Can every even Semiprime greater than 6 be represented as the sum of 2 distinct primes?

Can we represent every even Semiprime number greater than 6 as the summation of 2 distinct prime numbers?

As far as I can do manually I found out that this is true. Please help me derive this using the Goldbach's conjecture or otherwise.

• Once you have a proof of that, one might hope it could be easy to extend it to a proof of the full Goldbach conjecture... – Henning Makholm Aug 24 '16 at 10:10
• @HenningMakholm can this be derived from the Goldbach's Conjecture? Sorry I will edit my question excluding the word proof. – slhulk Aug 24 '16 at 10:12
• Yes, $10=7+3$..... You should probably change "any" to "every" though – barak manos Aug 24 '16 at 10:16
• x @slhuk, it's not exactly the Goldbach conjecture, but it's close -- generally, by experiment, the number of ways a given even number can be expressed as a sum of two primes appears grow towards infinity, but we cannot even prove that it's always $\ge 1$. Proving that for numbers of the particular form $2p$ it is $\ge 2$ would appear to be of the same order of difficulty. – Henning Makholm Aug 24 '16 at 10:17
• @HenningMakholm if the number of pairs appear to grow towards infinity is it always the case that 2 distinct pairs exist for every number greater than 6? can this be derived using GC? – slhulk Aug 24 '16 at 10:23

Ignoring the Goldbach Conjecture, here is my take on the question. I assume that for "even semiprime" you mean $2p$, where $p$ is an odd prime.

Note that $2p\equiv$ {0, 2, 4} (mod 6). We want $2p=n+k$, where $n,k$ are primes. Now we know that all primes greater than 3 have the form $6m±1$, $m\in \mathbb{N}$. In other words, $n\equiv$ {1, 5} (mod 6) and $k\equiv$ {1, 5} (mod 6).

Adding $n+k=2p$, there are three cases:

$n+k\equiv$ 0 (mod 6) $\rightarrow$ this is the sum of $1+5$ ($n\equiv$ 1 and $k\equiv$ 5 or vice versa)
$n+k\equiv$ 2 (mod 6) $\rightarrow$ this is the sum of $1+1$ (both $n$, $k\equiv$ 1)
$n+k\equiv$ 4 (mod 6) $\rightarrow$ this is the sum of $5+5$ (both $n$, $k\equiv$ 5)

Therefore all possible cases are covered. All we have to do is verify that our choice of $n$ and $k$ are prime. If $2p$ is large enough, there should be enough choices that there will be at least one pair of primes to choose for $n$ and $k$.