Am I wrong? Is my proof clear?
Is this already a known result? If so, are there any simple implications or corollaries of this proof? (if that makes sense) I know some about groups, rings and fields, but not enough to argue with them.
Suppose $n,k\in\mathbb{N}$ such that $n=2k$, and $k>2$.
$\exists x,y\in\mathbb{N}$ such that $x+y=k-1$.
I claim that if $2x+1$ and $2y+1$ are prime then $n$ is the sum of two primes.
$$\begin{align*} n&=2k,\\ &=2(x+y+1),\\ &=4xy+2(x+y+1)-4xy,\\ &=4xy+2x+2y+2-4xy,\\ &=(4xy+2x+2y+1)+1-4xy,\\ &=(4xy+2x+2y+1)+1-((2x+1)-1)((2y+1)-1),\\ &=(2x+1)(2y+1)+1-((2x+1)-1)((2y+1)-1). \end{align*}$$
Let $m,l\in\mathbb{N}$ such that $m=2x+1$ and $l=2y+1$.
Then $$n=ml+1-(m-1)(l-1)$$ and
$
\forall p_{1},p_{2}\in\mathbb{N}$ such that $p_{1}$ and $p_{2}$ are prime, if
$m=p_{1}$ and $l=p_{2}$
$$\begin{align*}
n&=2k,\\
&=ml+1-(m-1)(l-1),\\
&=p_{1}p_{2}+1-(p_{1}-1)(p_{2}-1),\\
&\therefore n=p_{1}+p_{2}
\end{align*}$$
