gpt 6 astra ultra attempt on Math-420

Exact variable accounting for fixed-variable Hilbert's Tenth Problem over the integers

Definitions and mathematical statement

For a fixed nonnegative integer $n$, let $\mathrm{H10}_{{\mathbb{Z}}}(n)$ be the decision problem whose input is an arbitrary finite integer-coefficient polynomial in at most $n$ variables and whose yes-instances are those with an integer zero: \[ f\in{\mathbb{Z}}[x_1,\ldots,x_n],\qquad \exists (a_1,\ldots,a_n)\in{\mathbb{Z}}^n:\ f(a_1,\ldots,a_n)=0. \] Determine the least $n$ for which this problem is undecidable, meaning no algorithm halts with the correct answer on every input. Coefficients and degree are unrestricted and part of the finite input. The number of variables is the parameter being minimized; fixing degree instead, allowing only nonnegative solutions, or replacing ${\mathbb{Z}}$ by ${\mathbb{Q}}$ changes the problem. To identify the threshold, both lower-variable decidability and undecidability at that threshold are needed.

Formulation and concept references: [S1].

Short English statement

What is the smallest fixed number of variables for which deciding existence of an integer root of an arbitrary integer polynomial becomes algorithmically impossible?

Research attempt

GPT-6 Astra Ultra; 27 September 2026. Status: UNRESOLVED.

This is a threshold question, so an undecidability construction in a small number of variables is only one side of the answer. I prove the elementary lower bound, formulate a precise height-bound obstruction, and audit the variable accounting in several natural reductions. A recent September 2026 manuscript must also be distinguished from the older bound quoted in the formulation source.

The zero- and one-variable problems are decidable.

With no variables the input is a constant integer and the answer is whether it equals zero. For one variable, first handle the zero polynomial and nonzero constants. Write a remaining polynomial as $x^r g(x)$ with $g(0)\ne0$. If $r>0$, zero is already a root. Otherwise, if $a\in\mathbb Z$ is a root, the equation \[ g(a)=c_m a^m+\cdots+c_1a+c_0=0 \] shows that $a$ divides the nonzero integer $c_0$. There are finitely many signed divisors, and evaluating at all of them decides whether an integer root exists. The divisor enumeration can be slow but terminates, which is the relevant criterion here. Consequently the least undecidable variable count $n_{\min}$ satisfies $n_{\min}\ge2$.

For comparison, linear equations in any finite number of variables are decidable. The equation $a_1x_1+\cdots+a_nx_n=b$ is soluble exactly when $\gcd(a_1,\ldots,a_n)$ divides $b$, with the all-zero coefficient case treated separately. Necessity follows by divisibility and sufficiency by the extended Euclidean algorithm. This does not settle $\mathrm{H10}_{\mathbb Z}(n)$ for that $n$, because its input degree is unrestricted. Degree bounds and variable bounds cannot be interchanged in a threshold claim.

Decidability is equivalent to a computable witness bound.

Fix $n$. Integer solubility is always semidecidable: enumerate the boxes $[-H,H]^n$ for $H=0,1,2,\ldots$ and evaluate the polynomial exactly. A solution eventually appears if one exists. The algorithm may run forever otherwise.

There is a total decision algorithm for this fixed $n$ if and only if there is a total computable function $B(f)$ such that every soluble input has a zero $a$ with \[ \max_i|a_i|\le B(f). \tag{1} \] A bound gives an algorithm by searching the finite box. Conversely, given a decision algorithm, return zero on a negative input. On a positive input, enumerate until a witness is found and return its height. The promised positive answer ensures termination of this second phase. This proves the equivalence without any complexity claim or assumption that a minimal solution is easy to find.

Therefore an undecidability proof rules out all computable bounds of the form $1$, not just polynomial or exponential bounds. Very large solutions in a finite sample do not establish undecidability: they can at most refute a particular proposed numerical bound. Conversely, searching large boxes without finding a root gives no negative certificate unless a valid bound has already been proved.

Variable accounting for elementary encodings.

Over the integers a finite system $f_1(x)=\cdots=f_r(x)=0$ is equivalent to the single equation \[ f_1(x)^2+\cdots+f_r(x)^2=0. \tag{2} \] The proof uses nonnegativity of integer squares. This step adds no variables, but can double the degree. It does not eliminate any auxiliary variables used to define the $f_i$ in the first place. The same encoding over $\mathbb C$ fails, for example at $(1,i)$ in $U^2+V^2=0$; a field change is not innocuous.

