If p is a prime congruent to 9 mod 16, can 4 divide the class number of Q(p^(1/4))? When $p$ is a prime $\equiv9\bmod16$, the class number, $h$, of $\mathbb Q(p^{1/4})$ is known to be even. In 
[Charles J. Parry,
A genus theory for quartic fields.
Crelle's Journal 314 (1980), 40--71]
it is shown that $h/2$ is odd when 2 is not a fourth power in $\mathbb Z/p\mathbb Z$. Does this still hold
when 2 is a fourth power?
Some years ago I gave an (unpublished) proof that this is true
provided the elliptic curve $y^2=x^3-px$ has positive rank, and in particular that it is
true on the B. Sw.-D. hypothesis. It's known that the above curve has positive rank for
primes that $\equiv5$ or $7\bmod16$, but to my knowledge $p\equiv9\bmod16$ remains untouched. But perhaps
there's an elliptic-curve free approach to my question?
 A: Let $p \equiv 1 \bmod 8$ be a prime number, let $K = {\mathbb Q}(\sqrt[4]{p})$, and let $F$ be the quartic subfield of the field of $p$-th roots of unity. An easy exercise involving Abhyankar's Lemma shows that $FK/K$ is an unramified quadratic extension, hence the class number of $K$ is always even.
The field $KF$ has the quartic subfield $L = {\mathbb Q}(\sqrt{u})$, where $u$ is the fundamental unit of $k = {\mathbb Q}(\sqrt{p})$. An routine application of the ambiguous class number formula to $L/k$ shows that $L$ has odd class number (there are two ramified primes, one infinite and the other one above $2$; clearly $-1$ is not a norm residue at the infinite prime). 
Now I claim that if $p \equiv 9 \bmod 16$, the class number of $KF$ is odd. By class field theory, this implies that the $2$-class number of $K$ must be $2$. An application of the ambiguous class number formula to $KF/L$ shows that the $2$-part of the ambiguous class group has order
$$ h = \frac{2}{(E:H)}, $$
where  $E$ is the unit group of $L$ and $H$ its subgroup of units that are norms from all completions of $KF$: in fact, only the two prime ideals above $p$ are ramified in $KF/L$. Thus it is sufficient to show that $E \ne H$. I will show that $\sqrt{u}$ is a quadratic nonresidue modulo the primes $\mathfrak p$ above $p$. But if $u = T + U \sqrt{p}$ (replace $u$ by $u^3$ in order to guarantee that $T$ and $U$ are integers), then $(\sqrt{u}/{\mathfrak p})_2 = (u/\mathfrak p)_4 = (T/p)_4 = (T^2/p)_8 = (-1/p)_8 = -1$
because $p \equiv 9 \bmod 16$; here we have used the congruence $T^2 \equiv -1 \bmod p$.
The reason why the case $(2/p)_4 = -1$ is easier is because in this case, the ideal above
$2$ ramified in $K$ generates a class of order $2$ in the $2$-class group, whereas this prime generates a class with odd order if $(2/p)_4 = +1$, which means that there is no strongly ambiguous ideal class in this case.
Edit. Paul Monsky has kindly written up this argument, filled in all the details, and made it available here. Thanks!
