This is the problem:
Does there exist positive integer $d$ ( which is not a perfect square ) such that the length of the least period in the simple continued fraction of $\sqrt{d}$ is $3$?
Consider the following theorem
Theorem : If the positive integer $d$ is not a perfect square, the simple continued fraction of $\sqrt{d}$ has the form $\sqrt{d} = [a_0;\overline{a_1,a_2,\cdots,a_{r-1},2a_o}]$ with $a_o = \lfloor d \rfloor$. Here $r$ denotes the length of the least period in the expansion of $\sqrt{d}$. Where $\lfloor x \rfloor$ denotes the greatest integer function/ floor function of $x$.
We want to solve for non-square $d$ where$\sqrt{d} = [a_0;\overline{a_1,a_2,2a_o}]$, and $a_o = \lfloor d \rfloor$. Since $d$ is a positive integer, $a_o = \lfloor d \rfloor \ge 1$, and $a_1 , a_2$ are positive integers by definition. Please note that the converse of the above theorem is not true, for example, consider $[1;\overline{1,1,2}] = \sqrt{10}/2$ and $[0;\overline{1,1,0}] = \sqrt{2}/2$. I calculated first few continued fractions for $\sqrt{d}$, $$\begin{array}{c|c|c} \sqrt{d} & \text{Continued fraction} & r\\ \hline
√2 & [1;\bar{2}] & 1 \\
√3 & [1;\overline{1,2}] & 2 \\
√5 & [2;\bar{4}] & 1\\
√6 & [2;\overline{2,4}] & 2\\
√7 & [2;\overline{1,1,1,4}] & 4\\
√8 & [2;\overline{1,4}] & 2\\
√10 & [3;\bar{6}] & 1\\
√11 & [3;\overline{3,6}] & 2\\
√12 & [3;\overline{2,6}] & 2\\
√13 & [3;\overline{1,1,1,1,6}] & 5\\
√14 & [3;\overline{1,2,1,6}] & 4\\
√15 & [3;\overline{1,6}] & 2\\
√17 & [4;\bar{8}] & 1\\
√18 & [4;\overline{4,8}] & 2\\
√19 & [4;\overline{2,1,3,1,2,8}] & 6 \\
√20 & [4;\overline{2,8}] & 2\\
√21 & [4;\overline{1,1,2,1,1,8}] & 6\\
√22 & [4;\overline{1,2,4,2,1,8}] & 6\\
√23 & [4;\overline{1,3,1,8}] & 4\\
√24 & [4;\overline{1,8}] & 2\\ \end{array}$$
As we can see for $1< d \le 24, r \ne 3$. Also, on a side note, observe that there does not exist two consecutive intergers $d$ and $d+1$ such that both $\sqrt{d}$ and $\sqrt{d+1}$ have $r=1$, moreover there are infinitely $\sqrt{d}$ such that the length of there least period is $1$ or $2$, $\sqrt{n^2+1} = [n;\overline{2n}]$, $\sqrt{n^2+2} = [n;\overline{n,2n}]$ and
$\sqrt{n^2-1} = [n-1;\overline{1,2(n-1)}]$ , where $n \in \mathbb{N}$ .Even for $r=4$, we have $\sqrt{n^2-2} = [n-1; \overline{1,n-2,1,2(n-1)}]$, $n>2$. Now i have a hunch that no such $d$ exists for which $\sqrt{d}$ have $r=3$. Any hints on how to prove this ? In general does there exist a number $m$ such that $r\ne m $ for any $\sqrt{d}$ ?
Best Answer
Yes there are infinitely many. And it is not difficult to find them.
