Units in Real Quadratic Fields

Let $d$ be a nonsquare positive integer, and set

$$\mathbb{Q}(\sqrt{d}) = \{a + b\sqrt{d} : a, b\in\mathbb{Q}\}$$

$$\mathbb{Z}[\sqrt{d}] = \{a + b\sqrt{d} : a, b\in\mathbb{Z}\}.$$

Then $\mathbb{Q}(\sqrt{d})$ is a real quadratic field and $\mathbb{Z}[\sqrt{d}]$ is a ring. There is a homomorphism called norm:

$$N : \mathbb{Q}(\sqrt{d})^* \rightarrow \mathbb{Q}^*, \qquad N\lef... ...rt{d}\right) = \left(a+b\sqrt{d}\right)\left(a-b\sqrt{d}\right) = a^2 - b^2 d.$$

Definition 3.1   An element $x\in R$ is a unit if there exists $y\in{}R$ such that $xy=1$.

Proposition 3.2   The units of $\mathbb{Z}[\sqrt{d}]$ are exactly the elements of norm $\pm 1$ in $\mathbb{Z}[\sqrt{d}]$.

Proof. Suppose $u\in\mathbb{Z}[\sqrt{d}]$ is a unit. Then

$$1 = N(1) = N(u u^{-1}) = N(u)\cdot N(u^{-1}).$$

Since $N(u), N(u^{-1})\in\mathbb{Z}$, we have $N(u) = N(u^{-1}) = \pm 1$ $\qedsymbol$

Thus Fermat's challenge amounts to determing the group $U^+$ of units in $\mathbb{Z}[\sqrt{d}]$ of the form $a+b\sqrt{d}$ with $a, b \geq 0$.

Theorem 3.3   The group $U^+$ is an infinite cyclic group. It is generated by $p_m+ q_m\sqrt{d}$, where $\frac{p_m}{q_m}$ is one of the partial convergents of the continued fraction expansion of $\sqrt{d}$. (In fact, if $m$ is the period of the continued fraction of $\sqrt{d}$ then $n=m-1$ when $m$ is even and $2n-1$ when $m$ is odd.)

The theorem implies that Pell's equation always has a solution! Warning: the smallest solution is typically shockingly large. For example, the value of $x$ in the smallest solution to $x^2 - 1000099y^2 = 1$ has 1118 digits.

The following example illustrates how to use Theorem 3.3 to solve Pell's equation when $d=61$, where the simplest solution is already quite large.

Example 3.4   Suppose $d=61$. Then

$$\sqrt{d} = [7, \overline{1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14}],$$

which has odd period $n=11$. Thus the group $U^+$ is generated by

$$x = p_{21} = 1766319049$$

$$y = q_{21} = 226153980.$$

That is, we have

$$U^+ = \langle u \rangle = \langle 1766319049 + 226153980\sqrt{61}\rangle,$$

and $x = 1766319049$, $y=226153980$ gives a solution to $x^2 - dy^2 = 1$. All the other solutions arise from $u^n$ for some $n$. For example,

$$u^2 = 6239765965720528801 + 798920165762330040\sqrt{61}$$

leads to another solution.

Remark 3.5   To help with your homework, note that if the equation

$$x^2 - dy^2 = n$$

has at least one (nonzero) solution $(x_0, y_0)\in\mathbb{Z}\times \mathbb{Z}$, then it must have infinitely many solutions. This is because if $x_0^2 - dy_0^2 =n$ and $u$ is a generator of the cyclic group $U^+$, then for any integer $i$,

$$N(u^i(x_0+ y_0\sqrt{d})) = N(u^i)\cdot N(x_0+y_0\sqrt{d}) = 1\cdot n = n,$$

so

$$x_1 + y_1\sqrt{d} = u^i(x_0 + y_0\sqrt{d})$$

provides another solution to $x^2 + d y^2 = n$.