It's known 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, 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.