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 $a_1=a_2=\cdots=a_n$?

Question

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

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?

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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 $b_{ij}=0$?

Answer 1

$\def\C{\mathbb{C}}$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.