Continued fraction of the square $\sqrt{n}$ of the form $[a,\overline{k,b}]$

Question

Continued fraction of the square $\sqrt{n}$ for a fixed $k >2 $ can we say there are infinitely man $n$ such that $\sqrt{n} =$ $[a,\overline{k,b}]$?

If so would the proof go by the route of contradiction? Could one claim one is the largest and contradict it by finding a larger one?

Answer 1

Looking at the expansions of the small square roots, it appears that if $k$ divides $2a$ the expansion of $a^2+\frac {2a}k$ is of the form $[a,\overline {k,2a}]$. Go through the continued fraction expansion and see if this works.

Added: given $k$ we can find infinitely many examples where the expansion is $[a,\overline{k,2a}]$. Choose any $a$ so that $k$ divides $2a$. Then $\sqrt{a^2+\frac {2a}k}=[a,\overline{k,2a}].$

The proof is just to compute it. $$\sqrt{a^2+\frac {2a}k}=a+\sqrt{a^2+\frac {2a}k}-a\\ =a+\frac 1{\frac 1{\sqrt{a^2+\frac {2a}k}-a}}\\ =a+\frac 1{\frac {\sqrt{a^2+\frac {2a}k}+a}{(a^2+\frac {2a}k)-a^2}}\\ =a+\frac 1{\frac {\sqrt{a^2+\frac {2a}k}+a}{\frac {2a}k}}\\ =a+\frac 1{\frac k{2a}\left(\sqrt{a^2+\frac {2a}k}+a\right)}\\ =a+\frac 1{k+\frac k{2a}\left(\sqrt{a^2+\frac {2a}k}-a\right)}\\ =a+\frac 1{k+\frac 1{\frac {2a}{k\left(\sqrt{a^2+\frac {2a}k}-a\right)}}}\\ =a+\frac 1{k+\frac 1{\frac {2a\left(\sqrt{a^2+\frac {2a}k}+a\right)}{2a}}}\\ =a+\frac 1{k+\frac 1{2a-\left(\sqrt{a^2+\frac {2a}k}-a\right)}}$$ If you choose $a=\frac k2$ you will get an expansion $[a,\overline{2a,2a}]$ which is not the minimal one. If that bothers you, prohibit that case.

Answer 2

If you’re merely asking whether there are infinitely many $n$ for which the expansion of $\sqrt n$ is of that form, that’s easy, since the expansion of $\sqrt{n^2+2n}$ is $[n,\overline{1,2n}]$. Notice that $n^2-2n$ is just one less than a square.