Are there infinitely many $p$ such that $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ has no integer solutions? $\frac{p_k+p_{k+1}}{2} = m^2 - n^2$ with $p_k$ as the $k$-th prime has no integer solutions for $9$ values of $k \leq 25$, $17$ values of $k \leq 50$, and $33$ values of $k \leq 100$. This means that roughly $33\%$ of each sample consists of such primes. Though this is a small sample, the ratio of primes with this property to those which do not seems to be consistent. Are there infinitely many primes with this property, and if so, what proportion of primes have it?

Write $m^2-n^2=(m+n)(m-n)$  The two factors on the right have the same parity, so any number that is equivalent to $2 \bmod 4$ cannot be expressed this way.  Because you divide the sum by $2$, this is any pair of primes that sum to a number that is equivalent to $4 \bmod 8$.  For example, the sequential primes $5,7$ give $\frac 12(5+7)=6$, which cannot be expressed as $m^2-n^2$ because it only factors into $1,6$ or $2,3$, and neither pair has the same parity.  $\frac 12(7+11)=9$, which we can factor as $1,9$ or $3,3$ and $m=5,n=4$ works for the first and $m=3,n=0$ for the second.  $\frac 12(11+13)=12$, which we can factor as $2,6$ getting $m=4,n=2$.  If the left side is odd, both factors will be odd, while if the left side is a multiple of $4$ there will be at least one factorization with both factors even.
