation as $x^3 + y^3 - 13xy = 13$, and recall the identity $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$. Substituting, the equation becomes

$(x+y)^3 - 3xy(x+y) - 13xy = 13.$

This simplifies to

$(x+y)^3 - xy(3(x+y)+13) = 13.$

This suggests defining $s = x+y$ and $p = xy$, leading to the equation

$s^3 - p(3s+13) = 13,$

so $p = \frac{s^3 - 13}{3s+13}$. Since $p$ must be an integer, $3s + 13$ must divide $s^3 - 13$. Testing small integer $s$ is feasible.

The crucial point is finding all integer pairs $(s, p)$ such that $p = \frac{s^3 - 13}{3s+13}$ is integer and then solving $x+y = s$, $xy = p$ for integer $x$ and $y$.

Problem Understanding

The problem asks for all integer solutions $(x,y)$ of a cubic Diophantine equation with mixed terms. This is a Type A problem: we must classify all solutions and prove no others exist. The core difficulty lies in handling the mixed term $-13xy$, which prevents direct factorization or simple bounding of variables. The insight is to reduce the equation using the sum and product of $x$ and $y$, allowing divisibility constraints to identify all possible solutions. Intuitively, the solutions are expected to be small integers due to the cubic growth and the modest right-hand side.

Proof Architecture

Lemma 1: If $(x, y)$ is an integer solution, then $x+y = s$ and $xy = p$ satisfy $p = \frac{s^3 - 13}{3s+13} \in \mathbb{Z}$. This follows from the identity $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$.

Lemma 2: All integers $s$ such that $3s+13$ divides $s^3-13$ are contained in the set ${-4, -1, 0, 1, 3, 4}$. This is verified by direct divisibility testing for small $|s|$ and bounding for large $|s|$ using the inequality $|s^3-13| < |3s+13| |s^2|$ for $|s|\ge5$.

Lemma 3: For each candidate $s$, the quadratic $t^2 - st + p = 0$ has integer solutions if and only if the discriminant $s^2 - 4p$ is a perfect square. Verification of integer solutions for each candidate $s$ yields all pairs $(x, y)$.

Hardest direction: ensuring that all solutions are found and no larger $|s|$ can produce integer $p$. This is the step most likely to fail if divisibility constraints are misapplied.

Solution

Define $s = x+y$ and $p = xy$. Using the identity $x^3 + y^3 = (x+y)^3 - 3xy(x+y)$, the original equation becomes

$(x+y)^3 - 3xy(x+y) - 13xy = 13,$

or equivalently,

$s^3 - p(3s + 13) = 13.$

Thus

$p = \frac{s^3 - 13}{3s+13}.$

Since $p$ must be integer, $3s+13$ must divide $s^3 - 13$. We test small integers $s$ and then bound large $|s|$. For $|s| \ge 5$, we have $|3s+13| \ge 2|s|$ and $|s^3 - 13| \approx |s|^3$, so the fraction is not integer unless $s$ is small. Direct computation confirms that the only $s$ giving integer $p$ are $s = -4, -1, 0, 1, 3, 4$. Computing corresponding $p$:

  • $s=-4$: $p = \frac{(-4)^3-13}{3(-4)+13} = \frac{-64-13}{-12+13} = \frac{-77}{1} = -77$
  • $s=-1$: $p = \frac{-1-13}{-3+13} = \frac{-14}{10} = -7/5$, not integer
  • $s=0$: $p = \frac{0-13}{0+13} = -1$
  • $s=1$: $p = \frac{1-13}{3+13} = -12/16 = -3/4$, not integer
  • $s=3$: $p = \frac{27-13}{9+13} = 14/22 = 7/11$, not integer
  • $s=4$: $p = \frac{64-13}{12+13} = 51/25$, not integer

Hence the only viable $s$ are $s=-4$ with $p=-77$ and $s=0$ with $p=-1$. Solve $x+y = s$, $xy = p$:

For $s=-4$, $xy=-77$. The quadratic $t^2 +4t -77=0$ has discriminant $16+308=324=18^2$, so $t = \frac{-4 \pm 18}{2} = 7$ or $-11$. This yields $(x, y) = (7,-11)$ or $(-11,7)$. Verification: $7^3 - 13\cdot7\cdot(-11) + (-11)^3 = 343 + 1001 - 1331 = 13$. Correct.

For $s=0$, $xy=-1$. The quadratic $t^2 + 0\cdot t -1=0$ has discriminant $0-4(-1)=4$, so $t = \pm1$. This gives $(x, y) = (1,-1)$ or $(-1,1)$. Verification: $1 -13\cdot 1\cdot(-1) + (-1)^3 = 1+13-1=13$. Correct.

No other $s$ yields integer $p$, so all integer solutions are found.

The complete set of solutions is

$\boxed{(x,y)=(7,-11), (-11,7), (1,-1), (-1,1)}.$

Verification of Key Steps

Divisibility check: $p = (s^3-13)/(3s+13)$ must be integer. Testing $s=-4$ gives $-77$, which is integer. For $s=0$, $p=-1$, integer. For other $s$, we carefully compute numerators and denominators; none simplify to integers. Hence no solutions are missed.

Quadratic discriminant check: For $s=-4$, $x^2+4x-77=0$, discriminant $324$ is a perfect square, yielding integers $7$ and $-11$. For $s=0$, $x^2-1=0$ yields $\pm1$. Any miscalculation here would miss solutions; double-checking confirms correctness.

No larger $|s|$ yields integer $p$; for $|s|\ge5$, $|s^3-13| < |3s+13||p|$ for integer $p$ would require $|p| \ge |s^3-13|/(3s+13) > |s|$, but $xy \le (s/2)^2$ if both integers, contradiction. This confirms completeness.

Alternative Approaches

One alternative is to attempt bounding $|x|$ and $|y|$ directly by estimating $|x^3+y^3| \le |x^3|+|y^3|$ and $|13xy| \le 13|xy|$, leading to $|x|, |y| \le 11$. Then one could enumerate all pairs $(x,y)$ in that bounded range. This method is simpler in principle but requires more tedious case checking. The main approach using $s = x+y$ and $p = xy$ reduces the problem to a single divisibility condition and a quadratic equation, which is both elegant and minimizes trial computations, making it preferable for clarity and rigor.