Even numbers sum of two primes We don't know if the Goldbach conjecture is true, but do we we know some type of even numbers which can be expressed as sum of two prime numbers (excluding the trivial sums of two prime numbers) ?
Edit : I am searching an infinite set $S$ of even number for which we can prove that every $ s \in S$ can be expressed as sum of two prime number (the Goldbach conjecture says that $S=2\mathbb{N}\setminus\{0,2\}$ works). For sure, $S=\{p+q, p, q, \text{prime} >2\} $ works. For example (but i don't believe it's possible), $S$ could explicitly be given by $\{2P(n), n\in \mathbb{N} \} $, where $P$ is polynomial. 
 A: A very small set of very Goldbachian numbers (Explanation by Pomerance):
Let $n = 2\cdot p_1\cdot p_2\cdot \dots$ and $\hat{p}$ be the smallest prime not dividing $n$. If $\hat{p}^2 \geq \frac{n}{2}$ then
$n$ is the sum of two primes (in a maximum number of ways).
By Bertrand's Postulate, there exists a prime number $q \in (\frac{n}{2},n-2)$. Since $q$ is prime, $n-q$ is coprime to $n$. If $n-q$ is composite then $n-q$ must be a product of primes not dividing $n$. However, if the square of the smallest prime not dividing $n$ is larger than $\frac{n}{2}$, no such composite number $n-q$ exists. So $n-q$ is prime and $q$ is prime and therefore $n$ is a sum of two primes.
The following numbers satisfy Goldbach for every possible choice of a prime number in the interval $(\frac{n}{2}, n-2)$:
\begin{align}
n&&\text{factors of }n&&\text{min } \hat{p}\perp n&&\hat{p}^2\not<\frac{n}{2}\\
12&&2^2\cdot 3&&5&&25\not<6\\
18&&2\cdot 3^2&&5&&25\not<9\\
24&&2^3\cdot 3&&5&&25\not<12\\
30&&2\cdot 3\cdot 5&&7&&49\not<15\\
36&&2^2\cdot 3^2&&5&&25\not<18\\
42&&2\cdot 3\cdot 7&&5&&25\not<21\\
48&&2^4\cdot 3&&5&&25\not<24\\
60&&2^2\cdot 3\cdot 5&&7&&49\not<30\\
90&&2\cdot 3^2\cdot 5&&7&&49\not<45\\
210&&2\cdot 3\cdot 5\cdot 7&&11&&121\not<105\\
\end{align}
