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?

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

Keith Conrad's papers are great:

Pell Equation I: This goes through a number of interesting examples and discussion.

Pell Equation II: This goes through the theory and proves the existence of the solution.

Of course, these only show that a solution exists. Not that there is a fundamental unit generating all solutions.

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.

| cite | improve this answer | |

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.