# Does $\sum\limits_{k=1}^n\frac{a_i-a_k}{a_i+a_k}\cdot\frac{a_j-a_k}{a_j+a_k}=0$ for all $i\neq j$ imply $a_1=a_2=\cdots=a_n$?

$$\forall i,\forall j\neq i,\quad\sum_{k=1}^n\frac{a_i-a_k}{a_i+a_k}\cdot\frac{a_j-a_k}{a_j+a_k}=0.$$

We can't have two different $$a_i=0$$ because of the denominators; but we can allow one $$a_i=0$$, if the terms $$k=i$$ and $$k=j$$ are excluded from the sum.

For $$n=3$$, these equations are easy to solve: \begin{align*} (1,2):\quad&\frac{a_1-a_3}{a_1+a_3}\cdot\frac{a_2-a_3}{a_2+a_3}=0\\ (1,3):\quad&\frac{a_1-a_2}{a_1+a_2}\cdot\frac{a_3-a_2}{a_3+a_2}=0\\ (2,3):\quad&\frac{a_2-a_1}{a_2+a_1}\cdot\frac{a_3-a_1}{a_3+a_1}=0. \end{align*} Indeed we just get $$a_1=a_2=a_3$$.

For $$n=4$$, the first of $$6$$ equations is

$$(1,2):\quad\frac{a_1-a_3}{a_1+a_3}\cdot\frac{a_2-a_3}{a_2+a_3}+\frac{a_1-a_4}{a_1+a_4}\cdot\frac{a_2-a_4}{a_2+a_4}=0.$$

(For the other $$5$$, just permute the indices.) I multiplied to clear the denominators, then added equations $$(1,2)$$ and $$(3,4)$$ to get

$$4(a_1a_2-a_3a_4)^2=0$$

and thus

$$a_1a_2=a_3a_4,\quad a_1a_3=a_2a_4,\quad a_1a_4=a_2a_3.$$

These imply that $$a_1^2=a_2^2=a_3^2=a_4^2$$; and we can't have $$a_i=-a_j$$, again because of the denominators. So $$a_1=a_2=a_3=a_4$$.

Does this continue for $$n\geq5$$?

If the variables are non-negative real numbers, then we can arrange them in order, $$a_1\geq a_2\geq a_3\geq\cdots\geq a_n\geq0$$; equation $$(1,2)$$ is then a sum of non-negative terms, so each term must vanish, which gives $$a_2=a_3=\cdots=a_n$$. Then equation $$(2,3)$$ has only its first term remaining, which gives $$a_1=a_2$$.

What if some of the variables are negative, or complex numbers?

We might define $$b_{ij}=\dfrac{a_i-a_j}{a_i+a_j}$$ to simplify the equations to $$\sum_kb_{ik}b_{jk}=0$$. Collecting these into an antisymmetric matrix $$B$$, we see that the system of equations is just saying that

$$BB^T=-B^2=B^TB=D$$

is some diagonal matrix. But I don't think this tells us enough about $$B$$ itself.

The defining equation for $$b_{ij}$$ can be rearranged to

$$a_j=\frac{1-b_{ij}}{1+b_{ij}}a_i$$

so in particular

$$a_3=\frac{1-b_{13}}{1+b_{13}}a_1=\frac{1-b_{23}}{1+b_{23}}a_2=\frac{1-b_{23}}{1+b_{23}}\cdot\frac{1-b_{12}}{1+b_{12}}a_1;$$

cancelling $$a_1$$,

$$(1+b_{31})(1+b_{23})(1+b_{12})=(1-b_{31})(1-b_{23})(1-b_{12});$$

expanding,

$$2b_{12}b_{23}b_{31}+2b_{12}+2b_{23}+2b_{31}=0.$$

In this process I divided by some things that might be $$0$$, but this resulting cubic equation is valid nonetheless.

I think we can dispense with $$a_i$$ now. In summary, we need to solve the system of equations \begin{align*} \forall i,\forall j,\quad&b_{ij}+b_{ji}=0\\ \forall i,\forall j,\forall k,\quad&b_{ij}b_{jk}b_{ki}+b_{ij}+b_{jk}+b_{ki}=0\\ \forall i,\forall j\neq i,\quad&\sum_kb_{ik}b_{jk}=0. \end{align*} Is the only solution $$b_{ij}=0$$?

$$\def\C{\mathbb{C}}$$This answer solves the system of equations$$\begin{gather*} \sum_{k = 1}^n \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} = 0 \quad (\forall i ≠ j) \tag{*} \end{gather*}$$ in $$\C$$ and the italic letter $$i$$ is not the imaginary unit $$\mathrm{i}$$.

