Some Proofs

The rest of this lecture is devoted to proving most of Theorem 3.3. We will prove that partial convergents to continued fractions contribute infinitely many solutions to Pell's equation. We will not prove that every solution to Pell's equation is a partial convergent, though this is true.¹

Fix a positive nonsquare integer $d$.

Definition 4.1   A quadratic irrational $\alpha=a+b\sqrt{d}$ is reduced if $\alpha>1$ and if the conjugate of $\alpha$, denoted by $\alpha'$, satisfies $-1<\alpha'<0$.

For example, the number $\alpha=1+\sqrt{2}$ is reduced.

Definition 4.2   A continued fraction is purely periodic if it is of the form $[\overline{a_0,a_1,\ldots,a_n}]$.

The continued fraction $[\overline{2}]$ of $1+\sqrt{2}$ is purely periodic.

Lemma 4.3   If $\alpha$ is a reduced quadratic irrational, then the continued fraction expansion of $\alpha$ is purely periodic. (The converse is also true, and is easy to prove.)

Proof. The proof can be found on pages 102-103 of Davenport's book. $\qedsymbol$

Lemma 4.4   The continued fraction expansion of $\sqrt{d}$ is of the form

$$[a_0, \overline{a_1, \ldots, a_{n-1}, 2 a_0}].$$

Proof. Let $a_0$ be the floor of $\sqrt{d}$. Then $\alpha=\sqrt{d} + a_0$ is reduced because $\alpha>1$ and $\alpha'=-\sqrt{d}+a_0$ satisfies $-1<\alpha'<0$. Let $[a_0, a_1, a_2, \ldots ]$ be the continued fraction expansion of $\sqrt{d}$. Then the continued fraction expansion of $\sqrt{d}+a_0$ is $[2a_0, a_1, a_2, \ldots]$. By Lemma 4.3, the continued fraction expansion of $\sqrt{d}+a_0$ is purely periodic, so

$$[2a_0, a_1, a_2, \ldots] = [\overline{2a_0, a_1, a_2, \ldots, a_{n-1}}],$$

where $n$ is the period. It follows that $a_n = 2 a_0$, as claimed. $\qedsymbol$

The following proposition shows that there are infinitely many solutions to Pell's equation that arise from continued fractions.

Proposition 4.5   Let $p_k/q_k$ be the partial convergents of the continued fraction expansion of $\sqrt{d}$, and let $n$ be the period of the expansion of $\sqrt{d}$. Then

$$p_{kn-1}^2 - d q_{kn-1}^2 = (-1)^{kn}$$

for $k=1,2,3,\ldots$.

Proof. ²By Lemma 4.4, for $k\geq 1$, the continued fraction of $\sqrt{d}$ can be written in the form

$$\sqrt{d} = [a_0, a_1, a_2, \ldots, a_{kn-1}, r_{kn}]$$

where

$$r_{kn} = [2a_0, \overline{a_1, a_2, \ldots, a_{n}}] = a_0 + \sqrt{d}.$$

Because $\sqrt{d}$ is the last partial convergent of the continued fraction above, we have

$$\sqrt{d} = \frac{r_{kn} p_{kn-1} + p_{kn-2}}{r_{kn} q_{kn-1} + q_{kn-2}}.$$

Upon substituting $r_{kn} = a_0 + \sqrt{d}$ and simplifying, this reduces to

$$\sqrt{d}(a_0 a_{kn-1} + q_{kn-2} - p_{kn-1}) = a_0 p_{kn-1} + p_{kn-2} - d q_{kn-1}.$$

Because the right-hand side is rational and $\sqrt{d}$ is irrational,

$$a_0 a_{kn-1} + q_{kn-2} = p_{kn-1},$$    and $$\quad a_0 p_{kn-1} + p_{kn-2} = d q_{kn-1}.$$

Multiplying the first of these equations by $p_{kn-1}$ and the second by $-q_{kn-1}$, and then adding them, gives

$$p_{kn-1}^2 - d q_{kn-1}^2 = p_{kn-1}q_{kn-2} - q_{kn-1} p_{kn-2}.$$

But

$$p_{kn-1}q_{kn-2} - q_{kn-1} p_{kn-2} = (-1)^{kn-2} = (-1)^{kn},$$

which proves the proposition. $\qedsymbol$