On the Pell-like $px^2-qy^2 = 1$ for prime $p,q$

Question

Given any prime of form $p_n = u^2+nv^2$ for non-zero integers $u,v$. Consider,

\begin{aligned} &p_2x^2-2y^2 = 1\\ &p_3x^2-3y^2 = 1\\ &p_7x^2-7y^2 = 1\\ &p_{11}x^2-11y^2 = 1\\ \end{aligned}

Question 1: Are these four families of Pell-like eqns always solvable in non-zero integers $x,y$?

Ex.

Let $n=11$, and since $53=u^2+11v^2$, then $53x^2-11y^2=1$ is solvable. (Other $p_{11}$ will do.)

Question 2: Is there another $n$?

Let $n=13$, and since $157=u^2+13v^2$, but $157x^2-13y^2=1$ is NOT solvable in the integers.

Have these questions been answered in the literature already?

Answer 1

There is no solution for $$ 401 x^2 - 11 y^2 = 1. $$

j

agy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 
401 0 -11

  0  form            401           0         -11  delta     -6
  1  form            -11         132           5


           0          -1
           1          -6

To Return  
          -6           1
          -1           0

0  form   -11 132 5   delta  26
1  form   5 128 -63   delta  -2
2  form   -63 124 9   delta  14
3  form   9 128 -35   delta  -3
4  form   -35 82 78   delta  1
5  form   78 74 -39   delta  -2
6  form   -39 82 70   delta  1
7  form   70 58 -51   delta  -1
8  form   -51 44 77   delta  1
9  form   77 110 -18   delta  -6
10  form   -18 106 89   delta  1
11  form   89 72 -35   delta  -2
12  form   -35 68 93   delta  1
13  form   93 118 -10   delta  -12
14  form   -10 122 69   delta  1
15  form   69 16 -63   delta  -1
16  form   -63 110 22   delta  5
17  form   22 110 -63   delta  -1
18  form   -63 16 69   delta  1
19  form   69 122 -10   delta  -12
20  form   -10 118 93   delta  1
21  form   93 68 -35   delta  -2
22  form   -35 72 89   delta  1
23  form   89 106 -18   delta  -6
24  form   -18 110 77   delta  1
25  form   77 44 -51   delta  -1
26  form   -51 58 70   delta  1
27  form   70 82 -39   delta  -2
28  form   -39 74 78   delta  1
29  form   78 82 -35   delta  -3
30  form   -35 128 9   delta  14
31  form   9 124 -63   delta  -2
32  form   -63 128 5   delta  26
33  form   5 132 -11   delta  -12
34  form   -11 132 5
minimum was   5rep 0 1 disc   17644 dSqrt 132.83071934  M_Ratio  145.8182
Automorph, written on right of Gram matrix:  
930589980  -691078313
1056608089  1136194800
 Trace:  2066784780   gcd(a21, a22 - a11, a12) : 1
=========================================
jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$

Answer 2

$$ 977 x^2 - 7 y^2 = 1$$ is not possible in integers. $$ 23^2 + 7 \cdot 8^2 = 977. $$

jagy@phobeusjunior:~/old drive/home/jagy/Cplusplus$ ./indefCycle
Input three coefficients a b c for indef f(x,y)= a x^2 + b x y + c y^2 
977 0 -7

  0  form            977           0          -7  delta    -11
  1  form             -7         154         130


           0          -1
           1         -11

To Return  
         -11           1
          -1           0

0  form   -7 154 130   delta  1
1  form   130 106 -31   delta  -4
2  form   -31 142 58   delta  2
3  form   58 90 -83   delta  -1
4  form   -83 76 65   delta  1
5  form   65 54 -94   delta  -1
6  form   -94 134 25   delta  5
7  form   25 116 -139   delta  -1
8  form   -139 162 2   delta  81
9  form   2 162 -139   delta  -1
10  form   -139 116 25   delta  5
11  form   25 134 -94   delta  -1
12  form   -94 54 65   delta  1
13  form   65 76 -83   delta  -1
14  form   -83 90 58   delta  2
15  form   58 142 -31   delta  -4
16  form   -31 106 130   delta  1
17  form   130 154 -7   delta  -22
18  form   -7 154 130

Answer 3

1: Non-solvable equations for primes 2 and 3

For prime 2, $$97X^2-2Y^2=1$$ has no solutions and $97=5^2+2(6)^2$.

For prime 3, $$457X^2-3Y^2=1$$ there are no solutions and $$457 = 5^2+3(12)^2$$

This should complete your list. You can check that a solution for either case means $$X^2-194Y^2=97$$ or $$X^2-1371Y^2=457$$ has a solution (just multiplying by a constant). This site can help you verify the non-solvability.

2: Explicit results for prime 2

The situation for $$px^2-2y^2=1$$ is known. I have just posted a solution here.

The results, quoting from there, is as follows:

If $\beta\neq 0$ is an integer and $2\beta^2+1=\alpha^2p$ for some integer $\alpha$ and prime $p$, then the equation $$px^2-2y^2=1$$ is solvable and one of the solutions is given by \begin{align*} x&=\alpha\\ y&=\beta \end{align*} This is easily checked: $$px^2-2y^2=p\alpha^2-2\beta^2=2\beta^2+1-2\beta^2=1$$ (Note: The rest are obtained by composition with fundamental solutions.)

Moreover, all solvable primes must occur in such a way. To solve this we can use the representation theorems of Binary Quadratic Forms, which is provided in the link.

Note: In fact, this parametrized solution is a consequence of the representation theorems. It is not found by random guessing.

Therefore you are looking for the intersection: \begin{align*} 2\beta^2+1 &=\alpha^2 p\\ u^2+2v^2 &= p \end{align*} where $\alpha,\beta,u,v$ are integers. For your question you need $u^2+2v^2 = p$ implies $2\beta^2+1=\alpha^2p$ for some integers $\alpha,\beta$. This is not always true. For example, the equation I put on at the top $$97X^2-2Y^2=1$$ has no solutions and $97=5^2+2(6)^2$.

There is something similar for odd primes too, so that you can deduce conditions for the forms. However it is probably quite tricky to work out the details. (similar to what I did in my post)