Consider the family of pure cubic number fields
$K = {\mathbb Q}(\sqrt[3]{m})$ for $m = a^3 \pm 3$.


**Proposition**. If $4 \mid a$ and $m$ is cubefree, then the 
class number of $K$ is even.


**Proof.** Let $\omega = \sqrt[3]{m}$; the element $\alpha = a - \omega$
 has norm $\pm 3$. Since $3$ is completely ramified, the element
 $\varepsilon = \alpha^3/3$ is a unit.


If $m = a^3 + 3$, then $\varepsilon = 1 - 3a^2\omega + 3a\omega^2$.
 If $4 \mid m$, then $\varepsilon \equiv 1 \bmod 4$, hence
 $K(\sqrt{\varepsilon}\,)/K$ is an unramified quadratic extension.


Experiments seem to suggest that if $m = a^3+3$ and $a \equiv 2 \bmod 4$,
then $h$ is also even, but there is no explanation, class field theoretic
or otherwise. In fact, the class number is even for all cubefree values
of $m$ for $a = 2, 4, \ldots, 2 \cdot 88$, but is odd for $a = 2 \cdot 89$.


This cannot be an accident; the parity of the class number in the
case $m = a^3 - 3$ for $a \equiv 2 \bmod 4$ shows a more typical
(i.e. more random) behaviour in that the class number is odd quite 
often.


Question: *How can this behaviour in the case $m = 8a^3+3$ be explained?*


My first guess would be that, for fields in this family, there is
a family of ideals ${\mathfrak a}$ such that ${\mathfrak a}^2$ is
principal, but I can't seem to find anything in this direction.


 **Edit.** Dror's comment made me look at the family of elliptic curves $y^2 = x^3 - m$.
These have rank $\ge 1$, and by the parity conjecture rank $\ge 2$. An inequality due
to Billing now shows that $K$ has even class number. For details, see this
[pdf](http://www.rzuser.uni-heidelberg.de/~hb3/publ/pcub2.pdf) file.
Actually, Paul Monsky stumbled across something similar for pure quartic fields; 
see [here](https://mathoverflow.net/questions/33711/if-p-is-a-prime-congruent-to-9-mod-16-can-4-divide-the-class-number-of-qp1-4). 



Class number parity in pure cubic number fields 
Billing (Beiträge zur arithmetischen Theorie der ebenen 
kubischen Kurven vom Geschlecht Eins, R. Soc. Scient. Uppsala (4) 11, 
Nr. 1. Diss. 165 S. Uppsala 1938; see Ian Connell's Handbook for elliptic curves for
a modern presentation of the result) proved the following result:


Let $f(x) = x^3 + ax^2 + bx + c \in {\mathbb Z}[x]$ be irreducible, 
and consider the elliptic curve $E: y^2 = f(x)$. Let $K$ be the
cubic number field generated by a root $\alpha$ of $f$, and let
$E\_K$ be its unit group. Write $({\mathcal O}\_K: {\mathbb Z}[\alpha]) =: m\_f^2$. 
Then
$$ r \ \le \ r\_2(K) + r\_E(K) + 2n\_+ + n\_-, $$
where $r$ is the Mordell-Weil-rank of $E({\mathbb Q})$, 
$r\_2(K)$ is the $2$-rank of the ideal class group of $K$,
$r\_E(K)$ is the ${\mathbb Z}$-rank of the unit group $E\_K$ of $K$,
$n\_+$ is the number of primes $p \mid m\_f$ that split in $K$,  

and $n\_-$ is the number of primes $p \mid m\_f$ that decompose 
as $p {\mathcal O}\_K = {\mathfrak p}{\mathfrak p}'$ or as
$p {\mathcal O}\_K = {\mathfrak p}^2 {\mathfrak p}'$.


If ${\mathbb Q}(\sqrt[3]{m})$ is a pure cubic field with 
$m \not\equiv \pm 1 \bmod 9$ cubefree, then the index is trivial,
and $n\_+ = n\_- = 0$ provides us with the bound
$$ r \ \le \ r\_2(K) + 1. $$


On the other hand, the parity conjecture (see the article by
Liverance pointed out by Dror) implies that the Mordell-Weil
rank of $E$ is even for squarefree values of $m = 8b^3 + 3$, 
and the family of nontorsion points 
$$ P\_b\Big( \frac{2b^3+1}{b^2}, \frac{3b^3+1}{b^3} \Big) $$
shows that $r \ge 1$. Thus the parity conjecture implies $r \ge 2$,
and Billing's bound finally gives $r\_2(K) \ge 1$.


