Why are the continued fractions of non-square-root numbers ($\sqrt[a]{x}$ where $a>2$) not periodic? Ok, so it is quite amazing how the continued fractions for $\sqrt[2]{x}$ are always periodic for all whole numbers of $x$ (and where $x$ is not a perfect square): Here is a link I suggest at looking at: http://mathworld.wolfram.com/PeriodicContinuedFraction.html ...

The following is in the simple continued fraction form:
$$\sqrt x = a_1+ \cfrac{1}{a_2 + \cfrac{1}{a_3 + \cfrac{1}{a_4 + \cfrac{1}{a_5 + \cfrac{1}{a_6 + \cfrac{1}{a_7 + \cdots}}}}}}$$
For example:
The continued fraction of $\sqrt 7$ is as follows: 
$[2;1,1,1,4,1,1,1,4,1,1,1,4,\ldots]$ the period in this case is $4$... and the continued fraction can be written as:
$$[2;\overline{1,1,1,4}]$$

Some other examples:
$\sqrt2 = [1;\overline{2}]$ Period is 1
$\sqrt3 = [1;\overline{1,2}]$ Period is 2
$\sqrt{13} = [3;\overline{1,1,1,1,6}]$ Period is 5
$\sqrt{97} = [9;\overline{1,5,1,1,1,1,1,1,5,1,18}]$ Period is 11

There are many other sources which show this... but why does this not work for others, such as cubic roots? I have written a java program to compute the continued fraction for the nth root of the numbers between $1$ and $100$ (excluding perfect squares,etc...) for the first 100 terms. 
Here are the results:
$\sqrt[2]{x}$: http://pastebin.com/ZcasfRyP
$\sqrt[3]{x}$: http://pastebin.com/XG9UF8hR
$\sqrt[4]{x}$: http://pastebin.com/Edp307SE
$\sqrt[5]{x}$: http://pastebin.com/9SwwPqUa

As you can see no period...
So why is it periodic for square roots, but not for others? An extension of the question: https://math.stackexchange.com/questions/1898902/periodic-continued-fractions-of-non-square-root-numbers-sqrtax-where-a 
Kind Regards
Joshua Lochner
 A: The easiest way to show this is the contrapositive: if a continued fraction is periodic then its value is a quadratic.
To see this, note that if we have $y = [a_1; a_2, a_3, \ldots, a_n, x]$ for reals $x,y$ and integers $a_i$, then we can write $y=\dfrac{mx+n}{px+q}$ for some integers $m, n, p, q$.  (This is easy to prove by induction - note that if $z=[a_2; a_3, \ldots, a_n, x]$ $= \dfrac {m'x+n'}{p'x+q'}$, then $y=a_1+\dfrac1z$ $=a_1+\dfrac{p'x+q'}{m'x+n'}$ $=\dfrac{a_1m'x+a_1n'+p'x+q'}{m'x+n'}$ $=\dfrac{(a_1m'+p')x+(a_1n'+q')}{m'x+n'}$ is clearly of the required form.)
Now, if the continued fraction for $x$ is periodic then we have $x=[a_1; a_2, a_3, \ldots, a_n, x]$, so by the above there are integers with $x=\dfrac{mx+n}{px+q}$.  But now multiply both sides by $px+q$ to get $px^2+qx=mx+n$, or $px^2+(q-m)x-n=0$, so that $x$ is a solution to this quadratic equation.
To answer the question about patterns in the continued fractions of other numbers: to the best of my knowledge, nothing is known about the continued fractions of e.g. cube roots — not even whether their coefficients are bounded! — though it's known that they can't grow too quickly: this is a corollary of Roth's Theorem, which bounds the so-called irrationality measure of algebraic numbers (how well they can be approximated by rationals).  On the other hand, there is a special class of (transcendental) numbers for which more information on the continued fraction representation is known: certain Liouville numbers have continued fractions with complicated recursive (almost fractal) structures; see http://mathworld.wolfram.com/LiouvillesConstant.html and some of the references there for a little more information on this.
A: The idea is quite simple. Suppose that $$x=[a_0;\overline{a_1 , \dots , a_k}]$$ is a real number whose continued fraction is periodic. Then it satisfies the equation
$$x- a_0= \frac{1}{a_1+\frac{1}{\frac{\dots}{a_{k-1}\frac{1}{a_k+(x-a_0)}}}}$$
which can be transformed by a quadratic equation. with integer coefficients. For example, the number $x=[2; \overline{1,4}]$ satisfies
$$x-2= \frac{1}{1+\frac{1}{4+\frac{1}{x-2}}}$$ which is equivalent to the quadratic equation
$$(x-2)(5x-9)=4x-7$$
So, all periodic numbers must be quadratic.
A: It's because periodicity of a simple continued fraction amounts to a quadratic equation.  I will illustrate this by means of an example:
$$
7 + \cfrac 1 {2 + \cfrac 1 {3+\cfrac 1 {9 + \cdots\vphantom{\dfrac 1 1}}}}
$$
and assume $2,\,3,\,9$ repeats.  We have
$$
7,\  \overbrace{2,3,9,}\  \overbrace{2,3,9,}\  \overbrace{2,3,9,}\  \overbrace{2,3,9,}\  \ldots
$$
This is
$$
-2 + \left(9 + \cfrac 1 {2 + \cfrac 1 {3+\cfrac 1 {9 + \cdots\vphantom{\dfrac 1 1}}}} \right) \tag 1
$$
so that $9,\ 2,\ 3$ repeats right from the beginning.
Let $x = {}$the continued fraction in $(1)$, with $9,\ 2,\ 3$ repeating.
Then we get
$$
x = 9 + \cfrac 1 {2 + \cfrac 1 {3 + \cfrac 1 x}}
$$
Then, since $\dfrac 1 {3+\cfrac 1 x} = \dfrac x {3x+1}$, we have
$$
x = 9 + \cfrac 1 {2 + \cfrac x {3x+1}}.
$$
Now multiply the numerator and denominator of the fraction after $9+\cdots$ by $3x+1$, getting
$$
x = 9 + \dfrac{3x+1}{7x+2}
$$
so
$$
x = \frac{66x+ 19}{7x+2}.
$$
Multiply both sides by $7x+2$:
$$
x(7x+2) = 66x+19.
$$
And there you have a quadratic equation.
