$\forall {p_1\in\mathbb{P}, p_1>3},\ \exists {p_2\in\mathbb{P},\ p_3\in\mathbb{P}};\ (p_1 \neq p_2) \land (p_1\neq p_3) \land (p_1 = \frac{p_2+p_3}{2})$
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.
Best Answer
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"
Of course those are two big "if"s.