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Recently I came up with a positive solution $((x,y)\neq (\pm 1;0))$ to this diophantine equation $$ x^2-\left(w^2(2^{n-2}p)^2+2^n(2^{n-2}p)\right)y^2=1,\qquad n\geq 2, $$ where all variables are in $ \mathbb{Z}^*$

Is it a known case in terms of the theory? As one can read, for instance, in the notes

  • Michel Waldschmidt, Pell’s equation, author pdf.

In fact the solutions are rather elementary ideas.

Thank you.

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  • $\begingroup$ Can you fix the equation? There's an unpaired bracket, and also $2^n(2^{n-2}p)) = 2^{2n-2}p$, but it may be that's not what you meant. That means you could also write the last term on the left side as $p(2^{n-1}y)^2$. And what is $w$? A parameter? I can't comment on the solution itself. $\endgroup$
    – David Roberts ♦
    Commented Nov 9, 2020 at 5:11
  • $\begingroup$ You see that's a Pell like equation where say $$x^2-(w^2p^2+2^n(p))y^2=1$$, all variables as i said are nonzeoro relative integers, and for example for $n=2$ we get $x^2-(w^2p^2+4p)y^2=1$. There are similar equations known in the literature i attached an example file. $\endgroup$
    – Toni Mhax
    Commented Nov 9, 2020 at 5:39
  • $\begingroup$ The link to the pdf is broken for me, 404 error. Could you check it? $\endgroup$
    – asymptotic
    Commented Nov 9, 2020 at 14:59
  • $\begingroup$ ok fixed thank you. $\endgroup$
    – Toni Mhax
    Commented Nov 9, 2020 at 19:18
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    $\begingroup$ thanks relative integer is from entier relatif i'll edit $\endgroup$
    – Toni Mhax
    Commented Nov 10, 2020 at 1:11

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