3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Oct 28 '16 at 18:21
  • 1
    $\begingroup$ 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. $\endgroup$ – Gerhard Paseman Oct 28 '16 at 18:26
7
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 '16 at 18:08
6
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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). $\endgroup$ – Gerry Myerson Oct 29 '16 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 '16 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.