$p >2$ is a prime, any facts about congruence relation between the class number of $Q(\sqrt p)$ and $Q(\sqrt{-p})$?

Question

Let $p$ be an odd prime. This is a question about the class number of $Q(\sqrt p)$ and $Q(\sqrt{-p})$,which we denote by $h(p)$ and $h(-p)$ respectively. While doing my research on number theory I came cross an obstacle to smooth which I have to know some relation between $h(p)$ and $h(-p)$. I only know some basic facts about algebraic number theory.

With some efforts, I only find the following paper, which provides some useful results when $p \equiv 3 \pmod 4$. Click here. That paper discussed the relation between $h(p)$ and $h(-p)$ in congruence sense (modular $p$), which is enough for me. However, I didn't find any similar result when $p \equiv 1 \pmod 4$.

I am desperate to know any facts about relation of $h(p)$ and $h(-p)$ when $p \equiv 1 \pmod 4$. Hope some experts could help me, thank you very much!!!

Answer 1

This will be more of a long comment than an actual answer to your question.

There are some known results about $h(p)$ and $h(-p)$ modulo powers of two.

From "Class Numbers of Real Quadratic Number Fields" (1974) by Ezra Brown it is known that $h(p)$ is odd for all $p$.

Ezra Brown also has a nice survey "The Power of 2 Dividing the Class-Number of a Binary Quadratic Discriminant" (1972) where he mentions several results on $h(p)$ modulo powers of 2, including the following (when $p\equiv 1\pmod{4}$):

$$4|h(-p)\iff p\equiv 1 \pmod{8}$$

$$ 8|h(-p) \iff p=x^2+y^2 \text{ with } x+y\equiv \pm 1 \pmod {8}\hspace{1cm} \text{(due to Reichardt, 1970)}$$

$$h(-p)\equiv b+(p-1)/2 \pmod{8} \text{ where } p=a^2+b^2 \hspace{1cm} \text{(due to Brown, 1972)}$$

At the time of Brown's survey it didn't seem that there were any result for $h(-p)$ modulo 16. (But recently it was shown by Djordjo Milovic that there are infinitely many $p$ such that $16|h(-p)$.)

From the above result about $h(-p)$ modulo 4 and the fact that $h(p)$ is odd, it follows that (for $p\equiv 1\pmod{4}$) $$h(-p)\equiv h(p)\left(1-(-1)^{\frac{p-1}{4}}\right)\pmod{4}.$$

This relates $h(-p)$ and $h(p)$ by a congruence, but it is not nearly as strong as the congruence you mention in the $p\equiv 3\pmod{4}$ case which you linked; at the heart of that result was Zagier's formula:

(Zagier, 1974)

Let $p$ be a prime such that $p\equiv 3\pmod{4}$ and $h(p)=1$. Let $[b_0;b_1,b_2,\dots,b_n]$ be the negative continued fraction expansion of $\sqrt{p}$. Then $$ h(-p)=\frac{1}{3}\left(\sum_{i=1}^n b_i-3n\right). $$

The reason for $p\equiv 3\pmod{4}$ has to do with the fact that the fundamental unit of $\mathbb{Q}(\sqrt{-p})$ will be of norm 1 ( if $p\equiv 1\pmod{4}$ the the fundamental unit will be of norm -1). One might hope for an analog of Zagier's formula in the case that $p\equiv 1\pmod{4}$, but it doesn't seem that any is known.