Fix a large natural number $n$. Let $a_1,a_2,\dots,a_n$ be $n$ real numbers satisfying

$$ (1+a_i-a_j)(1+a_j-a_k)(1+a_k-a_i)=(1+a_j-a_i)(1+a_i-a_k)(1+a_k-a_j) $$ for any $1\le i,j,k\le n$.

How to find all solutions to the above system of equations? I can see that, obviously, a constant sequence is a solution. But I don't know if there are other solutions.


Fix $i, j, k$ and expand both sides of the equation. Lots will cancel and you should end up with$$(a_i-a_j)(a_j-a_k)(a_k-a_i)=0$$ so the two expressions are equal if and only if two of the $a$s are equal. You can check that putting two values equal makes the two sides equal very easily. The algebra says there are no other cases.

If any three of the $a_i$ are distinct, they will not give equality. So any sequence made up of at most two values will suffice.

[provided I got my algebra right - but putting everything on the same side I have terms of degree $3$ at most, and setting $a_i=a_j$ gives me zero so I know I have a factor $\pm (a_i-a_j)$ - I'll leave you to check the detail]

  • $\begingroup$ In Wolfram Alpha, input "factor (1+x-y)(1+y-z)(1+z-x)-(1+y-x)(1+z-y)(1+x-z)", then I got "-2 (x - y) (x - z) (y - z)". So you were right! $\endgroup$
    – Tony B
    Oct 7 '18 at 17:11
  • $\begingroup$ Note that it can be checked really fast if we rewrite it as: $(1 + a)(1 + b)(1 + c) = (1 - a)(1 - b)(1 - c)$ with $a +b + c = 0$ $\endgroup$ Jul 18 '19 at 19:13

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