gpt 6 astra ultra attempt on Math-60

Bateman-Horn conjecture

Definitions and mathematical statement

For fixed distinct nonconstant irreducible $f_1,\ldots,f_r\in{\mathbb{Z}}[x]$ of positive leading coefficient, let $d_i=\deg f_i$ and let $N_p$ count roots of $\prod_i f_i$ modulo $p$. Admissibility means $N_p<p$ for every prime. The Bateman--Horn prediction is \[ \#\{1\le n\le x:f_i(n)\text{ all prime}\} \sim\frac{C(f)}{\prod_i d_i}\frac{x}{(\log x)^r},\qquad C(f)=\prod_p\frac{1-N_p/p}{(1-1/p)^r}. \] Equivalently use $C(f)\int^x dt/\prod_i\log f_i(t)$, starting where all values exceed one. The polynomials are fixed before $x$ tends to infinity.

Formulation and concept references: [S1].

Short English statement

Do simultaneous prime values of fixed admissible polynomials follow the exact density predicted by Bateman and Horn?

Research attempt

Scope and checked context.

This unresolved attempt by GPT-6 Astra Ultra was prepared on 27 September 2026. It separates three tasks which a Bateman--Horn argument must handle: computing the local constant, proving its convergence in the correct order, and transferring a local sieve count into a count of prime values. The worked nonlinear case is the single polynomial $f(n)=n^2+1$. Even infinitude of its prime values is not proved here.

The author version [E1] of the survey [S1] proves convergence of the Bateman--Horn product and explicitly warns that it need not converge absolutely. Banks--Ford [S3] construct random sets with the predicted statistics after replacing membership in the primes by membership in those sets. That statement is not a proof about the actual primes. The inherited preprint [S2] is not invoked as a theorem. The elementary reductions below make no claim of historical novelty.

Finite local arithmetic is exact.

Let $F=\prod_i f_i$ and put $P(z)=\prod_{p\leq z}p$. Write \[ S(x,z)=\#\{1\leq n\leq x:\gcd(F(n),P(z))=1\}. \] For squarefree $d\mid P(z)$, let $N_d$ be the number of roots of $F$ modulo $d$. The Chinese remainder theorem gives $N_d=\prod_{p\mid d}N_p$, including $N_1=1$. Counting each residue class in an initial interval and using inclusion--exclusion yields \[ S(x,z)=x\prod_{p\leq z}(1-N_p/p) +O\left(\prod_{p\leq z}(1+N_p)\right). \tag{83.1} \] Indeed the error for each $d$ has magnitude at most $N_d$, and the sum of these errors over divisors is the displayed product. In particular, for fixed $z$ this is an unconditional density calculation.

Admissibility is also a finite test for a fixed tuple. Irreducibility of each nonconstant polynomial in $\mathbb Z[x]$ implies its coefficients have greatest common divisor one. Thus no reduction is the zero polynomial modulo any prime. If $D=\deg F$, then $N_p\leq D$ for every prime, so primes $p>D$ cannot violate $N_p<p$. One need only check the finitely many primes at most $D$. This certifies lack of a fixed prime divisor, not infinitude of simultaneous prime values.

The quadratic constant is conditionally convergent.

For $f(n)=n^2+1$ one has $N_2=1$. For an odd prime, \[ N_p=1+\chi_4(p),\qquad \chi_4(p)=\begin{cases}1&p\equiv1\pmod4,\\-1&p\equiv3\pmod4. \end{cases} \] For example a root of $x^2=-1$ gives an element of order four in the cyclic group $\mathbb F_p^\times$, and such an element exists exactly when $4\mid p-1$; there are then two roots. The local factors become \[ C_2=1,\qquad C_p=1-\frac{\chi_4(p)}{p-1}\quad(p>2). \tag{83.2} \] They are all positive. For $p\geq5$, Taylor's formula with a uniform remainder gives \[ \log C_p=-\frac{\chi_4(p)}p+O(p^{-2}). \tag{83.3} \] Thus convergence of the logarithmic product depends on cancellation between the two residue classes. The general convergence theorem in [E1, Section 5] supplies this cancellation in increasing prime order; its proof uses prime-ideal counting. Positivity alone does not supply it.

In fact this example proves that an absolute-convergence claim would be wrong: $|C_p-1|=1/(p-1)$ at every odd prime, and $\sum_p1/p$ diverges. Nor can one bound the logarithmic tail merely by $\sum_{p>z}p^{-2}$: the first-order term in $83.3$ remains. For a general tuple the corresponding expansion is \[ \log\frac{1-N_p/p}{(1-1/p)^r} =\frac{r-N_p}{p}+O_F(p^{-2}), \tag{83.4} \] once $p$ exceeds a fixed multiple of $D+r$. Linear prime patterns often have $N_p=r$ outside a finite exceptional set; nonlinear polynomials do not generally have that property. This is a precise obstruction to copying an absolutely convergent linear-pattern product argument.

Why removing small prime factors does not yet count primes.

For $n\leq x$, every composite $n^2+1$ has a prime factor at most $\sqrt{x^2+1}$. Taking that sieve level would therefore remove composites. It would also remove prime values at most $\sqrt{x^2+1}$, but these arise only for $n\ll\sqrt x$, an error negligible compared with $x/\log x$. Thus a sharp evaluation of $83.1$ at a level of order $x$ could be useful. The issue is that its error term is not small at that level. It contains a factor $3$ for each prime $p\equiv1\pmod4$ up to the level, so even this elementary inclusion--exclusion estimate offers no useful asymptotic.

