Class number parity in pure cubic number fields  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 file.
Actually, Paul Monsky stumbled across something similar for pure quartic fields; 
see here. 
 A: 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 $r_2(K) \ge 1$.
