Polignac's Conjecture

Polignac's Conjecture

Definitions and mathematical statement

Write $p_1<p_2<\cdots$ for the increasing sequence of primes. Polignac's assertion is \[ \forall k\in{\mathbb{Z}}_{>0}\ \forall N\ \exists n>N:\qquad p_{n+1}-p_n=2k. \] Consecutiveness means there is no other prime between the pair. Finding infinitely many pairs of primes at distance $2k$ without this condition would be a different and potentially weaker assertion for $k>1$.

Formulation and concept references: [S1].

Short English statement

Choose any positive even gap. Must that exact spacing recur between neighboring primes arbitrarily far along the prime sequence?

Research attempt

Current frontier.

The established bounded-gap theorem gives infinitely many neighboring prime pairs with a gap at most $246$; it does not specify each even gap. A preprint of Stadlmann, posted on 31 August 2026, reports the improved bound $240$ by combining distribution estimates. This recent result is recorded as a preprint, not independently verified here, and is not represented as a resolution of any prescribed-gap assertion. [E1][E2] Maynard's sieve explains how to find several primes among sufficiently many admissible translates; it need not make two specified translates simultaneously prime. [E3]

Attack route.

I separate three issues often conflated in this problem: obtaining prime endpoints, eliminating interior primes, and making the endpoint argument work for every fixed even gap. There are two ways to handle the middle issue. One can show that interior-prime exceptions are rarer than the expected endpoint pairs, or prearrange composite interior values using congruences. I develop both reductions, then test whether the elementary two-form Maynard variational problem can supply their missing endpoint input. Neither reduction assumes consecutiveness at the outset.

Attempt.

1. An endpoint-to-consecutive-gap estimate. Fix an even $h\geq2$. For $X>h$, define \begin{align*} P_h(X)&=\#\{p:\ p,p+h\text{ prime},\ p+h\leq X\},\\ G_h(X)&=\#\{p:\ p,p+h\text{ consecutive primes},\ p+h\leq X\},\\ T_{h,a}(X)&=\#\{p:\ p,p+a,p+h\text{ prime},\ p+h\leq X\} \quad(1\leq a<h). \end{align*} An endpoint pair is excluded from $G_h$ only if some intermediate translate is prime. The union bound therefore gives the exact inequalities \begin{equation}\label{wsref-0-31-31-35-33-3a-75-6e-69-6f-6e} 0\leq P_h(X)-G_h(X)\leq\sum_{a=1}^{h-1}T_{h,a}(X). \end{equation} For each fixed $a$, a three-dimensional upper-bound sieve gives \begin{equation}\label{wsref-0-31-31-35-33-3a-74-72-69-70-6c-65} T_{h,a}(X)\ll_h \frac{X}{(\log X)^3}. \end{equation} Here are the hypothesis checks. If $\{0,a,h\}$ is admissible, the three linear forms have nonzero discriminant factor $ah(h-a)$ and occupy fewer than $\ell$ residue classes for every prime $\ell$. Apply the standard prime-tuple upper bound, for example the $K=3$ instance of Theorem 1 in [E4]. If the triple is inadmissible, choose a prime $\ell$ at which its residues cover every class. One of $p,p+a,p+h$ must then equal $\ell$, so there are at most three candidate values of $p$. Thus this case is bounded as well. The number of shifts is fixed with $h$; no uniformity in growing $h$ is being claimed.

Consequently there is a constant $C_h$ such that for all sufficiently large $X$, \begin{equation}\label{wsref-0-31-31-35-33-3a-67-61-70-63-6f-75-6e-74} P_h(X)-C_h\frac{X}{(\log X)^3}\leq G_h(X)\leq P_h(X). \end{equation} This estimate isolates a quantitative input sufficient for the selected problem: \begin{equation}\label{wsref-0-31-31-35-33-3a-70-61-69-72-69-6e-70-75-74} \text{for every fixed even $h\geq2$,}\qquad \limsup_{X\to\infty}\frac{P_h(X)(\log X)^3}{X}=\infty. \end{equation} Indeed choose arbitrarily large $X$ for which the ratio exceeds $C_h+1$. Then $G_h(X)\geq X/(\log X)^3\to\infty$, which gives infinitely many consecutive gaps $h$. Any lower bound $P_h(X)\gg_h X/(\log X)^2$ along an unbounded sequence of $X$ would suffice. Mere infinitude of $P_h$ would not: an infinite but sufficiently sparse set could fit entirely inside the exception bound in (1153:gapcount).

