Question

If $\sqrt{N}$ be converted into a continued fraction, and if the penultimate convergents in the first, second, ....$k^{th}$ recurring periods be denoted by $n_1,n_2,.....n_k$ respectively.Show that $\frac{n_k+\sqrt{N}}{n_k-\sqrt{N}}={\left(\frac{n_1+\sqrt{N}}{n_1-\sqrt{N}}\right)}^k$.

Answer 1

$\sqrt{N}=\frac{(a_1+\sqrt{N})p_{2n}+p_{2n-1}}{(a_1+\sqrt{N})q_{2n}+q_{2n-1}}$

Therefore, $a_1{p}_{2n}+p_{2n-1}=N{q}_{2n},$

$a_1{q}_{2n}+q_{2n-1}=p_{2n}$

Hence, $\frac{p_{2n}}{q_{2n}}=\frac{(a_1+\frac{p_n}{q_n})p_{2n}+p_{2n-1}}{(a_1+\frac{p_n}{q_n})q_{2n}+q_{2n-1}}$

$=\frac{N{q}_{2n}+\frac{p_n}{q_{2n}}{p}_{2n}}{p_{2n}+\frac{p_n}{q_n}{q}_{2n}}$

Also we know that,

$n_2=\frac{1}{2}(n_1+\frac{N}{n_1})=\frac{n_1^2+N}{2n_1}$

$\therefore \frac{n_2}{\sqrt{N}}=\frac{n_1^2+N}{2n_1\sqrt{N}}$

By componendo and dividendo;-

$\frac{n_2+\sqrt{N}}{n_2-\sqrt{N}}=\frac{{(n_1+\sqrt{N})}^2}{{(n_1-\sqrt{N})}^2}$

By replacing ${}^{\prime }{2}^{\prime }$ with 'k' i.e. generalising;-

$\frac{n_k+\sqrt{N}}{n_k-\sqrt{N}}={\left(\frac{n_1+\sqrt{N}}{n_1-\sqrt{N}}\right)}^k$