How does \(p+q=2n\) fails? If,
\(\cfrac{p-q}{4}\longrightarrow\cfrac{n+1}{2}\) as primes\(\longrightarrow\)large and \(n\longrightarrow\infty\)
that insist on last admitted odd number, \(2n-1\) be prime is the only way to fail, then we may have a proof in the previous post "Silly Me, Goldbach" dated 24 Dec 2022, because the Prime Number Theorem indicates that it is not possible to fail that way.
Consider, from previously,
\(n_g+n_o=n+1\)
from the same post limits
\(n_o\) to about \([1,\cfrac{n+1}{2})\).
but does not guarantee that both \(p\) and \(q\) exist for any \(n\gt 2\), in
\(p+q=2n\)
\(p=2n-2n_o+1\) and \(q=2n-2n_g+1\)
\(p\) and \(q\) must be prime.
If we substitute for \(n_o\) in \(p\) for \(n_g\)
\(p=2n-2(n+1-n_g)+1\)
\(p=2n_g-1\) and similarly,
\(q=2n_o-1\)
Hey, Hey, Hey...a simpler way to find \(p\) and \(q\) and more importantly,
\(p\) and \(q\) are more relevant to the last odd number admitted as \(n\) increments. As,
\(\cfrac{p-q}{4}\longrightarrow\cfrac{n+1}{2}\), \(n_g\longrightarrow n\) as primes\(\longrightarrow\)large and \(n\longrightarrow\infty\).
In the worst case, where \(p\) and \(q\) are consecutive primes of increasing gap, ie. \(p-q\) increases most as \(n\) increases, Prime Number Theorem shows that \(p-q\) is roughly \(log(n)\) in the first \(n\) integers. That means,
\(p=2n_g-1\lt\lt 2n-1\)
where \(2n-1\) is the last odd number added. \(p\) will not fail because the last odd number added when \(2n\) increases to \(2(n+1)\) is not a prime number. \(p\) is prime number located away from this boundary as \(n\) increases. \(n_g\) and \(n_o\) are located in a small gap about \(\cfrac{n+1}{2}\).
Can \(p\) and \(q\) always be found? Is the only way to fail when \(p\) and \(q\) becomes too wide?
Maybe,
But PNT says average \(p-q\) when \(p\) and \(q\) are consecutive is of order \(log(n)\) as \(n\) increases and so strictly, \(p-q\lt n\).
Average! If the average fails then \((p-q)_{max}\) will fail. But not vice versa.
Goldbach's Conjecture will fail when the last consecutive primes \(p\) and \(q\) are greater than \(n\) apart.
\(p-q\gt n\)
not the average gap between consecutive primes but the largest. Unfortunately, Bertrand–Chebyshev theorem prove this wrong, there is a prime between \(n\) and \(2n\).