For the edge cases, $h=2$ permits no intermediate prime, so $P_2=G_2$. For $h=4$, the only nonconsecutive prime pair is $(3,7)$: otherwise an interior prime would produce three primes $p,p+2,p+4$, one divisible by $3$. Thus $P_4(X)-G_4(X)=1$ for $X\geq7$. The general quantitative input is a sufficient statement, not a claim that these easy edge cases require it.

2. Congruence preconditioning with no density assumption. A different reduction eliminates the interior primes in advance. For every even integer $a$ with $2\leq a\leq h-2$, choose a distinct prime $q_a>h$. Set \[ M=2\prod_{a\in\{2,4,\ldots,h-2\}}q_a. \] By the Chinese remainder theorem choose $0\leq r<M$ with \begin{equation}\label{wsref-0-31-31-35-33-3a-63-72-74} r\equiv1\pmod2,\qquad r\equiv-a\pmod{q_a}. \end{equation} The two linear forms \[ L_0(t)=Mt+r,\qquad L_h(t)=Mt+r+h \] are jointly admissible. If $\ell\mid M$, neither endpoint vanishes modulo $\ell$: for $\ell=q_a$ their residues are $-a$ and $h-a$, both nonzero because $0<a<h<q_a$; at $2$ both are odd. If $\ell\nmid M$, the two forms exclude at most two classes modulo $\ell$, and $\ell\geq3$. This also proves the primitivity of both forms.

If both endpoints are prime and $t$ is large enough, each interior value is composite. Odd $a$ gives an even integer larger than $2$. Even $a$ gives a multiple of its assigned $q_a$ larger than $q_a$. Therefore:

Infinitely many simultaneous prime values of $L_0,L_h$ imply infinitely many consecutive gaps of size $h$.

This is an unconditional implication with an unproved simultaneous-primality hypothesis. In the language of prime tuples, the two-form Dickson assertion for these explicit progressions would suffice. No asymptotic count is needed for this second reduction. The prime-tuple framework and its distinction from the presently known many-translates sieve are discussed in [E3][E4].

For a concrete check, at $h=6$ choose $q_2=7$, $q_4=11$. Then $M=154$ and $r=117$ work. At $t=1$ the endpoints are $271$ and $277$; their five interior values are composite, with $273$ divisible by $7$ and $275$ by $11$. One such instance checks the construction, not its infinitude.

There is also no hidden vanishing singular series. Writing $\nu_\ell$ for the number of roots of $L_0L_h$ modulo $\ell$, each local factor \[ (1-\nu_\ell/\ell)(1-1/\ell)^{-2} \] is positive. Outside finitely many primes $\nu_\ell=2$, and the factor is $1+O(\ell^{-2})$; hence their infinite product converges to a positive number. Positivity is a consistency check on a proposed asymptotic, not a proof that either endpoint is prime infinitely often together with the other.

3. A sharper obstruction in the basic two-form sieve. Let $\mathcal R=\{(s,t):s,t\geq0,\ s+t\leq1\}$ and let $F\in L^2(\mathcal R)$ be nonzero. Extend it by zero off $\mathcal R$, and write \begin{align*} I(F)&=\iint_{\mathcal R}|F(s,t)|^2\,ds\,dt,\\ J_1(F)&=\int_0^1\left|\int_0^{1-t}F(s,t)\,ds\right|^2dt,\\ J_2(F)&=\int_0^1\left|\int_0^{1-s}F(s,t)\,dt\right|^2ds. \end{align*} I claim the uniform estimate \begin{equation}\label{wsref-0-31-31-35-33-3a-4d-32} J_1(F)+J_2(F)\leq2\log2\,I(F). \end{equation} This is more informative than bounding each $J_i$ separately by $I$.

