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I am sure that this is well-known, but I looked around for the last half hour and couldn't see an answer. I just wondered whether it's possible to insist on taking all primes to be large in Vinogradov type results?

Thanks in advance for any help.

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Yes, I think $c=0$ works. –  Chris Gerig Jul 22 '13 at 17:48
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@Chris: I edited the title to exclude this trivial solution. –  GH from MO Jul 22 '13 at 18:11
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You missed the $n = 1$ case. –  user2357112 Jul 22 '13 at 23:39
    
@user2357112: I edited the title to forbid the $n=1$ case. –  GH from MO Jul 23 '13 at 3:35

1 Answer 1

up vote 16 down vote accepted

The usual proof of Vinogradov's result can be modified to show that every sufficiently large odd $n$ has $\asymp n^2/(\log n)^3$ representations as a sum of three primes with each prime exceeding $cn$, provided $c>0$ is sufficiently small. This gives (easily) a positive answer to your original question.

The best unconditional result of this sort seems to be by Baker and Harman ( R. Soc. Lond. Philos. Trans. Ser. A Math. Phys. Eng. Sci. 356 (1998), 763–780.). They show that every sufficiently large odd $n$ can be written as a sum of three primes from $[\frac{n}{3}-n^{4/7},\frac{n}{3}+n^{4/7}]$.

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Thank you very much, this is excellent. –  Brian Jul 22 '13 at 18:25
    
Sorry to bother you again. I was just thinking about the result that you mentioned and wondered whether it is true that we can (almost) choose where the $p_i$ come from. That is, given $\alpha _1, \alpha _2, \alpha _3 > 0$ with $\sum _{i=0}^2 \alpha _i = 1$, can we insist that the primes $p_1,p_2,p_3$ sum to $n$ and also satisfy $|p_i - \alpha _in| = o(n)$ for all $i$? –  Brian Jul 22 '13 at 18:34
    
@Brian: The general statement you propose certainly follows from Vinogradov's method and known cancellation results for exponential sums over primes in short intervals. Very briefly, Vinogradov's method reduces the problem of sums over three primes to sums over three integers. For example a result by Pan-Pan (Chinese Ann. Math. Ser. B 11 (1990), 138–147.) yields, I believe, the expected number of representations $n=p_1+p_2+p_3$ with $|p_i - \alpha _in| = O(n^{2/3+\epsilon})$. –  GH from MO Jul 22 '13 at 19:48

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