Can Every Irrational, but Non-Transcendental be Expressed as a Repeating Continued Fraction? So, as far as I understand any rational number can be expressed as a continued fraction of finite length
for example:
$$\frac {7}{17} = \cfrac {1}{2 + \cfrac{1}{2 + \cfrac{1}{3}}}$$
which for convenience I will express as $[0,2,2,3]$
Now I also understand that some irrationals can be written as infinite continued fractions
$$\sqrt{2} =1 + \cfrac{1}{2 +\cfrac{1}{2 + \cfrac{1}{\cdots}}}$$
expressible as $[1,\overline{2}]$
Now my question, can all irrational numbers that can be expressed as the root of a polynomial (algebraic irrational) be expressed as a repeating continued fraction?  How would we go about proving this if it is true?
 A: If a simple continued fraction repeats but does not terminate then it converges to a second-degree algebraic number. ("Simple" means all of the numerators are $1$.)
First, look at the case where the repeating part begins immediately:
$$
x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 {a_1 + _{\large\ddots}}}}}}
$$
Then you have
$$
x = a_1 + \cfrac 1 {a_2 + \cfrac 1 {a_3 + \cfrac 1 {\ddots \cfrac{}{a_n + \cfrac 1 x}}}}
$$
This reduces to a quadratic equation in $x$ with integer coefficients.
To see that reduction to a quadratic equation, start with
$$
a_{n-1} + \cfrac 1 {a_n + \cfrac 1 x} = a_{n-1} + \frac{x}{a_n x + 1} = \frac{ (a_{n-1} a_n + 1) x + a_n }{a_{n-1} x + 1}
$$
So you have
$$
a_{n-2} + \cfrac 1 {a_{n-1} + \cfrac{x}{a_n x + 1}} = a_{n-2} + \frac{a_n x + 1}{a_{n-1}(a_n x + 1) + x } = \text{etc.}
$$
Keep going until you have
$$
x = \frac{\text{a first-degree polynomial in } x}{\text{another first-degree polyonomial in } x},
$$
then multiply both sides by the denominator to get a quadratic equation.
If the repeating part doesn't start at the beginning, you can work on the repeating part, getting it to be the solution of a quadratic equation, then repeatedly rationalize denominators until you're done.
Here's a concrete example:
$$
x = 3 + \cfrac 1 {2 + \cfrac 1 {5 + \cfrac 1 {2 + \cfrac 1 {5 + \cdots}}}}
$$
with $2,5$ repeating.
One can write
$$
y = 3 + \cfrac 1 x = 3 + \cfrac 1 {2 + \cfrac 1 {5 + \cfrac 1 x}}
$$
So
$$
x = 2 + \cfrac 1 {5 + \cfrac 1 x} = 2 + \frac x {5x+1} = \frac{11x+2}{5x+1}
$$
$$
x = \frac{11x+2}{5x+1}
$$
$$
5x^2 + x = 11x+2
$$
$$
5x^2 - 10 x - 2 = 0.
$$
$$
x = \frac{10 \pm \sqrt{140}}{10} = \frac{5 \pm \sqrt{35}} 5.
$$
Since the value we seek is positive, we have
$$
x= \frac{5 + \sqrt{35}} 5.
$$