To convert $r$ nonnegative unknowns into integer unknowns, Lagrange's four-square theorem permits \[ x_i=y_{i1}^2+y_{i2}^2+y_{i3}^2+y_{i4}^2. \tag{3} \] Every right side is nonnegative, and every nonnegative integer has such a representation. Thus the direct substitution replaces $r$ unknowns by $4r$ and at most doubles the degree. This is a valid but not optimal conversion. In particular, a nine-variable theorem over the nonnegative integers does not become a nine-variable theorem over $\mathbb Z$ by changing the domain label. More efficient encodings must be proved and their witnesses counted explicitly.

There is a different bookkeeping issue in a universal polynomial $U(a,z_1,\ldots,z_m)$. If the reduction receives an input integer $a$ and substitutes it as a coefficient, the resulting equation has $m$ existential unknowns, not $m+1$. If $a$ remains existentially quantified, it is a different problem and the reduction loses the original input. Both errors can shift a purported threshold.

Why a computable pairing does not give a free compression.

Cantor pairing gives a computable bijection $\mathbb N^2\to\mathbb N$, but its inverse coordinates are not polynomials. One elementary obstruction is as follows. Suppose polynomials $P,Q\in\mathbb Q[t]$ take nonnegative values on all sufficiently large integers and satisfy \[ \frac{(P(t)+Q(t))(P(t)+Q(t)+1)}2+Q(t)=t. \tag{4} \] If both are constant, the left side is constant. Otherwise their eventual nonnegativity ensures that the leading coefficients of the nonconstant polynomials are positive, so $P+Q$ has degree $d=\max(\deg P,\deg Q)\ge1$. The left side has degree $2d\ge2$, contradicting $4$. Thus this particular computable coding does not provide polynomial decoding without further existential witnesses. The argument does not exclude all ingenious Diophantine encodings; it pinpoints a failure in one common shortcut.

The current upper-bound source must be stated carefully.

The older eleven-integer-variable theorem is recorded in [S1] and proved in Sun's earlier work [E1]. A newer manuscript, Sun's Ten unknowns for Hilbert's tenth problem over the integers, arXiv:2609.01594v2, dated 2 September 2026, states a ten-variable improvement. Its Theorem 1.1 represents every recursively enumerable set by $Q_{\mathcal A}(a,z_1,\ldots,z_{10})$ with $a$ a specialized parameter; Corollary 1.1 states undecidability with ten integer unknowns. I checked that statement and its variable/domain convention, but have not independently verified its full proof. Thus the established older bracket is $2\le n_{\min}\le11$, and the recent manuscript reports the improvement $n_{\min}\le10$. Even accepting that improvement does not identify the minimum.

Verification and outcome.

Exact finite tests check the signed-divisor root algorithm against bounded exhaustive evaluation on small polynomials, the integer sum-of-squares equivalence, and four-square coverage on a stated finite range. Those tests verify implementation examples, not the cited undecidability results. The first unresolved step toward the exact threshold is either a decision method for further fixed variable counts or a smaller-variable undecidability reduction with complete witness accounting. The proved elementary lower bound and equivalence $1$ do not bridge that gap. The requested minimum remains UNRESOLVED.

Sources

• S1 William Gasarch, Hilbert's Tenth Problem: Refinements and Variants $2021$, Notation 3.1 and Theorems 4.1-4.2. https://www.cs.umd.edu/users/gasarch/open/hilbert10openproblems.pdf. Formulation source.
• E1 Zhi-Wei Sun, Further results on Hilbert's Tenth Problem, arXiv:1704.03504. https://arxiv.org/abs/1704.03504. Earlier eleven-integer-variable theorem; consulted alongside the older bound in [S1] on 27 September 2026.
• E2 Zhi-Wei Sun, Ten unknowns for Hilbert's tenth problem over the integers, arXiv:2609.01594v2 (2 September 2026), Theorem 1.1 and Corollary 1.1. https://arxiv.org/html/2609.01594v2. Recent manuscript reporting a ten-variable improvement. The statement, parameter count, and integer domain were checked; its entire proof was not independently audited. Consulted 27 September 2026.