I was thinking about how a even number, $2n$, can be represented as the sum of two natural numbers in $n$ ways, and that $\forall a \in \Bbb N \Rightarrow \exists b \in \Bbb N:a+b=2n$, where $a,b<2n$. I thought, that if we consider that $Q(n)$ represents the amount of ways in which you can write $2n$ as the sum of two composite numbers, and we consider that for $a + b$ to be equal to $2n$, $2n-a=b$, such that, if both $a$ and $b$ are composite, $a$ has to be a divisor of $2n$, so that $b$ is also a composite number, and now if we consider the prime factorization of $2n$ as: $$2n=2(p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k})$$
We can now conclude, that we can write $Q(n)$ as a sum of the amount of distinct multiples up to $n$ of every distinct prime factor of $2n$, except for themselves, plus a term that represents weather $2n-1$ is composite or not, since we are not going to be able to reach that representation in another way, so, let $a(x)$ represent weather $x$ is prime or not, such that: $$a(x)= \begin{cases} 1, & \text{if $x$ prime} \\ 0, & \text{if otherwise} \end{cases}$$
Thus, $Q(n)$ can be written as: $$Q(n)= \sum_{i=1}^k (\lfloor\frac{n}{p_i}\rfloor-1)-\sum_{i=1}^k\sum_{j=1\;i \neq j}^k \lfloor\frac{n}{p_ip_j}\rfloor+1-a(2n-1)$$ Now, if we assume, that $2n$ cannot be expressed as the sum of two prime numbers, we conclude that every way of writing $2n$ as the sum of two numbers has to be in one of three forms only: $$ 2n= \begin{cases} prime + composite \\ composite + prime \\ composite + composite \end{cases} $$ And since $\forall a \in \Bbb N \Rightarrow \exists b \in \Bbb N:a+b=2n$, where $a,b<2n$, iff $2n$ is not expressible as the sum of two primes, we have that: $$Q(n)+\pi(2n)=n$$
And as we know, we can rewrite this as: $$\sum_{i=1}^k (\lfloor\frac{n}{p_i}\rfloor-1)-\sum_{i=1}^k\sum_{j=1\;i \neq j}^k \lfloor\frac{n}{p_ip_j}\rfloor+1-a(2n-1)=n-\pi(2n)$$ Thus, we can conclude, that Goldbach's conjecture is the same as: $$\sum_{i=1}^k (\lfloor\frac{n}{p_i}\rfloor-1)-\sum_{i=1}^k\sum_{j=1\;i \neq j}^k \lfloor\frac{n}{p_ip_j}\rfloor+1-a(2n-1) \neq n-\pi(2n)$$ Now, to extend this to a proof of Goldbach's conjecture I do not know how, so, if you can answer this, it'd be amazing, I'm just looking for an answer. And also, I'm sorry if the title isn't very accurate.
One of your earlier statements is
However, it's not true that if both $a$ and $b$ are composite that $a$ has to be a divisor of $2n$. For example, if $n = 17$, then note $a = 25$ and $b = 9$ gives $25 + 9 = 2(17) = 34$, but neither $a$ or $b$ are a divisor of $2n$.
With what you stated further on being based on this, it means the rest of your statements afterwards will not be accurate.