More fun in 2015, Part 14

We'll need to use the identities $\prod\limits_m w_m = 1$ and $\prod\limits_m (1 + w_m) = 2$ which can be verified by evaluating the polynomial $\prod\limits_m (x - w_m)= x^{2015} - 1$ at $x = 0$ and at $x = -1$ . For ease of notation I mean here for the indices of any products to run from $1$ to $2015$ , inclusive, unless otherwise specified.

Letting $X = \prod\limits_{m<n} (w_m+w_n)$ note that $\prod_m \prod_n (w_m + w_n) = \prod_{m=n} (w_m+w_n) \prod_{m < n} (w_m + w_n) \prod_{m>n}(w_m+w_n) = 2^{2015} X^2.$

On the other hand, $\prod_m \prod_n (w_m + w_n) = \prod_m w_m \prod_n (1 + w_n w_m^{-1}) = \prod_m w_m \prod_k (1+ w_k) = \prod_m 2w_m = 2^{2015}.$

Comparing these two expressions gives $X^2 = 1$ , from which we conclude $|X| = 1$ .

In response to Otto's challenge question: Let $w_m = e^\frac{2 \pi m i}{2015}$ . Whenever $m + n \neq 2015$ note that $w_m + w_n$ and its complex conjugate $w_m^* + w_n^* = w_{2015-m}+w_{2015-n}$ form a pair of distinct factors of the product $X = \prod\limits_{m<n} (w_m+w_n)$ . Since positive real factors do not contribute to the argument of a product, we can discard those pairs and write the argument of $X$ as $m(X) = m(\prod\limits_{m = 1}^{1007} (w_m + w_{2015-m})) = m(\prod\limits_{m = 1}^{1007} (w_m + w_m^*)) = m(\prod\limits_{m = 1}^{1007} Re(w_m)$ ). Thus $X$ is real and its sign is the product of the signs of the real parts of the first 1007 roots of unity. This, in turn, is $(-1)^k$ , where $k$ is the number of roots of unity that lie in the second quadrant of the complex plane. Drawing a picture reveals $k = \lceil \frac{1007}{2}\rceil$ = 554. So the sign of $X$ is $(-1)^{554}=1$ .

Following the above methods I found that if $w_1, \ldots, w_T$ are the $T^{th}$ roots of unity then the product $X = \prod\limits_{1 \leq m<n \leq T} (w_m+w_n)$ is equal to 1 when $T \equiv 1, 7$ (mod 8), is equal to -1 when $T \equiv 3, 5$ (mod 8), and is equal to 0 when $T \equiv 0, 2, 4, 6$ (mod 8).

Otto- thanks for a fantastic question with an interesting outcome!

Another approach to the problem (and the additional challenge) uses a special polynomial. For simplicity use the following shorthands

$\begin{aligned} N &:= 2015,\qquad z:=e^{i\frac{2\pi}{N}},\qquad Y:=\prod_{1\leq m<n\leq N}(w_m+w_n) \end{aligned}$

Before getting started, let's take a look at a polynomial that wil help a lot:

$\begin{aligned} p(x)&:=(-1)^N(x^N-1)=(-1)^N\prod_{1\leq m\leq N}(x-z^m)=\prod_{1\leq m\leq N}(z^m-x)\\\\ \text{(*)}\qquad \prod_{1\leq m \leq N}(z^m+z^n)&=p(-z^n)=(-1)^N((-1)^Nz^{Nn}-1)=1-(-1)^N \end{aligned}$

Now to the task. First, observe that the product $Y$ is fully symmetrical in $m,\:n$ :

$\begin{aligned} Y &= \prod_{1\leq m<n\leq N}(w_m+w_n) = \prod_{1\leq m<n\leq N}(w_n+w_m) \underset{m\leftrightarrow n}{=} \prod_{1\leq n<m\leq N}(w_m+w_n)\\\\ \Rightarrow\quad Y^2&=\prod_{1\leq m\neq n\leq N}(w_m+w_n)=\frac{ \displaystyle\prod_{1\leq m;\:n \leq N}(w_m+w_n) }{ \displaystyle\prod_{1\leq m=n\leq N}(w_m+w_n) } \end{aligned}$

Notice how the last expression does not depend on the order of the roots anymore? From now on, we assume the standard order of the roots $w_n=z^n$ . Let's tackle numerator and denominator separately using $(*)$ :

$\begin{aligned} \prod_{1\leq m;\:n\leq N}(w_m+w_n)&=\prod_{1\leq n\leq N}\left( \prod_{1\leq m\leq N}(z^m+z^n) \right)\underset{\text{(*)}}{=}\prod_{1\leq n\leq N}(1-(-1)^N) = (1-(-1)^N)^N\\\\ \prod_{1\leq m=n\leq N}(a_m+a_n)&=\prod_{0\leq m=n\leq N-1}(z^m+z^n)=\prod_{1\leq m\leq N}(2z)^m = 2^Nz^{ \sum_{m=1}^{N}m }=2^Nz^{ \frac{N}{2}(N+1) } = 2^Ne^{i\frac{2\pi}{N}\cdot\frac{N}{2}(N+1)} = 2^Ne^{i\pi(N-1)} = 2^N(-1)^{N+1} \end{aligned}$

With all that, we are nearly ready:

$\begin{aligned} Y^2=\frac{ (1-(-1)^N)^N }{2^N(-1)^{N+1}}=\begin{cases} 0,&N\text{ even}\\ 1,&N\text{ odd} \end{cases} \end{aligned}$

We get $\fbox{|Y|=1}$ for $N=2015$ . Regarding the sign of $Y$ , I'll refer to Paul Sherman's solution - mine was not nearly as neat!