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):
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.
Thank you for your help.
It is necessary to use the formula. http://www.artofproblemsolving.com/blog/101140
$$x^2-dy^2=-1$$
$$z^2-2dg^2=-1$$
It will give us a series of solutions using the Pell equation: $$p^2-2s^2=k$$
Let us use the fact that the following solution can be found knowing the previous, formula.
$$p_2=3p_1+4s_1$$
$$s_2=2p_1+3s_1$$
Using certain series of solutions you can find ratio formula.
$p^2-2s^2=1$ $;$ $(p_1;s_1) - (3;2);$ $d=s^2+1$
$p^2-2s^2=-1$ $;$ $(p_1;s_1) - (1;1) ;$ $d=s^2+4$
$p^2-2s^2=-7$ $;$ $(p_1;s_1) - (1;2) ;$ $d=s^2+1$
$p^2-2s^2=7$ $;$ $(p_1;s_1) - (3;1) ;$ $d=s^2+4$
$p^2-2s^2=17$ $;$ $(p_1 ; s_1) - (5;2) ;$ $d=s^2+9$
While it is difficult to say. Limited to any solution of these series or not. Clearly what is involved in this equation Pell.