Theorem: Continued Fraction Expansion of Irrational Square Root

Let $n \in \Z$ such that $n$ is not a square.
Then the continued fraction expansion of $\sqrt n$ is of the form:
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
or
:$\sqbrk {a_1 \sequence {b_1, b_2, \ldots, b_{m - 1}, b_m, b_m, b_{m - 1}, \ldots, b_2, b_1, 2 a_1} }$
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 denominator.

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:
:$a_0 = \floor {\sqrt n}$
:$\sqrt n = a_0 + \dfrac 1 {\alpha_0}$
For all $i \in \N: i > 0$, let:
:$a_{i + 1} = \floor {\alpha_i}$
:$\alpha_i = a_{i + 1} + \dfrac 1 {\alpha_{i + 1} }$
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:
{{begin-eqn}}
{{eqn | l = 1
      | o = <
      | r = \sqrt n + a_0
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \sqrt n - a_0
      | o = <
      | r = n - a_0^2
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a_0 - \sqrt n
      | o = >
      | r = -\paren {n - a_0^2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \tilde {\alpha_0}
      | o = >
      | r = -1
      | c = 
}}
{{end-eqn}}
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.
{{wtd|This gives us the fact that it repeats. We still need explain the pattern.}}