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.
