# Integers $d$ for which the negative Pell equation is soluble for both $d$ and $2d$?

Let $$\text{NPE}_d$$ denote the negative Pell equation: $$x^2-dy^2=-1$$ Where $$d$$ is a given positive nonsquare integer and integer solutions are sought for x and y.

we know that (in this paper archive):

Theorem : The equation $$\text{NPE}_d$$ has integer solutions if and only if there exist two integers $$a(d)=a$$ and $$b(d)=b$$ such that $$d=a^2+b^2$$ and there exists a Pythagorean triplet $$(A,B,C)$$ such that $$|aA-bB|=1$$ and in this case $$(Aa+Bb,C)$$ is a solution.

Obviously if $$\text{NPE}_d$$ has integer solutions then $$d$$ cannot be divisible by any prime $$p$$ such that $$p=3\mod 4$$.

My question: Is there any characterization for the integers $$d$$ for which $$\text{NPE}_d$$ and $$\text{NPE}_{2d}$$ have both integer solutions.

I used the characterization above, but I can't link the couple $$(a(d),b(d))$$ to $$(a(2d),b(2d))$$ because the theorem doesn't give us much information

The sequence of the elements $$d$$ for which $$\text{NPE}_d$$ is soluble is OEIS A031396.

I posted this question in math.exchange and does not receive any answer

I know the question is old, but is possible to give an exact characterization to $$d$$, at least if it's a prime number:

It is known that considering a prime $$p \equiv 1\mod 4$$, there is always a solution to $$x^2 - py^2 = -1$$ in integers, the proof is from Mordell "Diophantine Equations" pages 55-56 (thanks to @Will Jagy for pointing that out).

Furthermore Dirichlet proved that if the prime $$p \equiv 1 \mod 4$$ and $$p \equiv 5 \mod 8$$ or $$p \equiv 9 \mod 16$$, the equation $$x^2 - 2py^2 = -1$$ has still integer solutions (reference: https://www.forgottenbooks.com/en/download/ThePellEquation_10024828.pdf pag. 80)

So you can say that if $$p$$ is a prime and $$p \equiv 5 \mod 8$$ or $$p \equiv 9 \mod 16$$ both NPEₚ NPE₂ₚ have integer solutions

• Note that Mordell's proof can be easily adapted to this: primes $p,q \equiv 1 \pmod 4$ but mutual quadratic non-residues, that is Legendre symbol $(p,q) = -1$ then there are integer solutions to $x^2 - pq y^2 = -1$ The first two semiprimes where impossibility is a surprise are $205 = 5 \cdot 41$ and $221 = 13 \cdot 17$ both times mutual residues. Commented Jun 22 at 18:36
• It appears Whitford skips over that one. However, on page 80 he lists conditions 1,2,3,4, but the refers to conditions 3,4,5 . Perhaps he meant to include a condition 5 with the simple $(p,q) = -1,$ or meant that as condition 4 and the one with quartic residues moved to condition 5. Commented Jun 22 at 18:57

(Too long for a comment.)

The sequence of integers such that $x^2-py^2=-1$ is solvable is given by,

$$p = 1, 2, 5, 10, 13, 17, 26, 29, 37, 41, 50, 53, 58, 61, 65, 73, 74, 82, 85, 89, 97, 101,\dots$$

which is A031396, while that of $x^2-2q y^2 = -1$ is (if I did my code right),

$$q = 1, 5, 13, 25, 29, 37, 41, 53, 61, 65, 85, 101, \dots$$

and is not yet in the OEIS.

Point 1: The sequence $p$ does not contain non-square $q$ as a subset.

With limited data, it seems to be the case. But the first missing value is $q=221$, since $x^2-221y^2=-1$ is not solvable, while $x^2-2\cdot221y^2=-1$ is.

Point 2: There is an infinite number of intersections between $p$ and $q$.

Proof: We use the identities,

$$m^2-(m^2+1)\cdot 1^2 = -1$$

$$(2n+1)^2-2\cdot(2n^2+2n+1)\cdot 1^2 = -1$$

Equate,

$$m^2+1 = 2n^2+2n+1$$

and turns out to be a well-known Pell equation in disguise,

$$(2n+1)^2-2m^2 = 1$$

$$u^2-2v^2=1$$

with solutions $(u,v) = (3,2),\,(17,12),\,(99,70),\dots$ and proves there is an infinite number of $d$ such that

$$x^2-dy^2 = -1$$

$$x^2-2dy^2 = -1$$

is both solvable.

P.S. However, to characterize all $d$ seems to be difficult.

• The number 25 is in your list of values of $q$, so that is the first missing value. Commented May 16, 2015 at 16:16
• @KConrad: Oops, I was not being precise. I was focusing on whether there were missing non-square $q$. Commented May 17, 2015 at 1:19