Sums of Three Cubes Representation Conjecture

Sums of Three Cubes Representation Conjecture

Definitions and mathematical statement

For an integer $k$ set \[ R_k=\{(x,y,z)\in{\mathbb{Z}}^3:x^3+y^3+z^3=k\}. \] Since a cube is congruent to $0,1$, or $-1$ modulo $9$, the congruences $k\equiv4,5\pmod9$ preclude a representation. The basic conjecture is the converse: \[ k\not\equiv4,5\pmod9\quad\Longrightarrow\quad R_k\ne\varnothing. \] The catalogue also mentions the stronger refinement that $R_k$ is infinite for each eligible $k$; this is a separate strength of assertion. The variables may be negative or zero, and there is no requirement of positivity or pairwise coprimality. A solution whose coordinates are large relative to $k$ is a valid representation.

Formulation and concept references: [S1].

Short English statement

Is the obstruction modulo nine the only reason an integer cannot be written as three integer cubes? A stronger version asks for infinitely many representations.

Research attempt

Process and the two different assertions.

This attempt was generated by GPT-6 Astra Ultra on 27 September 2026. We derive an exact square-discriminant reduction, prove several infinite families directly, and use the reduction as a complete search within a specified height bound. The construction treats both the existence and infinitude questions where it applies. It does not supply either conclusion for every eligible integer.

The integer conjecture is stated in the introduction of [S1], but the main results of that paper concern a function field. In the accessible 2024 version, some results also explicitly assume a ratios conjecture. Neither changing the underlying field nor dropping such a hypothesis is permissible in an argument over $\mathbb Z$. The calculations below are unconditional integer identities and finite searches; they do not use an unproved analytic conjecture.

Local obstruction and elementary symmetries.

Cubes modulo $9$ are $0,1,-1$. The possible sums of three such residues are represented by $-3,-2,-1,0,1,2,3$, leaving exactly $4$ and $5$ modulo $9$ excluded. Every other residue occurs as a sum of three residue classes, so this particular modulus yields precisely the obstruction in the statement. This is not a proof that every compatible integer has a representation.

If $(x,y,z)\in R_k$, then $(-x,-y,-z)\in R_{-k}$, and $(ax,ay,az)\in R_{a^3k}$ for every integer $a$. Coordinate permutations preserve $R_k$. Scaling produces infinitely many different represented right-hand sides from one example; unless a further parameter varies within a fixed $R_k$, that observation alone is not the stronger infinitude assertion.

A complete algebraic reduction with parity retained.

Set $d=x+y$, $v=x-y$, and $s=k-z^3$. The identities \[ x=\frac{d+v}{2},\qquad y=\frac{d-v}{2},\qquad x^3+y^3=\frac{d(d^2+3v^2)}4 \tag{430.1} \] show exactly what is required. If $d\ne0$, any representation must satisfy $d\mid s$ and \[ D(d,z)=\frac{4(k-z^3)-d^3}{3d}=v^2\ge0, \qquad v\equiv d\pmod2. \tag{430.2} \] Conversely, suppose $d,z\in\mathbb Z$, $d\ne0$, $D(d,z)$ is an integer square $v^2$, and $v,d$ have the same parity. The values of $x,y$ in $430.1$ are then integers, and substitution gives $x^3+y^3+z^3=k$. The divisibility $d\mid s$ follows as well, since $(d^2+3v^2)/4$ is an integer for equal parities. It remains useful as an inexpensive preliminary search filter.

The sign of $d$ must be retained in $430.2$. Replacing it by $|d|$ while leaving the numerator unchanged would reject legitimate negative-sum pairs or create spurious candidates. Testing merely whether the rational number $D$ is nonnegative also fails: it must be an exact square of an integer. Finally, omitting parity would give half-integral $x,y$ and change the original problem.

The excluded case $d=0$ is simple and important. It forces $y=-x$ and $z^3=k$. Thus every perfect cube $k=a^3$ has the infinite family \[ (x,y,z)=(t,-t,a),\qquad t\in\mathbb Z. \tag{430.3} \] This includes $k=0$ and uses the fact that zero or negative coordinates are allowed. No condition of primitivity appears in the catalogue statement.

Infinite families without cancelling opposite coordinates.

