Squeeze Goldbach

 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 $2n$.