Are there primes $p$ and $q$ for which $p^2 - 2 q^2 = 5039$? This is the least prime $r$ for which I don't know whether $p^2 - 2 q^2 = r$ has a solution in primes.
The solutions of the Pell-type equation $x^2 - 2 y^2 = 5039$ are $x_n, y_n$ given by the recurrences $x_{n+4} = 6 x_{n+2} - x_n$ with initial values $x_0 = 71, x_1 = 209, x_2 = 217, x_3 = 1183$ and $y_{n+4} = 6 y_{n+2} - y_n$ with initial values $y_0=1,y_1=139,y_2=145,y_3=835$. Both $x_n$ and $y_n$ have lots of prime values. I haven't found any cases where they are both prime for the same $n$ (having tested up to $n=10000$). There are some "near misses", e.g. neither $x_{179}$ nor $y_{179}$ is prime but they have no small factors. Thus there doesn't appear to be any modular reason for solutions not to exist.
Heuristically, since $x_n$ and $y_n$ increase exponentially, each has probability $O(1/n)$ of being prime, so the probability of both being prime is $O(1/n^2)$, and since $\sum_n 1/n^2 < \infty$, we might expect finitely many $n$ with both $x_n$ and $y_n$ prime. So maybe there just happen to be none, but there's no way to actually prove that. Still, I thought I'd put this to MSE in the hope that there's something clever that I'm missing.
EDIT: I might mention that in order for $p^2 - 2 q^2 = r$ to have prime solutions, where $r$ is prime, either $q=2$ or $3$ (so $r + 8$ or $r + 18$ is the square of a prime) or $r \equiv 23 \mod 24$. Of the primes $\equiv 23 \mod 24$ less than $10000$, the only ones for which I haven't found prime solutions are $4079$, $5039$ and $7703$, but I can prove there are no prime solutions for $4079$ and $7703$ (all solutions to $x^2 - 2 y^2 = r$ in those cases have $x$ or $y$ divisible by $5$, $7$ or $11$).
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