It's [known](https://oeis.org/A035120) that the class number of $\mathbb{Q}(\sqrt{p})$ is $1$ for all primes $p<229$.


**Question**: What would it be like for conceptual explanations of $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$ (equivalently, $\mathbb{Q}(\sqrt{p})$ having odd conductor)?


To clarify the question:


* A conceptual explanation should treat the first few primes simutaneously, instead of a case-by-case analysis where a case is a single prime. (A case-by-case analysis with a finite number of cases that a priori covers the whole range of primes is allowed, e.g. the cases being p=1, 5 or 9 mod 12.)
* For "the first few primes of form $4k+1$", I mean such continuous primes up to a bound, e.g. $5,13,17$ but not $5,17,29$. The argument should be able to cover primes in such a way.
* To avoid trivialities, the conceptual explanations should cover at least $5, 13, 17$ and $29$.


An example of conceptual explanation would be like:


* By Example 2.9 of Masley's paper [Class numbers of real cyclic number fields with small conductor](http://www.numdam.org/article/CM_1978__37_3_297_0.pdf), the class number of such fields are odd.
* The Minkowski bound gives $h(\mathbb{Q}(\sqrt{p}))<3$ for $p<36$. Thus we have established $h(\mathbb{Q}(\sqrt{p}))=1$ for the first few primes of form $4k+1$: $5,13,17$ and $29$.
* This explanation also works for cyclic cubic fields of conductor $7$ and $13$.


Bonus for explanations that are not specialized on real quadratic fields, e.g. the explanation presented above.



Conceptual explanations of the class numbers for the first few $\mathbb{Q}(\sqrt{p})$ with odd conductor
We give a uniform approach to $p \leq 61$ by applying
analytic discriminant bounds to the Hilbert class field.
To be sure this is not entirely "conceptual", but then
some computation is needed even to deal with $p < 36$ using Minkowski.


If $p = 4k+1$ is prime then $K = {\bf Q}(\sqrt{p})$ has odd class number $h$,
so either $h=1$ or $h \geq 3$. If $h \geq 3$ then the Hilbert class field
$H\_K$ is a totally real field of degree $2h \geq 6$ and discriminant $p^h$.
We can now apply to $H\_K$ the known lower bounds on the discriminants of
totally number fields.


The Odlyzko bounds (see Table 4 on page 134 of
[his
1990 paper](http://www.numdam.org/article/JTNB_1990__2_1_119_0.pdf) give a lower bound of $7.941$ on the root-discriminant
$|{\rm disc}(F)|^{1/n}$ for any totally real field $F$ of degree $n \geq 6$.
Hence $p > 7.941^2 > 63$. So we have accounted for all $p \leq 61$.


If the zeta function $\zeta\_F$ satisfies the Riemann hypothesis,
the lower bound improves to $8.143$. Unfortunately this does not
account for any more primes because $8.143^2 = 66.3+$ and
$65$ is not prime. Since there exists a totally real sextic field
of discriminant $300125 = 5^3 7^4$ (see the
[LMFDB entry](https://www.lmfdb.org/NumberField/6.6.300125.1)),
such bounds can never get us past $300125^{1/3} = 66.95+$,
so the primes $p \in [73, 197]$ must be dealt with in some other way.


