On the solvability of the negative Pell equation $x^2-2py^2 = -1$

Question

Given prime $p=8n+1$. Then

$$x^2-2py^2 = -1\tag1$$

is not solvable for,

$$p_1= 17, 73, 89, 97, 193, 233, 241, 257, 281, 337, 353, 401, 433, 449, 577, 593,601, 617, 641,\dots$$

but is solvable for,

$$p_2= 41, 113, 137, 313, 409, 457, 521, 569, 761, 809, 857, 953, 1129, 1201, 1321, 1601,\dots$$

Compare to the primes of form $p=u^2+32v^2$ (A105389):

$$p_3 = 41, 113, 137, \color{brown}{257}, 313, \color{brown}{337}, \color{brown}{353}, 409, 457, 521, 569, \color{brown}{577}, \color{brown}{593}, 761, 809, 857, \color{brown}{881}, 953, \dots$$

(Also, $p_3$ has class number $h(-p)$ divisible by 8.)

Q: Is $p_2$ a subset of $p_3$?

(Equivalently, for $p=8n+1$, is it true that a necessary but not sufficient condition such that $(1)$ is solvable is that $p = u^2+32v^2$?)

I have checked that all solvable $p = 8n+1 \leq 18089$ has the form $u^2+32v^2$, but I don't know if all solvable $p$ have that form.

$\color{blue}{Edit}$: (In response to Jagy's answer.) The primes of form $p=u^2+64v^2$ (A014754) are,

$$p_4 = 73, 89, 113, 233, 257, 281, 337, 353, 577, 593, 601, 617, 881, 937, 1033, 1049, 1097\dots$$

but neither $p_1$ nor $p_2$ is a subset of $p_4$. However, the primes of form $p=u^2+64v^2=16n+9$,

$$p_5 = 73, 89, 233, 281, 601, 617, 937, 1033, 1049, 1097,\dots$$

as a result of Dirichlet, is unsolvable and so is a subset of $p_1$.

Answer 1

at least a start: for $p \equiv 1 \pmod 8,$ there is a trichotomy due to Dirichlet, pages 164-165 of Buell: exactly one of $$ A: \; 2 x^2 - p y^2 = 1, $$ $$ B: \; 2 x^2 - p y^2 = -1, $$ $$ C: \;2 x^2 - p y^2 = -2, $$ is solvable in integers. Your $(1)$ is the third choice $C$, as $y$ is then even. I'm afraid this material relates more clearly to $p = u^2 + 64 v^2,$ as the results are

If A is solvable, then C is not solvable and $p \equiv 1 \pmod {16}.$

If B is solvable, then C is not solvable and $p = u^2 + 64 v^2$

If $p \equiv 9 \pmod {16}$ and $2$ is not a fourth power, then C is solvable. Note $p \neq u^2 + 64 v^2$

These are the primes $9 \pmod {16}$ represented by $4 u^2 + 4uv + 17 v^2,$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./primego
Input three coefficients a b c for positive f(x,y)= a x^2 + b x y + c y^2 
4 4 17
Discriminant  -256

Modulus for arithmetic progressions? 
16
Maximum number represented?         1700
          p          mod 16
          17           1
          41           9
          97           1
         137           9
         193           1
         241           1
         313           9
         401           1
         409           9
         433           1
         449           1
         457           9
         521           9
         569           9
         641           1
         673           1
         761           9
         769           1
         809           9
         857           9
         929           1
         953           9
         977           1
        1009           1
        1129           9
        1297           1
        1321           9
        1361           1
        1409           1
        1489           1
        1657           9
        1697           1
........................

A little more: we can write $p = u^2 + 32 v^2$ if and only if $p \equiv 1 \pmod 8$ and there are four distinct roots to $$ z^4 - 2 z^2 + 2 \equiv 0 \pmod p. $$ We can write $p = u^2 + 64 v^2$ if and only if $p \equiv 1 \pmod 8$ and there are four distinct roots to $$ z^4 - 2 \equiv 0 \pmod p. $$ This is from the final table in Liu and Williams, about 1994.