Continued Fraction Expansion of Irrational Square Root

Theorem
Let $n \in \Z$ such that $n$ is not a square.

Then the continued fraction expansion of $\sqrt n$ is of the form:
 * $\left[{a_1 \left \langle{b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1}\right \rangle}\right]$

or
 * $\left[{a_1 \left \langle{b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1}\right \rangle}\right]$

where $m \in \Z: m \ge 0$.

That is, it has the form as follows:
 * It is periodic
 * It starts with an integer $a_1$
 * Its cycle starts with a palindromic section, either:
 * $b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1$
 * or:
 * $b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1$
 * which may be of length zero
 * Its cycle ends with twice the first partial quotient.

Proof
We use:
 * Expansion of Associated Reduced Quadratic Irrational to establish a series of reduced quadratic irrationals associated to $n$

and then use
 * Finitely Many Reduced Associated Quadratic Irrationals to assert this series must repeat (hence be periodic) due to the finite number of such irrationals.

Let:
 * $a_0 = \left \lfloor {\sqrt n} \right \rfloor$
 * $\sqrt n = a_0 + \dfrac 1 {\alpha_0}$

For all $i \in \N: i > 0$, let:
 * $a_{i + 1} = \left \lfloor {\alpha_i} \right \rfloor$
 * $\alpha_i = a_{i + 1} + \dfrac 1 {\alpha_{i + 1}}$

From here, we will prove:
 * $\alpha_0$ is a reduced quadratic irrational associated to $n$


 * If all $\alpha_i: 0 \le i \le k$ are all reduced quadratic irrationals associated to $n$, then so is $\alpha_{k + 1}$.

Since $\dfrac 1 {\alpha_0}$ is the fractional part of the irrational $\sqrt n$, we have:
 * $0 < \dfrac 1 {\alpha_0} < 1 \implies \alpha_0 > 1$

By simple algebra, we have:
 * $\alpha_0 = \dfrac {a_0 + \sqrt n} {n - a_0^2}$
 * $\tilde{\alpha_0} = \dfrac {a_0 - \sqrt n} {n - a_0^2}$

where $\tilde{\alpha_0}$ is the conjugate of $\alpha_0$.

Since $a_0$ is the floor of $\sqrt n$, we know that:
 * $a_0 - \sqrt n < 0 \implies \tilde{\alpha_0} < 0$

Since $n \in \Z \implies \sqrt n > 1$ and $\sqrt n > a_0$, we have:

Thus $\alpha_0$ is a reduced quadratic irrational.

Since $P = a_0$ and $Q = n - a_0^2 = n - P^2$, $Q$ clearly divides $n - P^2$.

Thus $\alpha_0$ is associated to $n$ as well.

We have that each $\alpha_i$ is a reduced quadratic irrational.

So, following the recurrence defined, each $a_i \ge 1$.

Also, by Expansion of Associated Reduced Quadratic Irrational, each $\alpha_{i + 1}$ is reduced and associated to $n$ since $\alpha_0$ is.

By Finitely Many Reduced Associated Quadratic Irrationals, 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.

This forces the continued fraction expansion:
 * $\sqrt n = a_0 + \cfrac 1 {a_1 + \cfrac 1 {a_2 + \ddots} }$

to be periodic.