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On leafing through some papers of John Nash (available online on his webpage) I found this intriguing little observation:

Noticing that with larger even numbers it seemed to become possible to find Goldbach decompositions with the smaller prime very very much smaller than the larger I thought of the simple conjecture that only a finite set of even integers would be such that they could not be expressed in the form of a sum of two primes where the size of the cube of the smaller prime would be less than the size of the larger prime.

Has anyone ever tackled this conjecture? Would it be an 'easy' conjecture?

Let me make the conjecture precise, for the sake of clarity. Let $N$ be an even number and let $s(N)$ stand for the smallest prime such that $N-s(N)$ is prime (if such a prime exists, i.e. if $N$ is expressible as the sum of two primes.) Define the following set: $$S_m = \lbrace N \vert N \text{ is even and } s(N)^m > N-s(n) \rbrace $$ With this notation, Nash's conjecture asks: Is $S_3$ finite or infinite?

Nash calculated the first member of $S_3$, which is 63274 = 293 + 62681.

What about other values of $m$? If indeed $S_3$ is infinite, is there an $m$ such that $S_m$ is finite?

(I'm tagging this as a reference request too since I know very little about the Goldbach literature and would be intrigued to read any related papers.)

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    $\begingroup$ The usual heuristics suggest that for large even integers $N$ the smallest prime $p$ such that $q=N-p$ is also prime should be bounded by some power of $\log N$, and thus that for each $m$ we have $p^m < q$ once $N$ is large enough. Of course a proof of such a result is well out of reach as long as we can't even prove it for $m=1$... $\endgroup$
    – Noam D. Elkies
    Dec 2, 2011 at 3:26
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    $\begingroup$ There is no $n$ in the quoted observation, so I don't know what you mean by $n=3$ or by "other values of $n$". $\endgroup$
    – Gerry Myerson
    Dec 2, 2011 at 4:49
  • $\begingroup$ The smaller prime is tabulated, and some references given, at oeis.org/A002373 $\endgroup$
    – Gerry Myerson
    Dec 2, 2011 at 4:51
  • $\begingroup$ @Gerry Apologies - edited to make it clearer. I meant $n$ to be the exponent of the smaller prime. $\endgroup$
    – Chuck
    Dec 2, 2011 at 13:57
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    $\begingroup$ There is most likely a literature on results of the form "For almost all even numbers 2N we can write 2N = p_1 + p_2 with p_1/p_2 < 1/F(N)". I'm pretty sure simply getting F(N) -> infty would be a simple application of the circle method, but I'd be surprised if one could get much better than F(N) > N^c for some small c, if that. $\endgroup$
    – Ben Green
    Dec 2, 2011 at 17:39

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