Definition: An irrational root \(\alpha\) of a quadratic equation with integer coefficients is called reduced if \(\alpha > 1\) and its conjugate \(\tilde{\alpha}\) satisfies \(-1 < \tilde{\alpha} < 0\). \(\Box\)
Solutions (since assumed real) of such quadratics can be written as $$\alpha = \frac{\sqrt{D} + P}{Q}$$ where \(D, P, Q \in \mathbf{Z}\) and \(D, Q > 0\). It is also possible (though not required) to ensure that \(Q\) divides \(D - P^2\). This is actually a necessary assumption for some of the stuff I do, is mentioned here and generally frustrated the heck out of me, so that. As an example for some enlightenment, notice $$\alpha = \frac{2 + \sqrt{7}}{4}$$ is reduced but \(4\) does not divide \(7 - 2^2\). However, if we write this as \(\frac{8 + \sqrt{112}}{16}\), we have our desired condition.
Definition: We say a reduced quadratic irrational \(\alpha\) is associated to \(D\) if we can write $$\alpha = \frac{P + \sqrt{D}}{Q}$$ and \(Q\) divides \(D - P^2\). \(\Box\)
Lemma 1: Transforming a reduced irrational root \(\alpha\) associated to \(D\) into its integer part and fractional part via $$\alpha = \lfloor \alpha \rfloor + \frac{1}{\alpha'},$$ the resulting quadratic irrational \(\alpha'\) is reduced and associated to \(D\) as well. (This is what one does during continued fraction expansion, and as I did with \(\sqrt{2}\) during my last post.)
Proof: Letting $$\alpha = \frac{\sqrt{D} + P}{Q}$$ and \(X = \lfloor \alpha \rfloor\), we have $$\frac{1}{\alpha'} = \frac{\sqrt{D} - (QX - P)}{Q}.$$
- Since \(\sqrt{D}\) is irrational, we must have \(\frac{1}{\alpha'} > 0\) and since \(\frac{1}{\alpha'}\) is the fractional part we know $$0 < \frac{1}{\alpha'} < 1 \Rightarrow \alpha' > 1.$$
- Transforming $$\alpha' = \frac{Q}{\sqrt{D} - (QX - P)} \cdot \frac{\sqrt{D} + (QX - P)}{\sqrt{D} + (QX - P)} = \frac{\sqrt{D} + (QX - P)}{\frac{1}{Q}\left(D - (QX - P)^2\right)},$$ we have \(P' = QX - P\) and \(Q' = \frac{1}{Q}\left(D - (QX - P)^2\right)\) and need to show \(Q' \in \mathbf{Z}\). But \(D - (QX - P)^2 \equiv D - P^2 \bmod{Q}\) and since \(\alpha\) is associated to \(D\), \(Q\) must divide this quantity, hence \(Q'\) is an integer.
- Since \(X = \lfloor\frac{\sqrt{D} + P}{Q}\rfloor\) is an integer and \(\alpha\) is irrational, we know \(X < \frac{\sqrt{D} + P}{Q}\) hence \(P' = QX - P < \sqrt{D}\) forcing \(\tilde{\alpha}' < 0\).
- Since \(\alpha > 1\) we know \(X \geq 1 \Leftrightarrow 0 \leq X - 1\). Thus \begin{align*}\tilde{\alpha} = \frac{P - \sqrt{D}}{Q} &< 0 \leq X - 1 \\ \Rightarrow Q &< \sqrt{D} + (QX - P) \\ \Rightarrow Q(\sqrt{D} - (QX - P)) &< D - (QX - P)^2 \\ \Rightarrow -\tilde{\alpha}' = \frac{\sqrt{D} - (QX - P)}{\frac{1}{Q}\left(D - (QX - P)^2\right)} &< 1 \end{align*} hence \(\tilde{\alpha}' > -1\) and \(\alpha'\) is reduced.
- Since \(Q' = \frac{1}{Q}\left(D - (P')^2\right)\), we know $$D - (P')^2 \equiv Q Q' \equiv 0 \bmod{Q'}$$ hence \(\alpha'\) is associated to \(D\).
