# How prove that $x_1 = x_{2000}$ implies $x_2 \ne x_{1999}$, where $x_{n+2}=\frac{x_{n}x_{n+1}+5x^4_{n}}{x_{n}-x_{n+1}}$?

Let $x_{1},x_{2},\cdots$ be real numbers, such that for $n \ge 1$: $$x_{n+2}=\dfrac{x_{n}x_{n+1}+5x^4_{n}}{x_{n}-x_{n+1}}$$

If $x_{1}=x_{2000}$, prove that $x_{2}\neq x_{1999}$.

my idea $$x_{n+2}x_{n}-x_{n+2}x_{n+1}=x_{n}x_{n+1}+5x^4_{n}\cdots (1)$$ $$x_{n+3}x_{n+1}-x_{n+3}x_{n+2}=x_{n+1}x_{n+2}+5x^4_{n+1}\cdots (2)$$ (2)-(1) $$x_{n+3}x_{n+1}-x_{n+2}x_{n}=x_{n+3}x_{n+2}-x_{n}x_{n+1}+5(x^4_{n+1}-x^4_{n})$$

Then following I can't

• What is your source? – Did Jun 30 '13 at 14:54
• If $x_{n+2}=f(x_n,x_{n+1})$ then $x_{n+1}=f(-x_n,x_{n+2})$ maybe this will help someone to prove it. Where did you get this problem from? It seams like a contest problem. – Paracosmiste Jun 30 '13 at 15:05
• If $x_{2}=x_{1999}$ then it follows that $x_3=-x_{2002}$. Does it help anyway to find some contradiction and conclude that $x_{2}\neq x_{1999}$? – Samrat Mukhopadhyay Jun 30 '13 at 18:14

First note that by the definition, $x_n\ne x_{n+1}$ for all $n\ge 1.$ Subtract $x_{n+1}$ from both sides of our recurrence relation to end up with $$x_{n+2}-x_{n+1}=\frac{5x_n^4+x_{n+1}^2}{x_n-x_{n+1}}.$$ So, for any $n\ge 1,$ $(x_{n+2}-x_{n+1})\cdot(x_{n+1}-x_{n})<0.$ Now, if $x_2>x_1,$ then $x_2>x_3$ and $x_4>x_3$ so iterating this process leads to $x_2>x_1=x_{2000}>x_{1999}.$ In the other case, namely $x_2<x_1,$ we have $x_2<x_3$ and $x_4<x_3$ etc. Finally, $x_{1999}>x_{2000}=x_1>x_2.$ In both cases, $x_2\ne x_{1999}.$