gpt 6 astra pro attempt on Math-130

Dickson's Conjecture

Definitions and mathematical statement

For $f_i(n)=a_in+b_i$ with integers $a_i\ge1$, define admissibility by the absence of a fixed prime divisor of $\prod_i f_i(n)$. Explicitly, \[ \forall p\in\mathcal P\ \exists n\in{\mathbb{Z}}:\quad p\nmid\prod_i(a_in+b_i). \] Dickson's assertion is $\forall X\ \exists n>X$ with all $a_in+b_i$ prime. The number of forms is finite and fixed for each assertion. Positivity for large $n$ follows from the positive slopes. The record does not require distinct forms or an asymptotic counting formula.

Formulation and concept references: [S1].

Short English statement

Must every locally admissible finite family of positive-slope linear forms produce simultaneous primes arbitrarily often?

Research attempt

Current frontier.

The formulation allows repeated forms and requires every form to be prime infinitely often [S1]. Dirichlet's theorem settles one primitive positive-slope form. Maynard's multidimensional sieve produces many primes in sufficiently large admissible tuples, not simultaneous primality of every member of any prescribed tuple [E1]. The other catalogue references are preserved; the listed proof-claim preprint is not used as a verified resolution.

Attack route.

Better distribution in progressions can improve sieving, but does not by itself distinguish the required factor-count parity. Averaging over coefficients needs a separate uniformity argument to recover every fixed tuple. We pursue an exact parity expansion after sieving to numbers with at most two prime factors, and test whether lower-order sign moments could be enough.

Attempt: local preprocessing.

Admissibility forces $\gcd(a_i,b_i)=1$ for each form. Delete repeated forms, which does not change the prime-producing set. Distinct remaining primitive coefficient pairs with positive slope cannot be proportional over ${\mathbb{Q}}$. Write $k$ for their number and \[ F(n)=\prod_{i=1}^k(a_in+b_i),\qquad \nu(p)=\#\{n\bmod p:F(n)=0\bmod p\}. \] Each form is nonzero modulo $p$ and has at most one root, so $\nu(p)\le k$. Once individual primitivity is known, admissibility therefore needs checking only for $p\le k$. For squarefree $d$, the Chinese remainder theorem gives \begin{equation}\label{wsref-0-72-36-32-3a-63-72-74} \#\{X<n\le2X:d\mid F(n)\} =\frac{\nu(d)}dX+O(\nu(d)),\qquad \nu(d)=\prod_{p\mid d}\nu(p). \end{equation} This follows by counting each residue class; it supplies no cancellation after a long sum over moduli.

An exact two-factor reduction.

Fix $1/3<\theta<1/2$ and let $z=X^\theta$. For large $X$, all $f_i(n)$ in $X<n\le2X$ lie in $[cX,CX]$ for fixed $c,C>0$. Define \[ \mathcal R_X=\{n\in{\mathbb{Z}}:X<n\le2X,\ p\nmid F(n) \text{ for all primes }p\le z\},\qquad R_X=|\mathcal R_X|. \] Every prime factor of each value on this set exceeds $z$. Three prime factors counted with multiplicity would give a value greater than $z^3>CX$. Hence \[ \Omega(f_i(n))\in\{1,2\},\qquad \symbf{1}_{\{f_i(n)\text{ prime}\}}=\frac{1-\lambda(f_i(n))}{2}. \] Prime squares have Liouville value $+1$ and are correctly excluded. Values at most one are absent for large $X$.

Let $P_X$ count simultaneous prime values in this interval and put \[ M_S(X)=\sum_{n\in\mathcal R_X}\prod_{i\in S}\lambda(f_i(n)), \quad S\subseteq[k],\qquad M_\varnothing=R_X. \] Every simultaneous-prime tuple is rough since $cX>z$. Expanding the exact indicator proves \begin{equation}\label{wsref-0-72-36-32-3a-65-78-61-63-74} P_X=2^{-k}\sum_{S\subseteq[k]}(-1)^{|S|}M_S(X). \end{equation}

A bias-aware sufficient estimate.

