Putnam 2006, Problem A-5

Let $n$ be a positive odd integer and let $\theta$ be a real number such that $\theta/\pi$ is irrational. Set $a_k = \tan(\theta + k\pi/n)$, $k = 1, 2, \ldots, n$. Prove that
$$\frac{a_1 + a_2 + \cdots + a_n}{a_1 a_2 \cdots a_n}$$
is an integer, and determine its value.

Solution:
Since the desired expression involves symmetric functions of $a_1, \ldots, a_n$, we start by finding a polynomial with $a_1, \ldots, a_n$ as roots. Note that
$$1 \pm i \tan \theta = e^{\pm i\theta} \sec \theta$$
so that
$$1 + i \tan \theta = e^{2i\theta}(1 - i \tan \theta).$$
Consequently, if we put $\omega = e^{2in\theta}$, then the polynomial
$$Q_n(x) = (1 + ix)^n - \omega(1 - ix)^n$$
has among its roots $a_1, \ldots, a_n$. Since these are distinct and $Q_n$ has degree $n$, these must be exactly the roots. If we write
$$Q_n(x) = c_nx^n + \cdots + c_1x + c_0,$$
then $a_1 + \cdots + a_n = -c_{n-1}/c_n$ and $a_1 \cdots a_n = -c_0/c_n$, so the ratio we are seeking is $c_{n-1}/c_0$. By inspection,
$$c_{n-1} = ni^{n-1} - \omega n(-i)^{n-1} = ni^{n-1}(1 - \omega)$$
$$c_0 = 1 - \omega$$
so
$$\frac{a_1 + \cdots + a_n}{a_1 \cdots a_n} = \begin{cases} n & n \equiv 1 \pmod{4} \\ -n & n \equiv 3 \pmod{4}. \end{cases}$$

**Remark:** The same argument shows that the ratio between any two odd elementary symmetric functions of $a_1, \ldots, a_n$ is independent of $\theta$.