# Solution to $(1+a_i-a_j)(1+a_j-a_k)(1+a_k-a_i)=(1+a_j-a_i)(1+a_i-a_k)(1+a_k-a_j)$

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]
• 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 $a +b + c = 0$ – Rybin Dmitry 2 days ago