If $ (x_1-x_2)(x_2-x_3)(x_3-x_1)=(x_1-x_3)(x_2-x_1)(x_3-x_2)$ then exist $ i\not =j$, suth that $ x_i=x_j$. It is true for n variables $ x_i$, when $ n$ is odd.

Rust wrote: If $ (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) = (x_1 - x_3)(x_2 - x_1)(x_3 - x_2)$ then exist $ i\not = j$, suth that $ x_i = x_j$. It is true for n variables $ x_i$, when $ n$ is odd. Could you please give a complete proof?I am waiting for it!

After discussing this problem for a long time,my friend(from Shanghai high school)told me that the equation indeed has solutions. For example $ (\frac{47}{7},1,\frac{127}{27},4,6,9,\frac{381}{61})$ then multiply all of them by$ 27*7*61$

hxy09 wrote: Rust wrote: If $ (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) = (x_1 - x_3)(x_2 - x_1)(x_3 - x_2)$ then exist $ i\not = j$, suth that $ x_i = x_j$. It is true for n variables $ x_i$, when $ n$ is odd. Could you please give a complete proof?I am waiting for it! Just by factoring out -1 three times we have $ (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) = - (x_1 - x_3)(x_2 - x_1)(x_3 - x_2)$, so if at the same time $ (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) = (x_1 - x_3)(x_2 - x_1)(x_3 - x_2)$ then $ (x_1 - x_2)(x_2 - x_3)(x_3 - x_1) = 0$, meaning two of $ x_1,x_2,x_3$ are equal. It seems Rust is completely right in this. This is a strange problem. Maybe the problem should be $ n > 3$?