TITLE: Is every prime the average of two other primes?
QUESTION [19 upvotes]: $\forall {p_1\in\mathbb{P}, p_1>3},\ \exists {p_2\in\mathbb{P},\ p_3\in\mathbb{P}};\  (p_1 \neq p_2) \land (p_1\neq p_3) \land (p_1 = \frac{p_2+p_3}{2})$
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.

REPLY [3 votes]: You can make a more expansive conjecture with the same degree of confidence:
Every integer greater than 3 can be expressed as the average of two primes.
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 $7+31=2∗19$. The Goldbach conjecture applies to even numbers only, but the average of two primes applies to every number - even, odd, prime - bigger than 3.
CSV of first 100,000: int, diff, p1, p2, type