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I am asking if this variant of the weak Goldbach Conjecture is already known.

Let $N$ be an odd number. Does there exist prime numbers $p_1$, $p_2$ and $p_3$ such that $p_1+p_2-p_3=N$? Ideally, can we find $p_1$, $p_2$ and $p_3$ so that they are small enough? For example, can we prove that for large enough $N$, we can find such a triplet that all of them are smaller than $N$?

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3 Answers 3

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Yes - the standard proof of Vinogradov's result by means of the circle method gives this result. You just need to examine an integral $$\int_{\mathbb{R}/\mathbb{Z}} (\widehat{f}(\alpha))^2 \widehat{f}(-\alpha) e(-\alpha N) d\alpha$$ instead of $$\int_{\mathbb{R}/\mathbb{Z}} (\widehat{f}(\alpha))^3 e(-\alpha N) d\alpha.$$ Here $\widehat{f}(\alpha) = \sum_n \Lambda(n) e(\alpha n) \eta(n/N)$, where $\eta$ is any weight supported in $\lbrack 0,1\rbrack$.

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    $\begingroup$ does $\int_{\mathbb{R}/\mathbb{Z}} (\widehat{f}(\alpha))^2 e(-\alpha N) d\alpha.$ prove Goldbach? $\endgroup$
    – JMP
    Commented Apr 15, 2015 at 12:21
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    $\begingroup$ It would, but that's hard. See terrytao.wordpress.com/2012/05/20/… $\endgroup$
    – H A Helfgott
    Commented Apr 15, 2015 at 13:31
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    $\begingroup$ Can your work can be adapted to prove the result for every $N$ when $p_1,p_2$ and $p_3$ are required to be less than $N$? $\endgroup$
    – gowers
    Commented Apr 16, 2015 at 7:21
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    $\begingroup$ It can, but it could be a little tricky - most of the weights I use are not compactly supported (though they decay very fast -- they are almost supported on a compact interval). $\endgroup$
    – H A Helfgott
    Commented Apr 16, 2015 at 14:37
  • $\begingroup$ I suppose that there is some large explicit $N$ such that all even integers below $N$ are a sum of two primes (by the way what is the current record?). Is it possible to use Harald's proof to significantly enlarge $N$ ? Rather obviously this does not constitute a method of proving Goldbach by induction :) $\endgroup$
    – Dr. Pi
    Commented May 5, 2015 at 1:57
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Harald's answer is perfect, but let me tell that using the circle method one can also prove similar results where $p_3$ runs through rather sparse sets (including sparse sets of primes). This is because exceptions to the Goldbach conjecture are relatively rare, i.e. $n+p_3$ will be a sum of two primes (less than $n$) for some odd $p_3<n$, even when $p_3$ is allowed to skip most values. See the works of van der Corput (1937), Tchudakoff (1938), Estermann (1938), Vaughan (1972), Montgomery-Vaughan (1975), etc.

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Harald Helfgott's proof of the weak Goldbach conjecture can be adapted to prove this result.

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  • $\begingroup$ You mean the result where you insist that $p_1$, $p_2$ and $p_3$ are less than $N$? $\endgroup$
    – gowers
    Commented Apr 15, 2015 at 20:44
  • $\begingroup$ the integral, @gowers, uses a function that only uses primes to a certain N, so the OP's question is implicit in the proof. $\endgroup$
    – JMP
    Commented Apr 16, 2015 at 23:27
  • $\begingroup$ Sorry, that wasn't what I meant to ask. I asked the question I actually wanted to ask in a comment on Harald's answer above, which he has now answered. (The question was whether one could do it for all N and not just sufficiently large N.) $\endgroup$
    – gowers
    Commented Apr 17, 2015 at 10:33

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