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Let $\cal D$ be the set of all $D=1,2,3,\dots$ such that $D\equiv 0,1 \mod(4)$ and $D$ is not a square. For $D\in\cal D$ let $\varepsilon(D)$ denote the smallest number $\varepsilon=\frac{a+b\sqrt D}2$ which is $>1$ and satisfies $a,b\in\mathbb Z$, $a^2-b^2D=4$ (Pell's equation). Let $\cal E$ denote the set of all such numbers $\varepsilon(D)$ with $D\in\cal D$. My question is about the differences $\varepsilon-\varepsilon'$. For $T>0$ let $M(T)$ denote the set of all pairs $(\varepsilon,\varepsilon')$ in ${\cal E}\times\cal E$ such that $0<\varepsilon-\varepsilon' < T$.

  1. Is it true that there exists $T>0$ such that $M(T)$ is finite?

  2. Is it true that $M(T)$ is always finite?

All hints to the literature on properties of the fundamental units $\varepsilon\in\cal E$ are welcome!

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2 Answers 2

up vote 3 down vote accepted

Note: The following answer assumes that the question means (all) fundamental units in its standard meaning as opposed to a litteral reading of the question in the body (see Impossible Hippopotamus's answer for details related to this, and also for a more complete answer in general).

For the second question, it is not true that $M(T)$ is always finite.

A way to answer this second question is via a result of Sprindžuk, The distribution of the fundamental units of real quadratic fields, Acta Arithmetica (1973/74). There it is proved, among other things, that the number of $\epsilon$, fundamental units of real quadratic number field, with $\epsilon \le x$ is $2x + O(x^{1/2})$, in fact something more precisely is established.

Thus for every $T_0 > 1/2$ there need to be infinitely many pairs at most $T_0$ apart, else one could not have about $2x$ of them in the interval $[1,x]$ (for $x$ ever larger).

So for $T_0> 1/2$ one has $M(T_0)$ is infinite.

For the first, I do not know. There are some ways to construct explictly sequences of fundamental units, but I did not come across one where the gap is arbitrarily small (but my knowledge here is certainly not at all exhaustive); I suspect however they could be used to give a different argument for the second question. See for example Degert, Über die Bestimmung der Grundeinheit gewisser reell-quadratischer Zahlkörper, Abh. Math. Sem. Univ. Hamburg (1958)

Somehow I would guess that $M(T)$ is in fact always infinite, based on the fact above that they are not too sparse globally, so somehow sometimes they should be arbitrarily close together. But this is a vague guess and might as well be wrong.

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The phrasing of your question is a little peculiar: one would usually impose the condition that $D$ is square-free. You also imposed the condition that the norm of $\varepsilon$ is $+1$, whereas the fundamental unit may have norm $-1$.

Under the assumption that $N(\varepsilon) = 1$, there is a trivial equality:

$$\varepsilon + \overline{\varepsilon} = a.$$

From the assumption that $\varepsilon > 1$, this yields the inequality

$$a - \frac{1}{a} > \varepsilon > a - \frac{1}{a-1}.$$

From this is follows that $|\varepsilon - \varepsilon'| > 1$ for distinct units of norm $+1$. Thus $M(1)$ is finite. For any $a > 2$, there exists a unique $\varepsilon$ with norm $+1$, given by solving the following equation for $b^2 D$:

$$a^2 - b^2 D = 4.$$

This specifies $\varepsilon$ uniquely. An estimate similar to the one given in quid's answer (in fact, it can be derived directly from that estimate) will show that the number of fundamental units which have norm $+1$ with $a \sim \varepsilon \le x$ is $x + O(x^{1/2})$, and thus that $M(T)$ is infinite for $T > 1$. The disparity between this answer and the one above is whether one allows units of norm $-1$ or not. (This estimate also shows that all but $O(x^{1/2})$ of the units $\varepsilon$ generated in this way for any $a \le x$ are fundamental units.)


If you wish to allow units of norm $-1$, then note that for every $a > 2$ there exists a unique unit $\varepsilon > 1$ of trace $a$ and norm $1$, and a unique unit $\epsilon > 1$ of trace $a$ and norm $-1$. We saw that above in the first case, and the second case is the same: simply solve the equation:

$$a^2 - b^2 D = -4.$$

In this case we will still have

$$\epsilon + \overline{\epsilon} = a,$$

and hence $0 = (\varepsilon + \overline{\varepsilon}) - (\epsilon + \overline{\epsilon})$, from which we get

$$|\varepsilon - \epsilon| = |\overline{\varepsilon} - \overline{\epsilon}| \sim \frac{1}{2a}.$$

From the reference in quid's answer, we know that for $x + O(x^{1/2})$ of the $a \le x$, either (and hence, because $O(x^{1/2}) + O(x^{1/2}) = O(x^{1/2})$, both) $\varepsilon$ and $\epsilon$ are fundamental units. Hence, allowing units to be of norm $-1$ and modifying your definition of $M(T)$ appropriately, one has $M(\delta) = \infty$ for any $\delta > 0$.

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