here is some problems I have in the following proof from Diophantine Analysis by Jorn Steuding with a little change:
A continued fraction denoted $\zeta$ is periodic if and only if it's a quadratic irrational,e.g. a continued fraction which is eventually periodic represents a numbers that is a root of a quadratic equation.
$\bbox[5px,border:2px solid #C0A000]\Longrightarrow $ this is known as Euler's continued fraction theorem and is easy to prove.
$\bbox[5px,border:2px solid #C0A000]\Longleftarrow $ this is known as Lagrange's continued fraction theorem, and the proof is as follows :
since $\zeta$ is quadratic irrational implies there exist a irreducible quadratic equation $P(x)=ax^2+bx+c$ with $a,b,c \in \mathbb{Z}$ and $a\ne 0$ such that :$$P(\zeta)=a\zeta^2+b\zeta+c=0$$ on the other hand since $\zeta$ is irrational therefore its continued fraction is infinite,e.g. $\zeta=[a_0;a_1,...]$
and we have: $$\zeta=\frac{\zeta_np_{n-1}+p_{n-2}}{\zeta_nq_{n-1}+q_{n-2}} \qquad(\small n\geq 2)$$
Substituting the above relation in the polynomial we get :
$$A_n\zeta_n^2 +B_n\zeta_n +C_n=0$$
where $$\large A_n=ap^2_{n-1}+bp_{n-1}q_{n-1}+cq^2_{n-1}$$ $$\large B_n=2ap_{n-1}p_{n-2}+b(p_{n-1}q_{n-2}+p_{n-2}q_{n-1})+2cq_{n-1}q_{n-2}$$ $$\large C_n=ap^2_{n-2}+bp_{n-2}q_{n-2}+cq^2_{n-2}$$
clearly $A_n,B_n,C_n \in \mathbb{Z}$ and $A_n \ne 0$ if $A_n = 0$ would imply that $\large \frac{p_{n−2}}{q_{n−2}}$is a root of $P(x)$ contradicting the irreducibility of $P$ Thus, $A_nx^2+B_nx+C_n$ is an irreducible quadratic polynomial with root $\zeta_n$.
here is the first problem I have, where did actually $\large \frac{p_{n−2}}{q_{n−2}}$ come from?
also $B^2_n− 4A_nC_n = (b^2 − 4ac) (p_{n−1}q_{n−2} − p_{n−2}q_{n−1})=\pm(b^2 − 4ac)...$
here is the second problem, how we can calculate $B^2_n− 4A_nC_n$?it looks hard to calculate and indeed takes much time, I guess there should be another way, notice that I know the fact $(p_{n−1}q_{n−2} − p_{n−2}q_{n−1})=(-1)^{n-1}$