4
$\begingroup$

Find a fundamental unit in the ring of integers $\mathbb Z[\frac{1+\sqrt{141}}{2}]$ of $\mathbb Q(\sqrt{141})$

I have different corollaries for different numbers, the most appropriate for $141$ is the one below.

enter image description here

I used an algorithm (don't know if you know this, but

$\beta_0=\sqrt{141}+\lfloor\sqrt{141}\rfloor, \quad\beta_{n+1}=\frac{1}{\beta_n-\lfloor\beta_n\rfloor}$

$a_n=\lfloor\beta_n\rfloor$

$p_n=p_{n-1}a_n+p_{n-2}, \quad q_n=q_{n-1}a_n+q_{n-2} $) to determine the continued fraction expansion of $141$

enter image description here

As you see the the sequence $(\beta_n)_n$ is periodic with period $t=3$ and thus $\sqrt{141}=[11;\overline{1,6,22}]$, and $\pm4$ doesn't appear in the sequence $r_0,\dots,r_{t-2}$, so according to the corollary can I then assume that the fundamantel unit is also the fundamental solution, i.e. $95+8\sqrt{141}$ ?

$\endgroup$
4
+100
$\begingroup$

Proposition 19 says we need to check a few things:

  • $141 > (\pm 4)^2 + \frac{1}{2}(|\pm 4| + 1)$
  • Then $x^2 - 141y^2 = \pm 4$ appears as solution to continued fraction expansion to $\sqrt{141}$.

Since $\pm 4$ does not appear in your remainders, $x^2 - 141 y^2 = \pm 4$ does not have a solution. Therefore you just solve Pell equation with $r = 1$, which is guaranteed to happen by Propoisition 2.20

Alternatively, Pell's equation says we can approximately find the square root of $141$:

$$ \left| \frac{x}{y} - \sqrt{141} \right| = \frac{4}{y^2}\cdot \frac{1}{\sqrt{\tfrac{4}{y^2} +141 }+ \sqrt{141}} \color{red}{\mathbf{<}} \frac{1}{2y^2}$$

The red inequality is not always true if $(4,141)$ is replaced with $(r,d)$, so we make up this condition as part of the theorem, so ensure the error is small enough to be a continued fraction.


The existence of Fundamental unit has to do with the Pigeonhole Principle. In fact, it was used in several places. Here is one.

Some integer $r \in \mathbb{Z}$ must have infinitely many solutions to the Generalized Pell Eq

$$ x^2 - dy^2 = r $$

Then we can take two such solutions (congruent mod $r$) and divide them and observe we still have element of $\mathbb{Z}[\sqrt{d}]$.

$$ x + y\sqrt{d} = \frac{x_1 + y_1\sqrt{d}}{x_2 + y_2\sqrt{d}} = \frac{(x_1 x_2 - d \cdot y_1 y_2)+ (x_1 y_2 + y_2 x_1)\sqrt{d}}{r} $$

This ratio is a solution to Pell equation $x^2 -d y^2 = 1$.

$\endgroup$
  • $\begingroup$ The Pigeonhole Principle and this solution of Pell's equation is due to Dirichlet $\endgroup$ – cactus314 Apr 7 '15 at 23:54
  • 1
    $\begingroup$ Another version of this discussion appears in Section 2.7 of Borevich + Shafarevich number theory book (PDF) in the Chapter on "Decomposable Forms", meaning that $x^2 - dy^2 = (x - \sqrt{d} y)(x + \sqrt{d}y)$ has complete factorization. Pell's equation can be thought of as a special case of Dirichlet Unit Theorem in Section 2.4 or part of Quadratic Forms in Section 2.7 $\endgroup$ – cactus314 Apr 8 '15 at 0:42
6
$\begingroup$

$\sqrt{141}=[11,\overline{1,6,1,22}]$. Yes $95+8\sqrt{141}$ is a fundamental unit.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.