It is well known that for any prime $pequiv 3$ (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.
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