The probabilistic independence conjecture for Goldbach events

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Fix a positive integer $N$ and let

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

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

A=\{(p_1,n_2,n_3)\in\Omega:p_1\text{ is prime},\ n_2,n_3\in\mathbb{N}\},

and

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 $A$ and $A'$ satisfy

\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.

References

Primary source

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

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A posted argument claims the conjecture is false for all sufficiently large odd inputs, but that counterexample has not been independently verified.

The conjecture asks whether the events that one coordinate is prime and that the other two coordinates are prime behave independently in random three-term decompositions of $N$ . Qiang Ma and Rui Zhang’s 2024 paper proposes a related Goldbach-average conjecture, but its retrieved abstract does not itself establish this exact event statement.

Known results

Vinogradov’s ternary Goldbach theorem gives the expected singular-series asymptotic for three-prime representations of sufficiently large odd integers.
Standard probabilistic heuristics explicitly require a Hardy–Littlewood singular-series correction, indicating that primality conditions are not naively independent.

Posted attempt

An unverified argument claims that, for odd $N$ , the exact ratio satisfies $\mathbf{P}(A\cap A')/[\mathbf{P}(A)\mathbf{P}(A')]=(1+o(1))\mathfrak G(N)$ , with $\mathfrak G(N)>1$ . It therefore claims a complete counterexample, rather than partial progress; no independent verification was retrieved.

Current status (as of August 2026): A complete counterexample is claimed for sufficiently large odd $N$ , but the exact conjecture remains unsettled because the claim is unverified.

Sources

CounterexampleThis solution needs a summarySee full solutionHide full solution

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

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 $N$ is odd throughout, and the counterexample respects that restriction.

Let

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

|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

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

Indeed, the measures

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

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

\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 $N$

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

where

\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

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

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

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

where

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,

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 $\mathfrak G(N)\to2C_2$ , while odd primes give $\mathfrak G(N)\to C_3$ . Thus

\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.