# Fundamental Solution of Pell's Equation

To find the solutions of Pell's equation $x^2 - dy^2 = 1$, one can look at the convergents of the continued fraction expansion of $\sqrt d$: If $(x, y)$ is a non-trivial solution, then $x \over y$ is a convergent of $\sqrt d$.

The smallest pair $(x,y)$ with $x > y > 0$ found by testing the convergents is called the fundamental solution.

My question is: how can we show that a fundamental solution must exist?

• Google "wikiproofs" for a detailed analysis. – Peter Apr 29 '18 at 18:46

Fix $d>0$ a squarefree integer. What are the solutions $(x,y) \in \mathbb{Z}^2$ to the Pell equation $x^2-dy^2=1$? We can rephrase this question as follows: what elements $a+b\sqrt{d} \in K:=\mathbb{Q}(\sqrt{d})$ of the order $R=\mathbb{Z}[\sqrt{d}]$ with $\text{Nm}_{K/\mathbb{Q}}(a+b\sqrt{d})=1$? This set is in bijection with the set of solutions of Pell's equation $G:=\{(a,b) \in \mathbb{Z}^2 \colon a^2-db^2=1\}$.
Now for any order $R \subseteq K$, $R^\times \cong \mu_R \times \mathbb{Z}^{r+s-1}$, where $\mu_R$ is the set of roots of unity in $R$. Then $R^\times \cong \mu_R \times \mathbb{Z}$. In this case, $G$ is index 1 (all units have norm $1$) or $2$ (there is a unit of norm $-1$) in $R^\times$. Therefore, $G= \pm \langle u \rangle \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}$.
Of course, this makes use of Dirichlet's Unit Theorem applied to $K=\mathbb{Q}(\sqrt{d})$. Since $d>0$, there are two real embeddings and no complex embeddings so that the unit group is of rank $r+s-1=2+0-1=1$ so that there is a single generator, namely the fundamental unit.