3

Is it true that for every integer number $N$ there exist prime numbers $p_1,p_2,p_3,p_4$ such that $N=p_1+p_2-p_3-p_4$ ?

M.H.Hooshmand
  • 2,537
  • 15
  • 26
  • I think that by using standard arguments like the Rusza-Plunnecke inequality (https://www.math.cmu.edu/~af1p/Teaching/AdditiveCombinatorics/Additive-Combinatorics.pdf) it is not difficult to show that the set $A=\mathcal{P}-\mathcal{P}$ (the difference set of primes) has a large density (since the set of primes is an additive basis of order $\leq 4$), hence $A-A$ is very likely to have density one. However, I fail to see how Vinogradov's theorem (only dealing with a sumset) would directly imply the claim. – Jack D'Aurizio Sep 07 '16 at 22:28
  • Another chance is probably to exploit the fact that under RH, for every $x$ large enough there is a prime in the interval $[x,x+\sqrt{x}\log(x)]$, hence the set $A-A$ has for sure a large density. – Jack D'Aurizio Sep 07 '16 at 22:39
  • The exceptional set of Goldbach's conjecture is very sparse ($\ll x^{1-c}$ for some $c>0$ by a well-known result of Montgomery and Vaughn). Therefore we can always find some large even number $Q$ such that $Q$ lies outside the exceptional set (so $Q = p_1+p_2$) and $Q - N$ is either $p+2$ or $p+3$. This works for both large and small values of $N$. – Erick Wong Sep 21 '16 at 22:03

1 Answers1

3

For large enough $N$ this is true. See Vinogradov's theorem.

If $N$ is even, apply the theorem to $N-3$. Otherwise, apply the thorem to $N-2$.

ajotatxe
  • 65,084
  • 2
    I am not sure I get it: since every odd number $\geq 7$ is the sum of three primes, why every large enough odd integer can be written as $p_1+p_2\color{red}{-} p_3$? – Jack D'Aurizio Sep 07 '16 at 22:26
  • @ajotatxe. Thanks for your answer. (1) It seems that $N-3$ and $N-2$ should be replaced by $N+3$ and $N+2$ (2) I think Jack is right. – M.H.Hooshmand Sep 08 '16 at 05:02
  • What do you mean "for large enough $N$"? This works if $N = 1$: $$\text{Let $p_1 = 2, \ p_2 = 7, \ p_3 = 3, \ p_4 = 5$}$$ $$1 = 2 + 7 - 5 - 3$$ If $1$ is large enough then this works for every positive integer. – Mr Pie Sep 18 '17 at 09:50
  • @user477343 Say you have a constant $p_n$. This would denote the $n^{th}$ prime number. – George N. Missailidis Sep 18 '17 at 11:18