# How to prove twin prime conjecture or Goldbach conjecture if we assume prime distribution is completely random?

How to prove twin prime conjecture or Goldbach conjecture if we assume prime distribution is completely random ?

If we assume that prime number distribution is COMPLETELY random (subject to 1/log(x) restriction), can we prove twin prime conjecture or Goldbach conjecture ?

My feeling is that, this will be trivial for twin prime conjecture, but how to give a rigorous proof ? Does "completely random" imply that there will be infinite twin primes ?

On the other hand, if prime number distribution is completely random, will Goldbach conjecture still hold true ?

For example, if we consider all numbers between 1 and 100, if we assume prime number completely random distribution, will Goldbach conjecture still hold true For this set of numbers ?

The reason I ask this question, because, people often said that "primes behave almost completely random except subject to 1/log(x) restriction".

• good question - if a sequence $a_n$ is such that $a_n\sim n\ln n$, which prime conjectures hold? – JMP Nov 9 '15 at 19:47
• Is there a precise statement of what it would mean for primes to be "completely random"? – Nate Eldredge Nov 9 '15 at 19:59
• Is there a solid mathematical definition for Completely Random? – barak manos Nov 9 '15 at 20:14

The random model for the primes is that $$X_n = \begin{cases} 1 \text{ if } n \ge 2 \text{ is prime } \\ 0 \text{ otherwise }\end{cases}$$ is a sequence of independent random variables with $$P[X_n = 1] = \frac{1}{\log n}, \qquad P[X_n = 0] = 1-\frac{1}{\log n}$$

• Under this model, the Goldbach conjecture is false with probability $1$ since $P[\sum_k X_{n-k}X_{n+k} = 0] =\prod_k (1-\frac1{\log (n-k)}\frac{1}{\log(n+k)})$

• Under this model, the twin prime conjecture is true with probability $1$, since $P[\forall n > N, X_n X_{n+2} = 0] = 0$

• Under this model, the Riemann hypothesis is true with probability $1$, since $\lim_{x\to \infty} P[|\sum_{n \le x} X_n - \sum_{n \le x} \frac{1}{\log n}| < x^{1/2} \log^2 x] = 1$

If primes are completely random, then there is a non-zero probability that the Goldbach conjecture is false. Actually, there is a non-zero probability that the Goldbach conjecture is false until someone proves otherwise, except that the probability is very close to zero indeed.

It's different with the twin prime conjecture. It's slightly hard to define what the probability of infinitely many twin primes would be, but you could calculate the probability that there are N twin primes <= $N^2$ and calculate the limit as N -> infinity.

And obviously primes are not completely random. For example, there are just two primes p where p is not 6k ± 1, and no primes at all of the form 6k or 6k + 4.

• How do you define the probability of a statement to be true or false? – Bruno Joyal Nov 9 '15 at 19:58
• If the primes are completely random, then it isn't a theorem, but a statement about random numbers. Which has a probability of being true or false. Obviously whether 3079 is a prime or not would be random with a probability other than 0 or 1 if primes were completely random. – gnasher729 Nov 9 '15 at 20:05
• And you can easily define probabilities according to our current knowledge. The probability that the 10^100th decimal of pi is 7 is 1/10th. If you don't agree, make an argument why it isn't. – gnasher729 Nov 9 '15 at 20:08

For the Goldbach conjecture: Obviously primes are not completely random. Define a set P of integers with the properties: If p < 210 then p is an element of P if and only if p is a prime. If p ≥ 210 then p is an element of P with probability (210 / 48 ln p) if p cannot be divided by 2, 3, 5 and 7, and not at all otherwise (there are 48 such numbers among any 210 consecutive integers). That has about the same distribution as the primes, and is kind of similar to primes. (Is there a better simple approximation than 1 / ln p for the probability that a random p is a prime? )

You could find the elements of that set up to 1000 and check if every number up to 1,000 is the sum of two elements, and repeat that experiment a million times with different random numbers. Or calculate the numbers a bit higher.