# Simple continued fraction of $\sqrt{d}$ with period of shortest length $3$

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 $$\sqrt{d}$$ ?

• According to oeis the least such is $41$. The tabulated sequence is here. $\sqrt {41}=[6; \overline {2,2,12}]$. – lulu Nov 8 '19 at 21:23
• Thank you :) . Are there infinitely many such numbers? @lulu – Sabhrant Nov 8 '19 at 21:31
• Move this to an answer, maybe? – Oscar Lanzi Nov 8 '19 at 21:33
• There appear to be infinitely many, see this – lulu Nov 8 '19 at 21:38
• There is also $\sqrt{n(n+1)}=[n;\overline{2,2n}]$. The case $n=1$ is "hidden" because $\overline{2,2}$ reduces to $\overline{2}$. – Oscar Lanzi Nov 8 '19 at 21:49

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 calcukating 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 $$r-1$$ $$2$$ digits in the block. When this is substituted into the previous equation this 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-substitutinging $$\sqrt{N}=x-(P_rk+1)$$ 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 eliminste $$P_{r-2}$$:

$$N=\dfrac{(kP_r+1)^2P_r+2(kP_r)P_{r-1}+P_r}{P_r}=(kP_r+1)^2+2kP_{r-1}+1$$

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, $$k\ge 0$$.

Just working numerically, $$41$$ is the least example, with $$\sqrt {41}=[6; \overline {2,2,12}]$$

here is a tabulated list of the periods of $$\sqrt d$$.

OEIS provides a list of $$d$$ for which the period is $$3$$, and that link provides a way of generating infinitely many examples.