Taking a fixed or slowly growing level controls the error more easily, but leaves integers with two large prime factors among the survivors. The implication "no small prime factor, hence prime" requires a level tied to the size of the values, not merely to the fact that the local product has stabilized numerically. A graph of partial products is therefore evidence about the proposed constant, not about the number of prime values.

A concrete weighted reduction, including prime powers.

Define \[ W(x)=\sum_{1\leq n\leq x}\Lambda(n^2+1),\qquad P_f(x)=\#\{n\leq x:n^2+1\text{ is prime}\}. \] There is no square among $n^2+1$ for $n\geq1$: it lies strictly between $n^2$ and $(n+1)^2$. If $n^2+1=p^k$ with $k\geq3$, then $p\leq(x^2+1)^{1/k}$. For each $p,k$ there is at most one positive $n$. Summing over $k\leq\log_2(x^2+1)$ bounds the number of these terms by $O(x^{2/3}\log(2x))$. Each has weight at most $\log(x^2+1)$, so \[ W(x)=\sum_{\substack{n\leq x\\n^2+1\ \mathrm{prime}}} \log(n^2+1)+O(x^{2/3}\log^2(2x)). \tag{83.5} \] This error is $o(x)$. Consequently the weighted assertion \[ W(x)\sim C(f)x \tag{83.6} \] would imply exactly the Bateman--Horn prediction in this case: \[ P_f(x)\sim\frac{C(f)}2\frac{x}{\log x}. \tag{83.7} \] To see this rigorously, call the prime-only weighted sum $B(x)$. Partial summation with $g(t)=\log(t^2+1)$ gives \[ P_f(x)=\frac{B(x)}{g(x)}+ \int_1^x\frac{B(t)g'(t)}{g(t)^2}\,dt. \] Under $83.6$, $83.5$ gives $B(t)=C(f)t+o(t)$ and in particular $B(t)=O(t)$. The integral is $O(x/\log^2x)+O(1)$, while $g(x)\sim2\log x$. This proves $83.7$. Conversely, applying partial summation in the other direction to $83.7$, and then $83.5$, gives $83.6$. The degree factor $1/2$ is therefore accounted for rather than absorbed into the constant.

The divisor expansion locates the missing arithmetic cancellation.

The identity $\Lambda(m)=-\sum_{d\mid m}\mu(d)\log d$ gives the finite, fully justified expression \[ W(x)=-\sum_{d\leq x^2+1}\mu(d)\log d\, \#\{n\leq x:n^2+1\equiv0\pmod d\}.\tag{83.8} \] For squarefree $d$, replacing the count by $xN_d/d$ introduces an error bounded termwise by $N_d\log d$. Summing those bounds up to $x^2+1$ does not produce $o(x)$. Truncating at a smaller $D$ leaves the exact remainder with $D<d\leq x^2+1$; it cannot be dropped simply because $\mu(d)$ has both signs. The desired cancellation involves the positions of polynomial roots in the incomplete interval $[1,x]$, not just their number in a complete residue system.

Verification and outcome.

The companion code checks root counts and $83.2$ at small primes, the Chinese-remainder multiplicativity and $83.1$ for a finite sieve, and the weighted identity $83.8$ on an explicit range. Its separate enumeration finds no proper-prime-power values for $1\leq n\leq200$; this finite outcome is not a proof of their absence at all heights. The uniform error estimate in $83.5$ is proved independently above. All finite identities use integer arithmetic or exact prime-exponent vectors; numerical logarithms are used only for optional displayed values, not equality certification.

The proved results are the finite local count, the failure of absolute convergence in a basic nonlinear example, and a weighted equivalence with a controlled prime-power error. The first unresolved estimate is $83.6$, or an equivalent evaluation of the full remainder in $83.8$. Neither the convergent constant nor admissibility provides that estimate. The Bateman--Horn conjecture remains UNRESOLVED.

Sources

• S1 S. L. Aletheia-Zomlefer, L. Fukshansky, and S. R. Garcia, Expositiones Mathematicae 38 $2020$, 430-479. https://www.sciencedirect.com/science/article/am/pii/S0723086918301178. Formulation source.
• S2 From Golomb to Bateman-Horn. https://www.preprints.org/manuscript/202512.2666. Further reference; not a proof endorsement.
• S3 Sets of integers satisfying Bateman-Horn statistics. https://arxiv.org/abs/2605.01155. Further reference; not a proof endorsement.

• E1 S. Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, The Bateman--Horn conjecture: Heuristics, history, and applications, arXiv:1807.08899, especially Section 5 and Theorem 5.4.3. https://arxiv.org/pdf/1807.08899. Author full text corresponding to [S1], inspected for convergence of the singular product and its possible failure of absolute convergence. Consulted 27 September 2026.
• E2 William Banks and Kevin Ford, Sets of integers satisfying Bateman--Horn statistics, arXiv:2605.01155v1 (1 May 2026). https://arxiv.org/html/2605.01155v1. Version-specific scope check of [S3]: the theorem concerns random replacement sets, not a proof of the conjecture for primes. The full proof was not independently audited or used here. Consulted 27 September 2026.