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.
Best Answer
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.