When is the equation $x^2-d^n y^2 = -1$ solvable?

Question

My goal is to prove or disprove that if $x^2-dy^2=-1$ is solvable, then $x^2-d^ny^2 = -1$ is solvable for every odd $n \geq 1$.

I do know that the former is solvable if and only if the continued fraction of $\sqrt{d}$ has an odd period, and I feel like it is the case that for all $n=2k+1$, $\sqrt{d^n}=\sqrt{d^{2k+1}}=d^k\sqrt{d}$ should also have an odd period, given that I am multiplying something with an odd period by an integer.

The thing is though, I'm not exactly sure how to prove that multiplying a continued fraction with an odd period by an integer gives me another continued fraction with an odd period, if that's even true.

Any suggestions to try out/consider would be immensely helpful as well!

André Nicolas · Accepted Answer · 2014-10-11T02:28:59.037

1

The conjectured result is not correct. For let $d=2$ and $n=3$. The equation $x^2-2y^2=-1$ has integer solutions but the equation $x^2-8y^2=-1$ does not. This is because for any odd $x$, we have $x^2\equiv 1\pmod 4$, and therefore $x^2-8y^2\equiv 1\pmod{4}$.

It is not hard to find infinitely many other counterexamples, such as $d=26$ or $d=82$. So the conjecture needs to be modified.

edited Oct 11 '14 at 02:28

answered Oct 10 '14 at 05:03

André Nicolas

507,029

Well spotted. Any counterexamples with $d$ odd? – Gerry Myerson Oct 11 '14 at 03:20
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That is the interesting problem. Presumably the answer is yes, the conjecture seems awfully strong. I am going to play with it, will "favourite" the question so as not to lose track. – André Nicolas Oct 11 '14 at 03:54
O. Perron. Die Lehre von der Kettenbruchen, page 108, proves that if $d$ is prime and 1 mod 4 then $d$ is not a counterexample. – Gerry Myerson Oct 11 '14 at 11:33

When is the equation $x^2-d^n y^2 = -1$ solvable?

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