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Let $k=\mathbb{Q}(\sqrt{d})$ -- $d$ is a positive square-free integer --
be a real quadratic field, and let $\varepsilon_k$ be a fundamental unit.
Let $(a,b)$ be the minimal solution to the Pell equation $x^2-dy^2=1$ and let $\varepsilon = a+\sqrt{d}b$.
Is it always true that $\varepsilon = \varepsilon_k^n$ for some $n \in \mathbb{N}$ ? If yes, could we be more precise about $n$ ?

Thank you ! Rony

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  • $\begingroup$ Yes: $n=1$ when $x^2-dy^2=-1$ has no solution, while $n=2$ when $x^2-dy^2=-1$ has a solution (and that solution gives the fundamental unit). Note that it's technically wrong to say "the" fundamental unit, since $\epsilon_k^{-1}$ is also a fundamental unit; in practice you can keep them sorted out. $\endgroup$ Mar 29, 2015 at 23:58
  • $\begingroup$ But in the case of $d=5$, $\varepsilon_k = \frac{1+\sqrt{5}}{2}$ is a solution to $x^2-5y^2=-1$ but the minimal solution to $x^2-5y^2=1$ is $(9,4)$ producing $\varepsilon = 9 +4\sqrt{5}$ which satisfies $\varepsilon = \varepsilon_k^6$, i.e. $n=6$, not $n=1$ ?! $\endgroup$
    – Rony Bitan
    Mar 30, 2015 at 6:40
  • $\begingroup$ Ah, you're right, my comment is incorrect when the algebraic integers/units involved can have denominator $2$ (e.g., $d\equiv1\pmod 4$). Hopefully someone can fix. $\endgroup$ Mar 30, 2015 at 6:50
  • $\begingroup$ I meant not $n=2$ ?! $\endgroup$
    – Rony Bitan
    Mar 30, 2015 at 6:57
  • $\begingroup$ Thank you. Do you have a reference to your above comment ? $\endgroup$
    – Rony Bitan
    Mar 30, 2015 at 7:18

1 Answer 1

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The integer $n$ is at most 6.

If we consider the group $O_K^\times$, where $K$ is the field $\mathbf{Q}(\sqrt{d})$, then:

  • the subgroup consisting of elements of norm $+1$ (rather than $-1$) has index at most 2, because it is the kernel of a homomorphism to $\{ \pm 1\}$;
  • the subgroup consisting of elements lying in $\mathbf{Z}[\sqrt{d}]$ has index either 1 or 3, because it is obviously everything if $d \ne 1 \bmod 4$, and if $d = 1 \bmod 4$, the group $(O_K / 2 O_K)^\times$ has order either 1 or 3, and a unit that is congruent to 1 modulo $2 O_K$ lies in $1 + 2 O_K \subseteq \mathbf{Z} + 2 O_K = \mathbf{Z}[\sqrt{d}]$.

Hence the subgroup of $O_K^\times$ corresponding to solutions of $a^2 - d b^2 = 1$ with $a,b \in \mathbf{Z}$ has index at most 6 in the full unit group; and the minimal positive integer solution of $a^2 - d b^2 = 1$ corresponds to a generator of this group.

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