gpt 6 astra ultra attempt on Math-1022

The additive basis conjecture

Definitions and mathematical statement

For a prime $p$ and an integer $n\ge1$, let $V=\mathbb F_p^n$. An additive basis here is a finite multiset $M$ of vectors for which every $v\in V$ is the sum of a submultiset of $M$. Each occurrence can be used at most once; equal vectors occurring several times remain separately available. The empty sum is allowed and equals the zero vector. This definition differs from an ordinary vector-space basis, where arbitrary field coefficients are permitted.

The conjecture asks for a positive integer $c(p)$ depending only on $p$ such that, for every $n$ and every sequence of vector-space bases $B_1,\ldots,B_{c(p)}$ of $V$, their multiset union is an additive basis. Writing $B_i=(b_{i1},\ldots,b_{in})$, the assertion is \[ \left\{\sum_{i=1}^{c(p)}\sum_{j=1}^n \epsilon_{ij}b_{ij}:\epsilon_{ij}\in\{0,1\}\right\}=V. \] The bases need not be distinct and may share vectors. All sums and the coefficients $0,1$ are interpreted in $\mathbb F_p$.

The constant must be independent of dimension and independent of the chosen bases. A bound which increases with $n$ would not settle this question. The source also suggests the stronger candidate $c(p)=p$; this is distinguished from the principal existence assertion, which allows any finite value depending on $p$. Saying that the union "contains" an additive basis is equivalent to saying that the union itself has the subset-sum property, since additional occurrences can always be left unused.

Short English statement

For each prime, do a fixed number of arbitrary vector-space bases supply every vector as a sum using each available occurrence at most once?

Research attempt: a sharp aligned case and the Fourier dimension loss

Scope and literature check.

GPT-6 Astra Ultra, 1 October 2026. Equal vector occurrences remain separately available. The target is a constant independent of $n$. The current version of Yang Yu's preprint [S3] proves that four bases suffice for $p=3$. This is a literature result, not a result of this attempt, and does not establish the proposed $c(p)=p$ for all primes. We prove elementary special cases and locate a dimension-dependent loss.

1. A necessary lower bound and its aligned counterpart.

Take $m$ copies of the standard basis. If $m<p-1$, every subset sum has first coordinate in $\{0,1,\ldots,m\}$, and misses the residue $p-1$. Hence any universal $c(p)$ must satisfy $c(p)\ge p-1$.

In dimension one, any $p-1$ nonzero vectors suffice, even when they are different. To see this, start with $A_0=\{0\}$ and successively put $A_j=A_{j-1}\cup(A_{j-1}+b_j)$. If $A_{j-1}$ is nonempty and proper, it cannot satisfy $A_{j-1}+b_j=A_{j-1}$: translation by $b_j\ne0$ is a cycle through all $p$ field elements. Therefore $|A_j|\ge\min(p,|A_{j-1}|+1)$, and $A_{p-1}=\mathbb F_p$.

The same proof handles arbitrary dimension when every given basis contains one nonzero multiple of each member of one fixed basis $(e_1,\ldots,e_n)$. Apply the one-dimensional argument independently to the $p-1$ occurrences in each coordinate direction, and take the union of the selected occurrences. This establishes the optimal constant $p-1$ for these aligned bases. For $p=2$ every single vector space basis already has the desired subset-sum property.

2. Fourier inversion supplies a general but inadequate bound.

Assume $p$ is odd. Select every occurrence in $m$ arbitrary bases independently with probability $1/2$, and let $X$ be the resulting sum. For $\xi\in\mathbb F_p^n$, let $\chi_\xi(v)=\exp(2\pi i\langle\xi,v\rangle/p)$. The product formula for the characteristic function is \[ \mathbb E\chi_\xi(X)= \prod_{i=1}^m\prod_{j=1}^n \frac{1+\exp(2\pi i\langle\xi,b_{ij}\rangle/p)}2. \] For nonzero $\xi$, at least one scalar product is nonzero in each basis, since a basis spans the whole space. Every corresponding factor has modulus at most $\rho=\cos(\pi/p)<1$; all other factors have modulus at most one. Consequently $|\mathbb E\chi_\xi(X)|\le\rho^m$.

Orthogonality of the characters, obtained by multiplying the geometric sums in the $n$ coordinates, gives for every $v$ \[ \Pr(X=v)=p^{-n}\sum_\xi \chi_\xi(-v)\mathbb E\chi_\xi(X) \ge p^{-n}\left(1-(p^n-1)\rho^m\right). \] Thus all vectors occur whenever $m>\log(p^n-1)/(-\log\rho)$. This is a genuine finite bound for every collection of bases, but grows linearly with dimension.

Verification and first remaining gap.

The probability calculation is over labelled occurrences, so coincident basis vectors cause no ambiguity. Fourier terms may be complex; the lower bound uses their absolute values and the real-valued inversion formula. The zero character contributes exactly one. The proof cannot drop the factor $p^n-1$: it counts all nonzero characters after a triangle inequality. A dimension-free conclusion would require stronger joint information about these characters or a different method. The aligned case avoids this loss because coordinates can be filled separately.

Final status

UNRESOLVED for general $p$ and arbitrary dimension. The aligned case is proved with optimal $p-1$ bases, and an explicit dimension-dependent general bound is derived. The known $p=3$ theorem is attributed to [S3]. Completion estimate: partial results only; no global percentage.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 3060 (OPG-150), OpenGarden collection. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Open Problem Garden, The additive basis conjecture. Original problem and discussion; the mathematical formulation below is expanded from the supplied source record. http://www.openproblemgarden.org/op/the_additive_basis_conjecture.
• S3 Y. Yu, The Permanent Rank of a Matrix (Part Three): Note on the Additive Basis Conjecture, arXiv:2510.01300, current version checked 1 October 2026. https://arxiv.org/abs/2510.01300.