Proof. Put \[ w_1(s,t)=\frac{1+s-t}{2},\qquad w_2(s,t)=\frac{1+t-s}{2}. \] They are positive in the interior and satisfy $w_1+w_2=1$. For $t<1$, \[ \int_0^{1-t}\frac{ds}{w_1(s,t)}=2\log2. \] Weighted Cauchy--Schwarz therefore gives \[ \left|\int_0^{1-t}F(s,t)\,ds\right|^2 \leq2\log2\int_0^{1-t}w_1(s,t)|F(s,t)|^2\,ds. \] Integrate in $t$ and repeat with $s,t$ interchanged. Their sum is (1153:M2); the boundary where a weight vanishes has measure zero. All integrands on the right are integrable because $0\leq w_i\leq1$. $\square$

In Maynard's basic simplex-supported setup, a level of distribution $\theta$ yields the leading weighted prime-count ratio \[ \frac{\theta}{2}\frac{J_1(F)+J_2(F)}{I(F)}. \] For two forms, (1153:M2) bounds it by $\theta\log2<1$ even if $\theta=1$. To force both forms prime by the nonnegative weighted-count criterion, that ratio must exceed $1$. Thus this particular two-form mean criterion cannot supply the missing input, even with a full level of distribution. The factor $\theta/2$, the support restriction, and the threshold are exactly those of Proposition 4.2 and its proof in [E3]. This does not exclude different weights, different support geometries, additional correlations, or genuinely different sieve arguments. No novelty claim is made for the variational inequality.

For example $F=1$ on $\mathcal R$ gives $I=1/2$, $J_1=J_2=1/3$, hence $(J_1+J_2)/I=4/3<2\log2$. Increasing the quality of this test function within the stated class cannot cross the required threshold.

Stress test.

The following exact enumeration, reproduced by the companion script, counts all endpoint pairs and neighboring gaps with upper endpoint at most $10^6$. The last column counts interior-prime incidences, with multiplicity, so it is the right side of (1153:union), not the number of bad pairs. These finite counts test the definitions and the union-bound direction; they say nothing about the limit in (1153:pairinput). In particular, the endpoint-pair and consecutive-gap columns are not interchangeable.

The CRT modulus grows with the chosen $h$, but that is allowed: Polignac quantifies over each fixed gap and does not demand estimates uniform in $h$. The sufficiently-large-$t$ qualification excludes a forced divisor being equal to the interior value itself. A theorem producing infinitely many prime values for each individual linear form is not enough, since the intersection of two infinite sets can be empty. Likewise, Bunyakovsky's single-polynomial assertion from UnsolvedMath problem 1292 cannot simply be applied to $L_0L_h$: that product is reducible. Both reductions still require a genuine two-prime correlation result, and at $h=2$ they necessarily include the twin prime problem.

Outcome.

Outcome: Conditional reduction.

Two complete implications have been established: the quantitative pair input (1153:pairinput) implies all the prescribed consecutive gaps, and an infinitude assertion for two explicit admissible arithmetic progressions implies any one prescribed gap without requiring a density. The elementary bound (1153:M2) explains why the most direct two-form version of the basic Maynard criterion cannot provide that input. Neither endpoint hypothesis has been proved. Thus this attempt has isolated consecutiveness rigorously, but has not resolved Polignac's conjecture.

Sources

• S1 MathOverflow contributors, What is the importance of Polignac's Conjecture? $2022$. https://mathoverflow.net/questions/416389/what-is-the-importance-of-polignac-s-conjecture. Catalogue formulation source.
• E1 D. H. J. Polymath, Variants of the Selberg sieve, and bounded intervals containing many primes, Research in the Mathematical Sciences 1 $2014$, article 12; arXiv:1407.4897, Introduction, the unconditional $H_1\leq246$ bound. https://arxiv.org/abs/1407.4897. Consulted 27 September 2026.
• E2 J. Stadlmann, Bounded gaps between primes, arXiv:2608.31126v1 (31 August 2026), Abstract, Introduction and the main $H_1\leq240$ theorem. https://arxiv.org/html/2608.31126v1. Consulted 27 September 2026; recent preprint, not independently audited here.
• E3 J. Maynard, Small gaps between primes, Annals of Mathematics 181 $2015$, 383--413; arXiv:1311.4600, Section 4, Propositions 4.1--4.3 and equation $4.4$. https://arxiv.org/html/1311.4600. Consulted 27 September 2026; simplex support and the $\theta/2$ factor checked.
• E4 T. Dubbe, Explicit bounds for prime $k$-tuplets, arXiv:2409.04261v1 (6 September 2024), Theorem 1, especially its $K=3$ case and its admissibility and nonzero-discriminant hypotheses. https://arxiv.org/html/2409.04261v1. Consulted 27 September 2026; used for the standard upper-bound sieve estimate, not a prime-tuple lower bound.