Eventually periodic regular continued fractions are quadratic irrationals

By increasing $N$ if necessary, we may still assume. [F1, F2, F3, F5, algebra] $a_{n+h}=a_n(n\ge N)$ with $N\ge 1$. Let $x=[a_N;a_{N+1},\dots ,a_{N+h-1},a_N,a_{N+1},\dots ]$ be the purely periodic tail. For $r\ge 1$, let $x_r$ be the finite continued fraction formed from $r$ copies of this period. By [F5], $x_r\to x$, and [F2] gives $x_{r+1}=\frac{x_r{p}_{h-1}+p_{h-2}}{x_r{q}_{h-1}+q_{h-2}},$ where the convergents are taken for the digit block $a_N,\dots ,a_{N+h-1}$. The denominators are positive by [F3], so a direct difference calculation shows that the displayed fractional-linear expression preserves the limit $x_r\to x$. Hence $x=\frac{x{p}_{h-1}+p_{h-2}}{x{q}_{h-1}+q_{h-2}}\mathrm{.}$ Clearing denominators yields the quadratic equation $q_{h-1}{x}^2+(q_{h-2}-p_{h-1})x-p_{h-2}=0.$

The discriminant of the quadratic in step 1.1 is. [step 1.1, F3, F4, algebra] $\mathrm{\Delta }=(q_{h-2}-p_{h-1}{)}^2+4q_{h-1}{p}_{h-2}=(p_{h-1}+q_{h-2}{)}^2-4(p_{h-1}{q}_{h-2}-p_{h-2}{q}_{h-1})\mathrm{.}$ Putting $m:=p_{h-1}+q_{h-2}>0$, fact [F4] gives $\mathrm{\Delta }=m^2-4(-1{)}^{h-2}\mathrm{.}$ If $h$ is odd, then $\mathrm{\Delta }=m^2+4$, which cannot be a square because $(n-m)(n+m)=4$ has no solution in integers with $m>0$. If $h$ is even, then $h\ge 2$, fact [F3] gives $p_{h-1}\ge p_1\ge 2$ and $q_{h-2}\ge q_0=1$, so $m\ge 3$; then $\mathrm{\Delta }=m^2-4$, which cannot be a square because $(m-n)(m+n)=4$ has no solution with $m\ge 3$. Therefore $\mathrm{\Delta }$ is not a square, so the root $x$ from step 1.1 is irrational.

Append the finite tails $x_r$ of step 1.1 after the prefix. [F1, F2, F5, step 1.1, algebra] $a_0,\dots ,a_{N-1}$. These are a subsequence of the convergents of $\alpha$, so [F5] makes their values tend to $\alpha$. By [F2] their values are $\frac{x_r{p}_{N-1}+p_{N-2}}{x_r{q}_{N-1}+q_{N-2}}\mathrm{.}$ The same positive-denominator difference calculation used in step 1.1 lets $r\to \mathrm{\infty }$ and gives $\alpha =\frac{x{p}_{N-1}+p_{N-2}}{x{q}_{N-1}+q_{N-2}}\mathrm{.}$ Clearing denominators and substituting the quadratic equation from step 1.1 shows that $\alpha$ also satisfies a quadratic equation over $\mathbb{Q}$.

Suppose that $\alpha$ were rational. Step 2.2 gives. [step 2.1, step 2.2, F4, algebra] $x(\alpha q_{N-1}-p_{N-1})=p_{N-2}-\alpha q_{N-2}\mathrm{.}$ If $\alpha q_{N-1}-p_{N-1}=0$, then also $p_{N-2}-\alpha q_{N-2}=0$, and eliminating $\alpha$ yields $p_{N-1}{q}_{N-2}-p_{N-2}{q}_{N-1}=0,$ contrary to [F4]. Therefore $x=\frac{p_{N-2}-\alpha q_{N-2}}{\alpha q_{N-1}-p_{N-1}}$ is rational, contradicting step 2.1. Hence $\alpha$ is irrational.

Steps 2.2 and 3.1 show that $\alpha$ is an irrational real root of a quadratic equation over $\mathbb{Q}$, which is exactly the definition of a quadratic irrational (Quadratic irrationals).