Thus \(\alpha'\) is both reduced and associated to \(D\). \(\Box\)
Lemma 2: There are finitely many reduced quadratic irrationals associated to a fixed \(D\).
Proof: As above write an arbitrary reduced irrational as \(\alpha = \frac{\sqrt{D} + P}{Q}\). Since \(\alpha > 1\) and \(\tilde{\alpha} > -1\), we know \(\alpha + \tilde{\alpha} = \frac{2P}{Q} > 0\) hence with the assumption \(Q > 0\) we have \(P > 0\). Since \(\tilde{\alpha} < 0\) we also have \(P < \sqrt{D}\). Also, since \(\alpha > 1\) by assumption we have \(Q < P + \sqrt{D} < 2\sqrt{D}\) thus there are finitely many choices for both \(P\) and \(Q\), forcing finitely many reduced quadratic irrationals associated to a fixed \(D\) (this amount is strictly bounded above by \(2D\)). \(\Box\)
Claim: The continued fraction expansion of \(\sqrt{D}\) is periodic whenever \(D\) is not a perfect square.
Proof: We'll use Lemma 1 to establish a series of reduced quadratic irrationals associated to \(D\) and then use Lemma 2 to assert this series must repeat (hence be periodic) due to the finite number of such irrationals.
Write \(a_0 = \lfloor \sqrt{D} \rfloor\) and \(\sqrt{D} = a_0 + \frac{1}{\alpha_0}\). From here, we will prove
- \(\alpha_0\) is a reduced quadratic irrational associated to \(D\).
- By defining \(a_{i+1} = \lfloor \alpha_i \rfloor\) and \(\alpha_i = a_{i + 1} + \frac{1}{\alpha_{i + 1}}\), \(\alpha_{i + 1}\) is also a reduced quadratic irrational associated to \(D\) (assuming all \(\alpha\) up until \(i\) are as well).
Since \(\frac{1}{\alpha_0}\) is the fractional part of the irrational \(\sqrt{D}\), we have $$0 < \frac{1}{\alpha_0} < 1 \Rightarrow \alpha_0 > 1.$$ By simple algebra, we have $$\alpha_0 = \frac{a_0 + \sqrt{D}}{D - a_0^2}, \qquad \tilde{\alpha_0} = \frac{a_0 - \sqrt{D}}{D - a_0^2}.$$ Since \(a_0\) is the floor, we know \(a_0 - \sqrt{D} < 0 \Rightarrow \tilde{\alpha_0} < 0\). Since \(D \in \mathbf{Z} \Rightarrow \sqrt{D} > 1\) and \(\sqrt{D} > a_0\), we have $$1 < \sqrt{D} + a_0 \Rightarrow \sqrt{D} - a_0 < D - a_0^2 \Rightarrow a_0 - \sqrt{D} > -(D - a_0^2) \Rightarrow \tilde{\alpha_0} > -1.$$ Thus \(\alpha_0\) is a reduced quadratic irrational. Since \(P_0 = a_0\) and \(Q_0 = D - a_0^2 = D - P_0^2\), \(Q_0\) clearly divides \(D - P_0^2\) so \(\alpha_0\) is associated to \(D\) as well.
Following the recurrence defined, since each \(\alpha_i\) is a reduced quadratic irrational, each \(a_i \geq 1\). Also, by Lemma 1, each \(\alpha_{i + 1}\) is reduced and associated to \(D\) since \(\alpha_0\) is. By Lemma 2, we only have finitely many choices for these, hence there must be some smallest \(k\) for which \(\alpha_k = \alpha_0\). Since \(\alpha_{i + 1}\) is determined completely by \(\alpha_i\) we will then have \(\alpha_{k + j} = \alpha_j\) for all \(j > 0\), hence the \(\alpha_i\) are periodic. Similarly, as the \(a_i\) for \(i > 0\) are determined completely by \(\alpha_{i - 1}\), the \(a_i\) must be periodic as well, forcing the continued fraction expansion $$\sqrt{D} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \ddots}}$$ to be periodic.\(\Box\)
Update: I posted this on
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