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We couldn't so far prove that every even integer greater than 2 is the sum of two primes.

Can something more be said about the following weaker form?

Let $r,p,q$ denote primes $$\forall r\exists p,q\bigl(4r=p+q\bigr)$$

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  • $\begingroup$ Here is a related summary, but I don't see much on the specific conjecture you have formulated. $\endgroup$
    – avs
    May 29 '19 at 17:58
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    $\begingroup$ Why $4r$ and not $2r$? Do you have a proof for $4r$? $\endgroup$ May 29 '19 at 18:02
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    $\begingroup$ As far as I know, Goldbach is not even proven for , lets say, the powers of $2$. $\endgroup$
    – Peter
    May 29 '19 at 18:51
  • $\begingroup$ I've only a proof for 2r ;) $\endgroup$
    – J.Ask
    May 29 '19 at 19:45
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    $\begingroup$ Aka p and q are equidistant 2r. Where goldbach would be equidistant r. For general r. $\endgroup$
    – user645636
    May 29 '19 at 20:21
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$$4r=2r+2r\land 4r=p+q\implies 2r-p=q-2r=d\\r\equiv 1\bmod 6\implies 2r\equiv 2\bmod 6\implies p\equiv q\equiv 5\bmod 6\\r\equiv 5\bmod 6\implies 2r\equiv 4\bmod 6\implies p\equiv q \equiv 1 \bmod 6$$

Only exceptions are if $4r-3$ is prime, then $(3,4r-3)$ is a pair of primes that work. A157978 in the OEIS is the sequence of exceptions.

Of course, none of that is really new. It all comes from the original Goldbach conjecture. If all of the following could be proven for any non exceptions, we could gaurantee Goldbach (via pigeonhole principle):

  • There are more primes congruent to 5 mod 6 between n and 2n, than composites of that form up to n for n congruent to 2 or 5 mod 6.
  • There are more primes congruent to 1 mod 6 between n and 2n, than composites of that form up to n for n congruent to 1 or 4 mod 6.
  • There are more primes congruent to 5 mod 6 between n and 2n, than composites up to n congruent to 1 mod 6, or more primes congruent to 1 mod 6 between n and 2n, than composites congruent to 5 mod 6 for n congruent to 0 or 3 mod 6.

Again not new but shows where the math came from.

Addendum:

A023212 might be interesting (to me) r values in the larger case. Because then $$p+q+1=\sigma(pq)\in\mathbb{P}$$

That is, the sum of a semiprime's proper divisors is prime.

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