Sifting changes the ratio of primes to semiprimes; first moments need not vanish. Choose a fixed $a\in(0,1)$ and define \[ C_S(X)=\sum_{n\in\mathcal R_X}\prod_{i\in S}(\lambda(f_i(n))+a). \] The proposed sign mean is $-a$, not asserted here as an arithmetic fact. Algebra yields \begin{equation}\label{wsref-0-72-36-32-3a-63-65-6e-74-65-72} P_X=2^{-k}\left((1+a)^kR_X+ \sum_{\varnothing\ne S\subseteq[k]}(-1)^{|S|}(1+a)^{k-|S|}C_S(X)\right). \end{equation}

\begin{notebookproposition} Suppose along an unbounded sequence of $X$ one has $R_X\to\infty$ and \[
 |C_S(X)|\le\eta^{|S|}R_X\quad(\varnothing\ne S\subseteq[k])
\] for fixed $a\in(0,1)$ and $0\le\eta<(2^{1/k}-1)(1+a)$. Then this tuple satisfies Dickson's conclusion. \end{notebookproposition}

The required lower bound for the rough set and the full set of centered moment estimates are both unproved. Neither is inferred from (r62:crt).

Stress test: every proper-subset moment can look random.

For $k\ge2$, put the uniform probability measure on sign vectors satisfying \[ \epsilon_1\cdots\epsilon_k=-(-1)^k. \] Every nonempty proper subset product has mean zero. Flip one coordinate in the subset and one outside it: this preserves the total product and reverses the subset product, pairing all vectors. Nevertheless the all-minus vector has probability zero. Perfect lower-order independence in this unbiased abstract model does not force the desired pattern. This is not an arithmetic counterexample or a refutation of the biased sufficient criterion; it identifies why its top-order information cannot simply be discarded.

The exponent $\theta>1/3$ is essential to the exact indicator. Without the exclusion of three factors, the same Liouville formula also counts products of three or five primes. Conversely, moving $\theta$ to $1/2$ cannot be inserted without rechecking the sieve estimates. The script tests (r62:exact) at $\theta=2/5$, $X=1000,10000$, for $(n,n+2)$ and $(n,n+2,n+6)$, using exact factorization and all subset sums. It also checks the sign-model obstruction. No limiting conclusion is drawn.

Exact finite output.

The computations produce the following counts; all prime counts agree with the full alternating moment expansion: \[ \begin{array}{c|c|r|r} X&\text{offsets of the forms}&R_X&P_X\\ \hline 1000&(0,2)&52&26\\ 1000&(0,2,6)&23&9\\ 10000&(0,2)&299&137\\ 10000&(0,2,6)&90&28 \end{array} \] These four finite identities neither prove the growth of $R_X$ nor certify the moment estimates along an unbounded sequence.

Outcome.

Outcome: Exploratory approach with unresolved gap.

The attempt gives an exact rough-set parity expansion, a bias-aware sufficient condition, and a finite model showing why all proper-subset moments can miss the target event. The analytic lower bound and full moment estimates remain missing. This is a precise obstruction analysis, not a new theorem on prime values. Resolving all tuples would in particular settle twin primes and the linear part of the earlier Schinzel entry.

Sources

• S1 Shaohua Zhang, Notes on Dickson's Conjecture, arXiv:0906.3850v3 $2009$. https://arxiv.org/html/0906.3850. Catalogue formulation source.
• S2 An exposition on the supersimplicity of certain expansions of the additive group of the integers. https://arxiv.org/html/2502.03915. Catalogue research/status reference; not a proof endorsement.
• S3 From Golomb to Schinzel. https://www.preprints.org/manuscript/202505.0651. Catalogue research/status reference; not a proof endorsement.

• E1 James Maynard, Small gaps between primes, Annals of Mathematics 181 $2015$, 383--413; arXiv:1311.4600, main admissible-tuple results. https://arxiv.org/abs/1311.4600. Several prime values in suitable large tuples, not all members of every fixed tuple. Consulted 27 September 2026.