Just Messing With Goldbach

Consider,

 $2n=p+q$

where $p$ and $q$ are both primes and $n$ is greater than $2$.

$2n-q=p$

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

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

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

When 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,

$2n-2n_o+1=p$

$p=2(n-n_o)+1$

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

Also,

$q=2(n-n_{oo})+1$ 

where $n\gt n_{oo}$ for positive prime.  This is because if we start over and swap $p$ and $q$ and arrived at an expression for $q$ in the form,

$q=2(n-n_o)+1$

Both $n_o$ and $n_{oo}$ are chosen such that $p$ and $q$ are primes given $n$.

And so,

$q+p=2(n+n-n_o-n_{oo})+2=2(n_j+1)$

where $n_j=2n-n_o-n_{oo}$ and both $q$ and $p$ are prime.

Consider,

$n_j=2n-n_o-n_{oo}=n$

$n=n_o+n_{oo}$

From previously,

$p=2(n-n_{o})+1$

$q=2(n-n_{oo})+1$

 where $n_{o}$ and $n_{oo}$ are selected to make $p$ and $q$ prime, and initially,

$p+q=2n$

 If we apply the condition,

$n=n_o+n_{oo}$

to $p$ and $q$, substituting values for $n$ into the expressions for $p$ and $q$ and create new numbers,

$p_1=2(n-n_{o})+1=2n_{oo}+1$

$q_1=2(n-n_{oo})+1=2n_o+1$

and so,

$p_1+q_1=2(n_{o}+n_{oo})+1=2(n+1)$ 

But are $p_1$ and $q_1$ prime?

Specifically, given,

$p=2(n-n_{o})+1$ is prime,

is $q_1=2n_o+1$ also prime?

Obviously,

$p+q_1=2(n+1)=2n_1$   where $n+1=n_1$

since we started with $p+q=2n$ where $p$ and $q$ are both prime and $n\gt 2$.  $q_1$ can be prime.

Equivalently, given

$q=2(n-n_{oo})+1$

$p_1=2n_{oo}+1=2(n+1)-q$

$p_1+q=2n_1$

so $p_1$ can also be prime, because we started with $p+q=2n$ where $p$ and $q$ are both prime and $n\gt 2$.

More importantly, $n_o$ and $n_{oo}$ are chosen such that, $p_1$ and $q_1$ are primes as are $p$ and $q$.

And so assuming $p+q=2n\implies p_1+q_1=2(n+1)$ where $p$, $q$, $p_1$ and $q_1$ are primes and $n\gt 2$. By induction $p+q=2n$ is true when we can start with,

$6=3+3$ where $n_o=n_{oo}=5$, $p_1=7$ and $q_1=7$

We start with $n=3$ and so $2n=6$.

This proof, does not suggest a way to find the next pair of primes such that,

$p_1+q_1=2(n+1)$

but it proves that such a pair exist provided $n_{}o$ and $n_{oo}$ can be found.

Unfortunately, the proof does show a way and it is haywire afterwards!  The expression, $p_1+q=2n_1$ suggest that one of the prime is retained to move on to $n+1$.  But as stated, $p+q=2n$ does not differential between $p$ and $q$.  Easily, we end up with $p=q$ and so, $n_{o}=n_{oo}$.

Do we spiral out of control?  If this prove is true, hopefully, depending on where the start is, to infinite prime!

Good morning.