Let,
\(S_n:\,\, p+q=2n\) \(p\neq q\) and \(p\) and \(q\) are both prime and \(n\gt 2\)
Consider \(n\) such that \(p=2n-1\) is prime and so,
\(2n-1+3=p+q=2n+2=2(n+1)\)
with \(q=3\), such that \(S_{n+1}\) is true.
Consider further that, at \(n\) we take a step \(q=5\), down the number line to \(2(n+2)\), obviously,
\(p+5=2n-1+5=2(n+2)\)
as such \(S_{n+2}\) is true.
Again from \(n\), we take a step \(q=7\),
\(p+7=2n-1+7=2(n+3)\)
as such \(S_{n+3}\) is true
and with a step \(q=11\)
\(p+11=2n-1+11=2(n+5)\)
as such for any prime \(p=2n-1\), \(S_{n+1}\), \(S_{n+2}\), \(S_{n+3}\), \(S_{n+5}\) are true.
In fact all steps of known prime are possible resulting in \(S_{n+a}\), where \(a=\left\lfloor\cfrac{p}{2}\right\rfloor\), being proven true.
We have then, \(S_{n+a}\) where \(a=\left\lfloor\cfrac{q}{2}\right\rfloor\) for all know prime \(p=2n-1\).
We starts \(S_{n}\) with \(n=2\) where \(p=2(2)-1=3\) is prime
\(2(2)-1+3=2(3)=p+q=2(n+1)\,\,\,\implies\,\,S_{3}\)
from previously results, we have also \(S_{4}\), \(S_{5}\), \(S_{7}\)
When we starts \(S_{n}\) with \(n=3\) where \(p=2(3)-1=5\) is prime
\(2(3)-1+3=2(4)=p+q=2(n+1)\,\,\,\implies\,\,S_{4}\)
from previously, we have also \(S_{5}\), \(S_{6}\), \(S_{8}\)
The skipped index in \(S_n\), are recovered when we start \(S_{n}\) with a later prime. Because for any prime \(p=2n-1\), \(S_{n+1}\), \(S_{n+2}\), \(S_{n+3}\), \(S_{n+5}\) are true.
The next consecutive odd (also a prime) gives \(S_{n+2}\), \(S_{n+3}\), \(S_{n+4}\), \(S_{n+6}\). From which we obtain \(S_{n+4}\). The rest of the indexes not not relevant, only the skipped index and the prime from which to reach the skipped index.
These skipped indexes are the result of odd numbers not being prime along the number line.
The size, \(i\) of the series \(S_{n}\)....\(S_{n+i}\) is the number of primes used to hop forward, which is the number of primes already discovered. Given the large size of this number, and the relatively small number of consecutive odd between consecutive primes, all skipped index can be recovered.
This means all \(S_{n}\) for \(n\gt 2\) can be proven true. That is to say,
Given \(p\neq q\) and, \(p\) and \(q\) are both prime
\(p+q=2n\)
is true for all \(n\gt 2\).
Goldbach's Conjecture proved!
