哥德巴赫想太多

Consider,

 $2n=p+q$

$2n-q=p$

let $q=\cfrac{a}{b}$

$2n-\cfrac{a}{b}=p$ ---(*)

$\cfrac{2nb-a}{b}=p$

If p is prime,

$b=2nb-a$ or

$b=1$

When $b=1$,

$2n-a=p$ is prime

we have instead,

$2n=p+a$

so,

$a=q$

interesting result, $\rightarrow$备用

or from $b=2n_ob-a$

$a=2n_ob-b=b(2n_o-1)$

$\cfrac{a}{b}=2n_o-1$

for a specific number $n_o$ and $b$ not zero.

sub. into (*) we have,

$2n-2n_o+1=p$

$2(n-n_o)+1$ is prime as $p$ is prime and $n\gt n_o$ as primes are positive.

Let $2(n_i-n_{oo})+1=q_i$ where $n_i\gt n_{oo}$ for positive prime.

where $n_i$ is paired with $n_{oo}$ to give $q_{i}$ as $n$ and $n_o$ is paired to give $p$.

In which case, we formulate,

$q_i+p=2(n-n_o)+1 + 2(n_i-n_{oo})+1$

and so,

$q_i+p=2(n+n_i-n_o-n_{oo})+2=2n_j$

where $n_j=n+n_i-n_o-n_{oo}+1$ which is just another integer and both $q_i$ and $p$ are prime.

Is $n_j$ strictly non negative?  Yes. And When we set $n+n_i=n_o+n_{oo}$ we start from $n_j=1$.

What conjecture?  Goldbach's.  All yours.  This is not prove of the conjecture.  It proves that the sum of two primes are even.

Note:  The trick is to realize that $n_o$ is not a running variable as $n$.