Class number/ quadratic field/divisibility If we take $n$ an even integer and greater than $5$, then $\mathbb Q(\sqrt{1 - 4k^n})$ are divisible by $n$, other than for $k = 13$ and $n = 8$. Why this is happened? If we take $n$ less than 5 (I mean for $n = 2$ or $n = 4$) what will happen? 
 A: Let $\alpha = \sqrt{1 - 4 k^n}$, and let $K = \mathbf{Q}(\sqrt{\alpha})$. Suppose that $k > 1$. I understand your question to be:  why is the class number of $K$ divisible by $n$?
The minimal polynomial of $\theta = \displaystyle{\frac{1 + \alpha}{2}}$ is
$$\theta^2 - \theta + k^n = 0.$$
Indeed, we even have $\mathbf{Z}[\theta]$ is the ring of integers of $K$.
Let $p$ be a prime dividing $k$. Then $p$ splits in $K$, as can be seen by factoring the polynomial above modulo $p$. Indeed, there is a unique such ideal which divides $\theta$, namely $\mathfrak{p} = (p,\theta)$. Equally, every prime dividing $\theta$ also divides $k$.
Since $\theta$ and $1 - \theta$ are co-prime, and since
$$\theta \overline{\theta} = \theta (1 - \theta) = k^n,$$
it follows that the exponent of $\mathfrak{p}$ in $\theta$ is $n$-times the exponent of $\mathfrak{p}$ in $k$. In particular,
there exists an ideal $\mathfrak{a}$ of norm $k$ such that
$\mathfrak{a}^n = (\theta)$.
Suppose that $\mathfrak{a}^m$ is principal. Then
$$k^m = N(\mathfrak{a}) = a^2 + ab + b^2 k^n = (a + b/2)^2 + b^2(k^n - 1/4) \ge k^n,$$
as long as $b \ne 0$. Yet if $b = 0$, then $(\theta^m) = \mathfrak{a}^{mn} = (a^n)$, and then $\theta = \pm a^n$ (the only units in $K$ are $\pm 1$), which is nonsense. Hence $k^m \ge k^n$, and thus $m \ge n$ (using the fact that $k > 1$), and thus the order of $\mathfrak{a}$ in the class group is exactly $n$. It follows that the class number is divisible by $n$ for any $n$.