There is also the elementary identity \[ (1+6t^3)^3+(1-6t^3)^3+(-6t^2)^3=2. \tag{430.4} \] Indeed the first two cubes sum to $2+6(6t^3)^2=2+216t^6$, and the last cube equals $-216t^6$. For positive integers $t$, the first coordinate is strictly increasing, so $430.4$ gives infinitely many distinct ordered triples in $R_2$. Multiplying all coordinates by any fixed nonzero integer $a$ proves infinitude for $k=2a^3$; negative $a$ gives the corresponding negative right-hand sides. The case $a=0$ has already been covered by $430.3$.

A separate identity is \[ (9t^4)^3+(3t-9t^4)^3+(1-9t^3)^3=1. \tag{430.5} \] Expansion of the middle cube gives $27t^3-243t^6+729t^9-729t^{12}$, while the other two give $729t^{12}$ and $1-27t^3+243t^6-729t^9$. All nonconstant terms cancel. Again positive $t$ gives distinct triples. For those parameters none of the pair sums is zero: they are $3t$, $9t^4+1-9t^3$, and $1+3t-9t^3-9t^4$, respectively. The second is positive for $t\ge1$, and the third is negative. Thus this verifies infinitude for $k=1$ without relying on an opposite pair; scaling gives the analogous additional family for each nonzero cube right-hand side.

These identities are exhibited as checked classical-style constructions, without a novelty claim. They solve restricted subfamilies, not a positive proportion or all of the eligible integers. In particular an integer need not be a cube or twice a cube merely because it avoids the two forbidden residue classes.

A finite search with a proof of completeness inside its box.

For a fixed integer bound $B\ge0$, a representation with $|x|,|y|,|z|\le B$ has $|d|\le2B$ and $|z|\le B$. Enumerate these pairs $(d,z)$. Handle $d=0$ by $430.3$. For $d\ne0$, test divisibility, the integrality and nonnegativity of $430.2$, its integer square root, and the parity condition. Recover $x,y$ using both signs of the root, and retain only those satisfying $|x|,|y|\le B$.

Necessity of every filter follows from $430.1$--$430.2$, so no bounded representation is missed. Sufficiency follows by substitution, so none is falsely accepted. This gives an exact decision procedure for the bounded question. All arithmetic, including the square-root test, can be done with integers by checking divisibility of $4(k-z^3)-d^3$ by $3d$, followed by $r^2=D$ for the integer square root $r$.

Increasing $B$ through the positive integers is a procedure that eventually finds a solution whenever one exists. It need not halt when no solution exists. To turn it into a universal existence proof, one needs a theorem ensuring success for every eligible $k$, perhaps through a justified height bound or an independent construction. The search itself supplies neither. A large unsuccessful search is not a counterexample, since the definition places no upper bound on the coordinates.

Why two further extensions do not follow.

First, imposing congruences on $z$ to make $d\mid k-z^3$ deals with only one filter. The resulting $D(d,z)$ must still be a nonnegative square with the correct parity. There is no demonstrated distribution law for these dependent discriminants. Treating the square test as an independent random event can motivate a heuristic, but does not prove that a compatible pair occurs.

Second, a rational parametrization of points on $x^3+y^3+z^3=k$ is insufficient unless integrality is controlled. If a rational triple has common denominator $b$, clearing denominators gives integer cubes summing to $b^3k$, not necessarily to $k$. When $b$ varies, the right-hand side varies as well. The scaling symmetry explains the exact loss and prevents a rational-solubility statement from being silently upgraded to integer solubility.

The infinitude refinement needs still more than a single successful bounded search. Permuting one triple gives at most six triples. Repeated enumeration finds the same finite set unless unbounded-height solutions actually exist. The parameter families $430.3$--$430.5$ provide that unboundedness in their restricted cases; the discriminant reduction alone does not.

Verification and outcome.

The verification script expands the displayed polynomial identities exactly and compares the discriminant search against direct enumeration of all bounded triples for a fixed range of right-hand sides. It also checks the full residue table modulo $9$. These are independent checks of algebra and finite search completeness, not evidence of a height bound for the conjecture.

UNRESOLVED. We prove infinitude for cube and twice-cube right-hand sides and give an exact bounded search for every $k$. The existence of a passing discriminant for every other eligible integer, and the stronger existence of infinitely many such representations, remain unproved.

Sources

• S1 T. Browning, J. Glas, V. Y. Wang, "Sums of three cubes over a function field," arXiv:2402.07146 $2024$; published as "Optimal sums of three cubes in F_q[t]," Math. Z. $2025$. https://arxiv.org/pdf/2402.07146. Formulation source.