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$\forall {p_1\in\mathbb{P}, p_1>3},\ \exists {p_2\in\mathbb{P},\ p_3\in\mathbb{P}};\ (p_1 \neq p_2) \land (p_1\neq p_3) \land (p_1 = \frac{p_2+p_3}{2})$

Now I'm not a 100% sure about this, but I vaguely remember proving this once, but I cannot recall how I did it right now.

It's also a bit like a weaker version of Goldbach's conjecture, where now those even numbers that a double a prime are the sum of two primes, with the added condition of the summed primes being different.

So I'm asking if someone can provide/link to a proof of this? Because I've been looking around wikipedia and google but I cannot find this statement anywhere.

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it fails at $p = 2$ and $3$. Do you have any evidences that the statement is true for other $p$? –  achille hui Jun 21 at 7:41
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By brute force, the statement is true for $3 < p_1 < 10^6$. It now sounds plausible that the statement is actually true. –  achille hui Jun 21 at 8:10
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What is the point of the unreadable symbolic fetishism in the question? It would be better, and much easier to just write "is every prime the average of two distinct primes?". –  Fredrik Meyer Jun 21 at 10:30
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@FredrikMeyer I did both didn't I? I put your preferred version in the title, and the symbolic one in the text. I don't see the problem tbh. The question is clear and the symbolic version just does alway with any possible ambiguity. –  user2520938 Jun 21 at 10:52
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"I vaguely remember proving this once, but I cannot recall how I did it right now." Maybe you almost wrote in a margin somewhere? –  Joshua Taylor Jun 21 at 12:58

4 Answers 4

This conjecture is very similar to Goldbach's conjecture, namely that every even number is the sum of two prime numbers. Here you are claiming that in the particular case that that even number is twice a prime number $p>3$, there is at least a second way to write it so (the first one being $p+p$). Goldbach's conjecture has resisted so far (a very long time) to attempts to prove it, even with sophisticated number theoretic means; my guess is that the conjecture asserting a second expression for numbers of the form $2p$ is not substantially easier to settle than Goldbach's conjecture (unless it happens to be false).

What you can say more precisely is that if Goldbach's conjecture and this conjecture are both true, then so is the somewhat stronger version of Goldbach's conjecture"

Every even number $n>6$ is the sum of two distinct primes.

Of course those are two big "if"s.

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As pointed out in the OP and the comments, this is likely to be true by analogy with Goldbach's conjecture and numerical computation. But nobody knows how to prove it. (I can't think of a method to approach the problem that wouldn't also solve Goldbach's conjecture. Of course, take that as a vacuously true statement if you like....)

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Goldbach's conjecture $\implies$ OP's conjecture, but OP's conjecture $\nRightarrow$ Goldbach's conjecture, so it might be easier to prove (I'm not the one who downvoted your answer by the way). –  barak manos Jun 21 at 8:36
    
@barakmanos Goldbach's Conjecture does not imply this conjecture though. Because Goldbach's Conjecture gives no restriction for the two summed primes. Goldbach's Conjecture is trivially true for even number of the form $2p, p\in\mathbb{P}$, because $2p = p + p$. In this conjecture $2p$ has be written as the sum of two different primes. –  user2520938 Jun 21 at 8:42
    
@user2520938: Goldbach's conjecture states that every even number is the sum of two prime numbers. OP's conjecture essentially says the same for some of the even numbers (namely, for every even number which is equal to a prime number multiplied by $2$). –  barak manos Jun 21 at 8:47
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@barakmanos You are still ignoring the condition that the two summed primes must be different, which is not mentioned in Goldbach's Conjecture. The two are incomparable. OP only takes a subset of all even numbers (weaker), but at the same time adds the condition that the summed primes be different (stronger). –  user2520938 Jun 21 at 8:55
    
The two are logically incomparable, but to the best of our knowledge, they are of equal difficulty. –  Greg Martin Jun 22 at 3:35

The Goldbach conjecture says that every even integer >= 4 is the sum of two primes.

Now quite obviously if p is a prime, then the even integer 2p is the sum of two primes, and if p is not a prime, then 2p is not the sum of two equal primes. We can remove the trivial cases 4 and 6. So I'll write the trivially equivalent modified Goldbach conjecture: Every even integer >= 8 that is not twice a prime is the sum of two different primes.

We now see that the conjecture that is discussed here has nothing to do with the Goldbach conjecture, because it says: Every even integer >= 8 that is twice a prime is the sum of two different primes. It discusses exactly those even integers that the modified Goldbach conjecture isn't concerned about.

(On the other hand, if a proof for my modified Goldbach conjecture is found, it is unlikely that it would rely on the fact that n/2 is composite or use it in any way, so quite likely a proof for the modified Goldbach conjecture would prove this conjecture as well.

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If the Goldbach's conjecture is True, Then OP's conjecture is True by ratiocination.
There's a series of problems of same kind, Goldbach's conjecture if one of them.

Goldbach's conjecture is equivalent to this problem:

Let $A = \{n:n \neq 2xy+x+y ,\ x,y \in \mathbb{N},\ 1 \le x \le y \}$,
Then the set of pairwise sums of elements of set A is the set of integers greater than 1.

One of the other problem in the same series:

Let $A = \{n:n \neq 3x^2+(6y−3)x−y,\ n \neq x^2+(6y−3)x+(y−1) ,\ x,y \in \mathbb{N},\ 1 \le x \le y \}$,
Then the set of pairwise sums of elements of set A is the set of nonnegative integers.

The problem above is for even number $N$ of the form $12n+10,n \in \mathbb{Z_{\ge 0}}$ ,that is $N = p_1+p_2,\ $ where $\ p_1 = 12i+5,p_2 = 12j+5,\ i,j \in \mathbb{Z},\ 0 \le i \le j,p_1,p_2 \in \mathbb{P} $.
There's problems of the same series for even number $N$ of the forms $12n+(2,4,6,8,12)$.

By ratiocination: problems in the series above can be solved by same method,then we can easy prove that any even number greater than 49 has at least two Goldbach pairs, this means OP's conjecture is True when $p>23$.

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