We seek continued fractions of the form
$\sqrt{N}=[a,\overline{b,c,2a}]$
First off, add $a$ to get a "pure" periodic expression. We shall call the quadratic surd $x$:
$x=a+\sqrt{N}=[\overline{2a,b,c}]$
We may then render
$x=2a+\dfrac{1}{b+\dfrac{1}{c+\dfrac{1}{x}}}$
and upon clearing fractions
$(bc+1)x^2+(b-c-2a(bc+1))x-(2ab+1)=0$
Now comes the sneaky part. If the above quadratic equation over the integers is to have a root $a+\sqrt{N}$, its other root must be $a-\sqrt{N}$ forcing the linear coefficient to be exactly $-2a$ times the quadratic one! Thereby $b=c$ above and the quadratic equation simplifies to:
$(b^2+1)x^2-2a(b^2+1)x-(2ab+1)=0$
This gives an integer radicand whenever $2ab+1$ is a multiple of $b^2+1$, in which cases the common factor of $b^2+1$ may be cancelled from the quadratic equation leaving the equation monic.
Suppose, for example, we drop in $b=2$. Then $2ab+1$ is to be a multiple of $5$ and $a$ can be any whole number one greater than a multiple of $5$. Putting $a=1$ results in the "trivial" solution $\sqrt{2}=[1,\overline{2}]$, as the period is reduced from three to one due to $b=c=2a$. But this equality is avoided for larger eligible values of $a$ and we get a series of period $3$ solutions. In all cases $N$ is one fourth the discriminant of the monic polynomial obtained after cancelling out the $b^2+1$ factor:
$a=6\to \sqrt{41}=[6,\overline{2,2,12}]$
$a=11\to \sqrt{130}=[11,\overline{2,2,22}]$
$a=5k+1\to \sqrt{25k^2+14k+2}=[5k+1,\overline{2,2,10k+2}]$
There are more families of solutions like this with other values of $b$. Just put in an even positive value for $b$ (why even?) and turn the crank. You must put $a>b/2$ to avoid the collapse we saw above with $\sqrt{2}$.
So much for a repeat petiod of $3$, what about larger periods?
Claim: For any positive whole numbers $r$ there are at least an infinitude of $\sqrt{N}$ continued fractions having repeat period $r$ where $N$ is a whole number, having the following form:
$\sqrt{N}=[kP_r+1;\overline{2,2,...,2,2(kP_r+1)}]$
$P_r$ is a Pell number defined by $P_0=0,P_1=P_{-1}=1,P_r=2P_{r-1}+P_{r-2}\text{ for } r\ge 2$, and $k$ is a whole number $\ge 0$ for $r=1$, $\ge 1$ otherwise. The number of $2$ digits before the final entries is $r-1$.
The proof bears some similarities to calculating the general solution for $r=3$ above. First add $kP_r+1$ to the expression to make a purely periodic fraction:
$x=kP_r+1+\sqrt{N}=[\overline{2(kP_r+1),2,2,...,2}]$
Then
$x=2(kP_r+1)+\dfrac{1}{[2,2,...,2,x]}$
By mathematical induction on $r$ and using the recursive relation defined for Pell numbers in the claim it is true that
$[2,2,...,2,x]=\dfrac{P_rx+P_{r-1}}{P_{r-1}x+P_{r-2}}$
with $r-1$ digits of $2$ in the block. When this is substituted into the previous equation the following is obtained:
$x=2(kP_r+1)+\dfrac{P_{r-1}x+P_{r-2}}{P_rx+P_{r-1}}$
$x=\dfrac{(2(kP_r+1)P_r+P_{r-1})x+2(kP_r+1)P_{r-1}+P_{r-2}}{P_rx+P_{r-1}}$
$(P_r)x^2-2(kP_r+1)P_rx-(2(kP_r+1)P_{r-1}+P_{r-2})=0$
Upon completing the square and back-substituting $\sqrt{N}=x-(P_rk+1)$ we obtain:
$N=\dfrac{(kP_r+1)^2P_r+2(kP_r+1)P_{r-1}+P_{r-2}}{P_r}$
Using the Pell number recursion to eliminate $P_{r-2}$:
$N=\dfrac{(kP_r+1)^2P_r+2(kP_r)P_{r-1}+P_r}{P_r}=(kP_r+1)^2+2kP_{r-1}+1$
thereby identifying $N$ as a whole number. For a full fundamental period $\ge 2$ the terminal element must not match the other elements, so in that case $k\ge 1$. Else (meaning a period of just $1$), $k$ may be any whole number, $k\ge 0$.