Progress:
A slightly stronger version of the original assumption is this:
Every even integer $n>2$ is a semiprime or sum of two even semiprime numbers.
I was wondering as to how this statement can be proved/disproved, though it may seem a bit trivial.
I would really appreciate any opinion or insight as well.
Regards
Best Answer
To summarize some of what I think the Comments are trying to say falsifies the "strong" version, let $n$ be any odd composite number. Then $2n$ is not an "even semiprime" (since $n$ is not prime).
But if $2n$ were the sum of two even semiprimes, we should have $n$ as a sum of two primes. Since this is possible only if $n-2$ is prime, it is pretty easy to find counterexamples. As barto and Wojowu point out, $n=27$ is an odd composite with $n-2$ also composite, so we cannot write $2n= 54$ as a sum of two even semiprimes. There are indeed infinitely many such counterexamples, so the strong version fails despite adding a caveat for "all sufficiently large integers".
On the other hand the weak version might hold for all sufficiently large integers. For example, the value $54$, despite not being "an even semiprime", can be expressed as the sum of two semiprimes:
$$ 54 = 15 + 39 = 3\cdot 5 + 3\cdot 13 $$
If $2n$ is not an even semiprime, then $n$ must be a composite. Suppose that $p$ is a prime factor of $n$, and that $n = pm$ for $m \gt 1$. Applying Goldbach's conjecture to $2m$ would give it as a sum of a pair of primes:
$$ 2m = q_1 + q_2 $$
and give $2n = 2pm = p q_1 + p q_2$ as a sum of a pair of semiprimes.
Therefore the "weak" version shown in the Question's title is implied by Goldbach's conjecture.