On the one hand, suppose $$(a_1, \cdots, a_n) \in \C^n$$ is a solution to ($$*$$). For any $$i, j, k$$,$$\begin{gather*} \frac{a_i - a_k}{a_i + a_k} - \frac{a_j - a_k}{a_j + a_k} = \frac{2(a_i - a_j) a_k}{(a_i + a_k) (a_j + a_k)}\\ = \frac{a_i - a_j}{a_i + a_j} · \frac{2(a_i + a_j) a_k}{(a_i + a_k) (a_j + a_k)} = \frac{a_i - a_j}{a_i + a_j} \left( 1 - \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} \right), \end{gather*}$$ thus$$\begin{gather*} \sum_{k = 1}^n \left( \frac{a_i - a_k}{a_i + a_k} - \frac{a_j - a_k}{a_j + a_k} \right) = \sum_{k = 1}^n \frac{a_i - a_j}{a_i + a_j} \left( 1 - \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} \right)\\ = n · \frac{a_i - a_j}{a_i + a_j} - \frac{a_i - a_j}{a_i + a_j} \sum_{k = 1}^n \frac{a_i - a_k}{a_i + a_k} · \frac{a_j - a_k}{a_j + a_k} \stackrel{(*)}{=} n · \frac{a_i - a_j}{a_i + a_j}. \tag{1} \end{gather*}$$ Define $$c_i = \dfrac{1}{n} \sum\limits_{k = 1}^n \dfrac{a_i - a_k}{a_i + a_k}$$ for all $$i$$, then (1) implies that $$\dfrac{a_i - a_j}{a_i + a_j} = c_i - c_j$$, i.e.$$\begin{gather*} (1 - c_i + c_j) a_i = (1 - c_j + c_i) a_j \quad (\forall 1\leqslant i, j \leqslant n). \tag{2} \end{gather*}$$ Note that for any $$1 \leqslant i < j < k \leqslant n$$ with $$a_i, a_j, a_k ≠ 0$$,$$\begin{cases} (1 - c_i + c_j) a_i = (1 - c_j + c_i) a_j\\ (1 - c_j + c_k) a_j = (1 - c_k + c_j) a_k\\ (1 - c_k + c_i) a_k = (1 - c_i + c_k) a_i \end{cases}$$ imply that$$(1 - c_i + c_j)(1 - c_j + c_k)(1 - c_k + c_i) = (1 - c_j + c_i)(1 - c_k + c_j)(1 - c_i + c_k),$$ which is simplified to be$$\begin{gather*} (c_i - c_j)(c_j - c_k)(c_k - c_i) = 0. \tag{3} \end{gather*}$$

Case 1: If $$a_{i_0} = 0$$ for some $$i_0$$, then $$a_i ≠ 0$$ for all $$i ≠ i_0$$ because of non-zero denominators in ($$*$$), and (2) implies that $$c_i = c_{i_0} + 1$$ for all $$i ≠ i_0$$. Thus for any $$i, j ≠ i_0$$, (2) implies that $$a_i = a_j$$.

Case 2: If $$a_i ≠ 0$$ for any $$i$$, then (3) implies that among any $$c_i, c_j, c_k$$, there are at least two equal to each other. Thus all $$c_i$$'s assume at most two values, and whenever $$c_i = c_j$$ for some $$i$$ and $$j$$, (2) implies that $$a_i = a_j$$.

To summarize, all possible $$(a_1, \cdots, a_n)$$'s (up to permutation) are of the form$$(\underbrace{a, \cdots, a}_{m \text{ copies of } a}, \underbrace{b, \cdots, b}_{n - m \text{ copies of } b})$$ where $$2 \leqslant m \leqslant n$$ (since $$n \geqslant 3$$), $$a, b \in \C$$ and $$a ≠ b$$. Now without loss of generality assume that $$a_1 = a_2 = a$$, then$$0 \stackrel{(*)}{=} \sum_{k = 1}^n \frac{a_1 - a_k}{a_1 + a_k} · \frac{a_2 - a_k}{a_2 + a_k} = \sum_{k = 1}^n \left( \frac{a - a_k}{a + a_k} \right)^2 = (n - m) \left( \frac{a - b}{a + b} \right)^2,$$ which implies that $$m = n$$. Therefore all $$a_i$$'s are equal.

On the other hand, it is easy to verify that $$(a_1, \cdots, a_n) = (a, \cdots, a)$$ $$(a \in \C^*)$$ are indeed solutions to ($$*$$). Therefore they are all the solutions.

• This could be simplified by using $b_{ij}$ instead of $a_i$. Jul 8 '20 at 15:17