  It is well known that for any prime $p\equiv 3$ (mod $4$), the class numbers
of the quadratic fields $\mathbb{Q}(\sqrt{p})$ and $\mathbb{Q}(\sqrt{-p})$,
$h(p)$ and $h(-p)$ respectively, are odd. It is natural to ask whether there is
a formula for $h(p)/h(-p)$ modulo powers of $2$. We show the formula $h(p)
\equiv h(-p) m(p)$ (mod $16$), where $m(p)$ is an integer defined using the
"negative" continued fraction expansion of $\sqrt{p}$. Our result solves a
conjecture of Richard Guy.
