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Does the period length $l(pq)$ of the continued fraction of $\sqrt{pq}$, for $p$ and $q$ primes, follow some type of divisibility property, say $$ l(pq) = c\frac{l(p)}{l(q)} \quad\text{or}\quad c\frac{l(q)}{l(p)}, $$ where $l(p)$ is the period of $\sqrt{p}$ and $c > 0$ is an absolute constant.

In the previous question, Franz Lemmermeyer mentioned that it depends on the squarefree kernel of the integer.

Note: The theorem on the upper bound of the length of period of continued fraction implies that $l(\ \cdot\ )$ is not multiplicative.

Thank you, Jerald Jetson.

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  • $\begingroup$ It should be easy enough to check this for a few primes. Did you? $\endgroup$
    – Gerry Myerson
    Commented Jul 22, 2010 at 0:06
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    $\begingroup$ Read the comments after Franz's answer, he revised that part of it. I will run some experiments. But bear in mind that sometimes your $l(pq)$ will be roughly the size of $l(p) \cdot l(q)$ which is consistent with some very simple bounds of Lagrange. But I bet that there are infinitely many instances of $$ p \cdot q = n^2 + 1 $$ which has about the shortest continued fraction period. That is, consider $$ \sqrt{10}, \; \sqrt{26}, \; \sqrt{65}, \; \sqrt{82}, \; \sqrt{122}, \; \sqrt{145}, \; \ldots \sqrt{901}, \; \ldots $$ where $901 = 17 \cdot 53$ was the first not divisible by 2 or 5. $\endgroup$
    – Will Jagy
    Commented Jul 22, 2010 at 0:21
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    $\begingroup$ Let me add the remark that the left hand side of the equation in your question is invariant under switching p and q, but the right hand side is not. Thus you can't really expect it to hold. $\endgroup$
    – Franz Lemmermeyer
    Commented Jul 22, 2010 at 8:06
  • $\begingroup$ @Franz, "$a$ or $b$" is equivalent to "$b$ or $a$", so the way I interpret it, the right side of the equation is certainly invariant under switching $p$ and $q$. $\endgroup$
    – Gerry Myerson
    Commented Jul 23, 2010 at 6:43
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    $\begingroup$ Ideas in this paper might be of use: math.princeton.edu/mathlab/jr02fall/Periodicity/periodmain.htm $\endgroup$
    – Joseph Malkevitch
    Commented Aug 6, 2010 at 16:49

1 Answer 1

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Let $p=2$, $q=5$, so $pq=10$. Then $\ell(p)=\ell(q)=\ell(pq)=1$, so if $c$ exists it must be 1. Now let $p=2$, $q=17$, so $pq=34$. Then $\ell(p)=\ell(q)=1\ne\ell(pq)$, so if $c$ exists it must not be 1. Contradiction.

EDIT: Here's a little table which, I think, shows how unlikely it is that there's any simple equation relating the period lengths of the square roots of primes $p$ and $q$ and their product. $$ \matrix{p&q&pq&\ell(p)&\ell(q)&\ell(pq)\cr2&5&10&1&1&1\cr2&17&34&1&1&4\cr2&13&26&1&5&1\cr2&7&14&1&4&4\cr13&89&1157&5&5&1\cr} $$

So, if $\ell(p)$ and $\ell(q)$ are both 1 then $\ell(pq)$ may or may not be 1; if $\ell(p)$ and $\ell(pq)$ are both 1 then $\ell(q)$ may or may not be 1; if $\ell(p)$ is 1 and $\ell(pq)$ is not 1 then $\ell(q)$ may or may not be 1; if $\ell(pq)$ is 1 and $\ell(q)$ is not 1 then $\ell(p)$ may or may not be 1.

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