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 \(n\).
