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Yes, every, not just even.

If a number is the average (or difference) of two primes, by doubling the number it has a partition of those two primes. So, for example, $(7+31)/2=19$ becomes $7+31=2*19=38$.

Since every $n*2$ is even, GC has an underappreciated claim upon every odd n > 3 (including every prime). No other conjecture is required.

CSV of first 100,000: int, diff, p1, p2, type

michaelmross
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  • Sure, the statement you've given is equivalent to GC, but I see no reason why it would engender a better understanding; in particular, why do you think that it is not already understood as having central importance? – Noah Schweber Mar 11 '17 at 17:08
  • I don't think it's generally understood. For example, see http://math.stackexchange.com/questions/28247/can-every-even-integer-be-expressed-as-the-difference-of-two-primes. However, I'll remove that last sentence. Thanks. – michaelmross Mar 11 '17 at 17:28
  • For strict equivalence, you should say every integer $>1$. –  Mar 11 '17 at 17:44
  • The equivalence of the two statements is interesting, but trivial to prove. Therefore I don't see any reason to change the original because there's really no tangible advantage in stating it this way. – Matt Samuel Mar 11 '17 at 18:52
  • Isn't it significant because it proves, also trivially, that there must be pairs of primes (albeit overlapping)? One between $3$ and $\frac{n}{2}$ and one between $\frac{n}{2}$ and $n$. – michaelmross Apr 03 '17 at 13:12
  • In fact this formulation is valid for every integer>1, as numbers 2 and 3 can be written in this form. I asked a very similar question last year and it was deleted I don't know why. I wonder how to recover the text for me. – Perspectiva8 Nov 20 '17 at 22:25

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