Let $S_k^\delta=\{a+b\le k\mid(a,b)\in\mathbb N_+^2\wedge a^2+b^2\in\mathbb P\wedge a^2+b^2+\delta\in\mathbb P\}$, where $\mathbb P$ is the set of primes.
If $\delta=2$ then the condition on $a^2+b^2$ is to be the lower of a twin prime, while if $\delta=-2$ the condition on $a^2+b^2$ is to be the higher of a twin prime. Only primes of form $4n+1$ is a sum of two squares but since the distribution of those as high/low twin prime is approximately uniform, I didn't expect any asymmetries among these two cases.
Reformulation:
But $S_{50,000}^2$ contain all odd integers $x$ such that $1375<x\leq 50,000$, while no odd integer of form $6n+27\;(n\in\mathbb N_+)\;$ is a member of $S_{50,000}^{-2}$.
Can anyone explain this asymmetry?
If $a^2+b^2-2$ and $a^2+b^2$ are prime, then $a^2+b^2\equiv 1\pmod 4$ and $a^2+b^2-1$ is a multiple of $6$ (with the exception $\{a,b\}=\{1,2\}$). Then $a^2+b^2\equiv 1\pmod{12}$. This implies that $a$ and $b$ are of opposite parity.
Assume WLOG that $a$ is odd. Then $a\equiv 3,9\pmod{12}$ and $b\equiv 2,4,8,10\pmod{12}$; or $a\equiv 1,5,7,11\pmod {12}$ and $b\equiv 0,6\pmod {12}$.
Thus, $a+b\equiv 1,5,7,11\pmod {12}$.