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Regarding to our hypothesis in https://math.stackexchange.com/questions/1918406/a-hypothesis-about-the-conjecture-every-even-number-is-the-difference-of-two-p , we guess that the following statements are equivalent (is it true?):

(a) For every integer number $N$ there exist prime numbers $p_1,p_2,p_3,p_4$ such that $N=p_1+p_2-p_3-p_4$;

(b) For every distinct non-zero integers $a, b$, at least one of the numbers $a, b$ and $a-b$ can be expressed as the difference of two primes.

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    $\begingroup$ Unlikely, as a) is true (courtesy Harald Helfgott and others) and b is not known. I see b as essentially equivalent to every even number is the difference between two primes. Gerhard "Otherwise It's Way Too Easy" Paseman, 2016.10.28. $\endgroup$ Oct 28, 2016 at 18:21
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    $\begingroup$ Actually, I spoke too quickly. I think it likely that a proof of a) will be found before a proof that a) and b) are equivalent, and that a) is likely to follow from work of Helfgott. Gerhard "It Can't Be That Easy" Paseman, 2016.10.28. $\endgroup$ Oct 28, 2016 at 18:26

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Statement a) is true for all $N$. This follows from the fact that the number of even integers which cannot be expressed as the sum of 2 primes is small. The best result in this direction is due to Pintz, who showed that there are $\mathcal{O}(x^{2/3})$ even integers below $x$, which cannot be written as a sum of two primes. In particular there is some $x_0$, such that for $x>x_0$ the number of exceptions is $<x^{3/4}$. Note however, that for the following argument much weaker estimates suffice.

Assume first that $N$ is even. If $N$ cannot be written as the difference of two Goldbach numbers $<x$, then for all $n<x-N$ we have that at most one of $n, n+N$ is Goldbach. Hence the number of Goldbach numbers below $x$ is at most $\frac{x-N}{4}+N$, which for $x>\max(x_0, 5N)$ gives a contradiction.

Next suppose that $N$ is odd. Then take $p_4=2$. If $N+2$ cannot be written in the form Goldbach minus prime, where the Goldbach number is bounded by $x$, then for each prime $p\leq x-N-2$ we have that $N+2+p$ is not Goldbach. Hence the number of even integers below $x$, which are not Goldbach, is at least $\pi(x-N-2)$. Again taking $x=\max(x_0, 4N)$ gives a contradiction.

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  • $\begingroup$ Indeed, the original estimates for Goldbach exceptions, due to van der Corput (1937), Tchudakoff (1938), Estermann (1938), suffice for this argument. $\endgroup$
    – GH from MO
    Oct 30, 2016 at 18:08
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Statement (a) is true for $N$ sufficiently large, and current technology is probably capable of showing it for all $N$'s. Specifically, choose $p_4=2$ for $N$ odd, and $p_4=3$ for $N$ even. Then it suffices to show that every sufficiently large odd number can be written as $p_1+p_2-p_3$ with primes $p_1,p_2,p_3$, and this was discussed in this earlier MO post.

Statement (b) seems to be out of reach. Specializing to $b=7$, it states that either $a$ or $a-7$ is a difference of two primes (because $7$ is not a difference of two primes). Specializing further that $a$ is odd but not of the form $p-2$ (with $p$ a prime), the conclusion is that $a-7$ is a difference of two primes. So (b) implies that almost every even number is a difference of two primes. Currently we know by the recent breakthroughs around the twin prime conjecture (Zhang, Maynard, Tao, Polymath8) that a positive proportion of the even integers can be written as a difference of two primes (in fact the lower density of such integers exceeds $1/354$ as proved here), but proving this for almost every even integer seems to be out of reach. On the other hand, assuming a generalized Elliott-Halberstam conjecture, statement (b) follows for any even integers $0<b<a$ such that $a\equiv 0\pmod{3}$ or $b\equiv 0\pmod{3}$ or $a\equiv b\pmod{3}$. Specifically, under these hypotheses, Polymath8b proved that infinitely many translates of $\{0,b,a\}$ contain at least two primes, hence in particular one of $a$, $b$ , $a-b$ is a difference of two primes.

To summarize, (a) is essentially known, while (b) seems to be out of reach.

Added. As Jan-Christoph Schlage-Puchta explained in his response, statement (a) is true for every $N$. In fact the original estimates for Goldbach exceptions, due to van der Corput (1937), Tchudakoff (1938), Estermann (1938), suffice for his argument.

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  • $\begingroup$ I take it that it follows that the two statements are not, in fact, equivalent (other than in the purely logical sense in which any two true statements are equivalent). $\endgroup$ Oct 29, 2016 at 6:52
  • $\begingroup$ @GerryMyerson: I agree. With current technology, it is not clear how to deduce (b) from (a). $\endgroup$
    – GH from MO
    Oct 29, 2016 at 9:29

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