The probabilistic independence conjecture for Goldbach events

From papers

Fix a positive integer NN and let

Ω={(a,b,c)N3:a+b+c=N and a,b,c1}.\Omega=\{(a,b,c)\in\mathbb{N}^3:a+b+c=N\text{ and }a,b,c\geq1\}.

Let F\mathcal{F} be the power set of Ω\Omega, and let P\mathbf{P} be the uniform probability measure, P(E)=#E/#Ω\mathbf{P}(E)=\#E/\#\Omega. Define

A={(p1,n2,n3)Ω:p1 is prime, n2,n3N},A=\{(p_1,n_2,n_3)\in\Omega:p_1\text{ is prime},\ n_2,n_3\in\mathbb{N}\},

and

A={(n1,p2,p3)Ω:n1N, p2,p3 are prime}.A'=\{(n_1,p_2,p_3)\in\Omega:n_1\in\mathbb{N},\ p_2,p_3\text{ are prime}\}.

Goldbach-event independence conjecture. The events AA and AA' satisfy

P(AA)P(A)P(A).\mathbf{P}(A\cap A')\sim\mathbf{P}(A)\mathbf{P}(A').

This proposes a probabilistic approach to Goldbach-type problems, interpreting prime selections in additive decompositions as approximately independent; the assertion is presented as a conjecture and remains open.

Progress summary

Open

No public discussion or published progress on this conjecture was found.

No publicly retrieved source discusses this conjecture, proves it, refutes it, or records partial progress.

Current status (as of August 2026): The conjecture appears open, with no recorded public activity or verified progress.

Sources & referencesView supporting material

Primary source

Qiang Ma and Rui Zhang, “Cancellation in sums over special sequences on GL_m and their applications”, arXiv:2411.06978 (2025).

Solutions 1

Counterexample

Counterexample: the conjecture fails uniformly for every sufficiently large odd NN.

The exact statement is Conjecture 1.5 of Q. Ma and R. Zhang, arXiv:2411.06978v3, subsequently published in the Canadian Journal of Mathematics, doi:10.4153/S0008414X25101090. The source assumes NN is odd throughout, and the counterexample respects that restriction.

Let

ΩN={(a,b,c)N3:a+b+c=N},ΩN=(N12)N22.\Omega_N=\{(a,b,c)\in\mathbb N^3:a+b+c=N\}, \qquad |\Omega_N|=\binom{N-1}{2}\sim\frac{N^2}{2}.

Write AN={a is prime}A_N=\{a\text{ is prime}\} and AN={b,c are prime}A'_N=\{b,c\text{ are prime}\}. Exact counting gives

AN=p<N\p prime(Np1),AN=#{(p,q):p,q prime, p+q<N}.|A_N|=\sum_{\substack{p<N\p\text{ prime}}}(N-p-1), \qquad |A'_N|=\#\{(p,q):p,q\text{ prime},\ p+q<N\}.

The prime number theorem alone yields

ANN22logN,ANN22log2N.|A_N|\sim\frac{N^2}{2\log N}, \qquad |A'_N|\sim\frac{N^2}{2\log^2N}.

Indeed, the measures

νN=logNNpN\p primeδp/N\nu_N=\frac{\log N}{N} \sum_{\substack{p\le N\p\text{ prime}}}\delta_{p/N}

converge weakly to Lebesgue measure on [0,1][0,1]; integrating 1x1-x and the indicator of x+y<1x+y<1 gives the two constants 1/21/2. Consequently

Pr(AN)1logN,Pr(AN)1log2N.\Pr(A_N)\sim\frac1{\log N}, \qquad \Pr(A'_N)\sim\frac1{\log^2N}.

On the other hand, the source's own equation (1.1), Vinogradov's ternary Goldbach theorem, gives for odd NN

ANAN=(1+o(1))G(N)N22log3N,|A_N\cap A'_N| =\frac{(1+o(1))\mathfrak G(N)N^2}{2\log^3N},

where

G(N)=N(11(1)2)N(1+1(1)3),\mathfrak G(N)= \prod_{\ell\mid N}\left(1-\frac1{(\ell-1)^2}\right) \prod_{\ell\nmid N}\left(1+\frac1{(\ell-1)^3}\right),

with products over primes. Therefore the true asymptotic is

Pr(ANAN)Pr(AN)Pr(AN)=(1+o(1))G(N).\boxed{\displaystyle \frac{\Pr(A_N\cap A'_N)} {\Pr(A_N)\Pr(A'_N)} =(1+o(1))\mathfrak G(N).}

Since NN is odd, the factor at =2\ell=2 equals 22. Hence, for every odd NN,

2C2G(N)C3,2C_2\le\mathfrak G(N)\le C_3,

where

C2=3 prime(11(1)2)=0.6601618,C3=23 prime(1+1(1)3)=2.3009615.C_2=\prod_{\ell\ge3\ \mathrm{prime}} \left(1-\frac1{(\ell-1)^2}\right)=0.6601618\ldots, \quad C_3=2\prod_{\ell\ge3\ \mathrm{prime}} \left(1+\frac1{(\ell-1)^3}\right)=2.3009615\ldots.

In particular,

2C22(1j11(2j)2)=2π212>1.2C_2\ge 2\left(1-\sum_{j\ge1}\frac1{(2j)^2}\right) =2-\frac{\pi^2}{12}>1.

Both endpoints are sharp: odd primorials give G(N)2C2\mathfrak G(N)\to2C_2, while odd primes give G(N)C3\mathfrak G(N)\to C_3. Thus

lim infNN oddPr(ANAN)Pr(AN)Pr(AN)=2C2>1,lim supNN oddPr(ANAN)Pr(AN)Pr(AN)=C3.\liminf_{\substack{N\to\infty\N\ \mathrm{odd}}} \frac{\Pr(A_N\cap A'_N)}{\Pr(A_N)\Pr(A'_N)} =2C_2>1, \qquad \limsup_{\substack{N\to\infty\N\ \mathrm{odd}}} \frac{\Pr(A_N\cap A'_N)}{\Pr(A_N)\Pr(A'_N)} =C_3.

The proposed asymptotic independence is therefore false throughout its intended odd-integer domain. No binary Goldbach assumption is needed.

0 endorsements
Shivam Patel ·