How can someone solve the negative Pell's equation $x^2-48y^2=-1$? Is there any general solution? Thank you.

  • $\begingroup$ When the Pell equation with $-1$ has a solution the standard continued fraction algorithm will find it. This one doesn't, as @MichaelRosenberg proves. $\endgroup$ – Ethan Bolker Jan 5 '18 at 16:39
  • $\begingroup$ why? I am trying to understand the reason $\endgroup$ – user504516 Jan 5 '18 at 16:39

If so then $x^2+1$ is divided by $3$, which is impossible.

Indeed, we can consider three cases:

  1. $x=3k$;

  2. $x=3k+1$ and

  3. $x=3k-1$, where $k$ is an integer number.

Easy to see that it's impossible.

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  • $\begingroup$ I think you are missing a few steps. Edit: Never mind! $\endgroup$ – Brazilian Cérebro Jan 5 '18 at 16:37
  • $\begingroup$ I can't understand your answer. Can you explain it a little more further or give me a source to read? $\endgroup$ – user504516 Jan 5 '18 at 16:38
  • $\begingroup$ @John IO added something. See now. $\endgroup$ – Michael Rozenberg Jan 5 '18 at 16:43
  • $\begingroup$ Because we can "find" an expression for $x^2$ by setting $x^2 \equiv -1 \pmod {48}$ (same as you would do with linear diophantine equations). But since 48 is a multiple 3, this is equivalent to saying that $x^2 \equiv -1 \pmod {3}$. You can quickly check with a remainder table that this is impossible $\endgroup$ – Francisco José Letterio Jan 5 '18 at 16:45

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