Eventually periodic continued fraction implies root of polynomial of degree 2

Question

How to prove that every irrational number with eventually periodic continued fraction expansion is a root of a polynomial of degree 2?

Answer 1

This is a classical result.

If you want to prove it yourself start with the case of a constant continued fraction and note that: \begin{equation} r=a+\frac{1}{b+\frac{1}{b+\frac{1}{b+\ldots}}}=a+\frac{1}{(b-a)+r} \end{equation} Now multiplying both sides by $(b-a)+r$ yields a quadratic expression (with integer coefficients) for $r$.

The general case is similar, and I hope you can do it on your own, using the special case I provided above.