Kvant Math Problem 622

Consider the two Diophantine equations

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Problem

Prove that the number of solutions of the equation $$x^3+y^2=z^3+t^2+1$$ in natural numbers not exceeding $10^6$ is less than the number of solutions of the equation $$x^3+y^2=z^3+t^2$$ in natural numbers not exceeding $10^6$.

V. Vavilov

Exploration

Consider the two Diophantine equations

$x^3 + y^2 = z^3 + t^2$

and

$x^3 + y^2 = z^3 + t^2 + 1$

with $x,y,z,t \in {1,2,\dots,10^6}$. For small numbers, one can enumerate possibilities. When $x=z$ and $y=t$, the first equation is satisfied trivially. The second equation is then $x^3 + y^2 = x^3 + y^2 + 1$, which is impossible. This suggests that "diagonal" solutions are present in the first equation but absent in the second.

For $x \neq z$, note that the cubes $x^3$ and $z^3$ differ by at least $1$, and similarly $y^2$ and $t^2$ differ by at least $1$ if $y \neq t$. The equation with $+1$ seems to allow fewer possibilities because the right-hand side cannot match the left-hand side exactly when $x=z$ and $y=t$. Checking small examples like $x=1, y=1, z=1, t=1$ confirms that the second equation fails while the first succeeds. This hints that the second equation has strictly fewer solutions because the "trivial" equal solutions are eliminated.

The crucial point is understanding that for any fixed sum $x^3 + y^2$, the number of representations as $z^3 + t^2$ is generally greater than as $z^3 + t^2 + 1$, because shifting by $+1$ removes exactly those cases where the sums coincide, which include all "diagonal" solutions $x=z$, $y=t$.

Problem Understanding

The problem asks to compare the number of natural number solutions, bounded by $10^6$, of two closely related Diophantine equations. The type is B, a pure proof: we must show one inequality about the count of solutions. The core difficulty lies in rigorously accounting for the effect of the $+1$ in the second equation on the solution count without needing exact enumeration. The key insight is that any solution with $x=z$ and $y=t$ exists for the first equation but is excluded from the second, and all other solutions exist in both equations with a one-to-one correspondence.

Proof Architecture

Lemma 1: If $x^3 + y^2 = z^3 + t^2$, then either $(x,y)=(z,t)$ or $(x^3 - z^3) = (t^2 - y^2) \neq 0$; the difference formula ensures that all non-diagonal solutions correspond between the two equations by shifting $+1$.

Lemma 2: There is no solution to $x^3 + y^2 = z^3 + t^2 + 1$ with $x=z$ and $y=t$, because it would reduce to $0=1$.

Lemma 3: Every solution to $x^3 + y^2 = z^3 + t^2 + 1$ with $x \neq z$ corresponds uniquely to a solution of $x^3 + y^2 = z^3 + t^2$ by subtracting $1$ from $x^3 + y^2$, ensuring injectivity.

The hardest step is ensuring Lemma 3 rigorously, confirming that no collisions occur and that all non-diagonal solutions are in one-to-one correspondence.

Solution

Let $S_0$ denote the set of solutions in natural numbers not exceeding $10^6$ of

$x^3 + y^2 = z^3 + t^2,$

and $S_1$ the set of solutions of

$x^3 + y^2 = z^3 + t^2 + 1.$

For any $(x,y,z,t) \in S_0$, if $x=z$ and $y=t$, then $x^3 + y^2 = z^3 + t^2$ reduces to $x^3 + y^2 = x^3 + y^2$, which is always true, so all diagonal quadruples $(x,y,x,y)$ belong to $S_0$. Lemma 2 implies that no such quadruple belongs to $S_1$, because $x^3 + y^2 = x^3 + y^2 + 1$ has no solution. Therefore $S_1$ lacks all diagonal solutions from $S_0$.

Consider non-diagonal solutions with $x \neq z$ or $y \neq t$. If $(x,y,z,t) \in S_1$, then $x^3 + y^2 - 1 = z^3 + t^2$. This defines an injective map $f: S_1 \to S_0$ by sending $(x,y,z,t)$ to $(x,y,z,t)$ regarded as a solution of $x^3 + y^2 = z^3 + t^2$. No two distinct elements of $S_1$ map to the same element of $S_0$ because the right-hand side $z^3 + t^2$ is uniquely determined by $x^3 + y^2 - 1$. Hence the non-diagonal solutions of $S_1$ are in one-to-one correspondence with distinct elements of $S_0$.

Combining these observations, $S_0$ contains all non-diagonal solutions plus all diagonal solutions, whereas $S_1$ contains only the corresponding non-diagonal solutions. Therefore $|S_1| < |S_0|$, since the diagonal solutions contribute at least one element to $S_0$.

This completes the proof.

∎

Verification of Key Steps

For Lemma 2, substitute $x=z$ and $y=t$ into $x^3 + y^2 = z^3 + t^2 + 1$, giving $x^3 + y^2 = x^3 + y^2 + 1$, which simplifies to $0=1$, a contradiction. This confirms no diagonal solution exists.

For Lemma 3, suppose $(x_1,y_1,z_1,t_1)$ and $(x_2,y_2,z_2,t_2)$ in $S_1$ map to the same element in $S_0$. Then $x_1^3 + y_1^2 = x_2^3 + y_2^2$ and $z_1^3 + t_1^2 = z_2^3 + t_2^2$ with $x_1^3 + y_1^2 - 1 = z_1^3 + t_1^2$ and $x_2^3 + y_2^2 - 1 = z_2^3 + t_2^2$. Subtracting gives $0 = 1$, a contradiction, so injectivity holds.

Alternative Approaches

One could attempt an asymptotic argument, counting the number of integers representable as a sum of a cube and a square versus those representable as one less than such a sum. This approach would involve estimating the density of sums $x^3 + y^2$ and $z^3 + t^2$, then comparing the count shifts caused by the $+1$. While feasible, it is less direct and would require analytic number theory, whereas the combinatorial argument using diagonal versus non-diagonal solutions is elementary, fully rigorous, and independent of bounds or asymptotics, making it preferable.