# 1997 JBMO Problems/Problem 2

## Problem

Let $\frac{x^2+y^2}{x^2-y^2} + \frac{x^2-y^2}{x^2+y^2} = k$. Compute the following expression in terms of $k$: $$E(x,y) = \frac{x^8 + y^8}{x^8-y^8} - \frac{ x^8-y^8}{x^8+y^8}.$$

## Solution

To start, we add the two fractions and simplify. \begin{align*} k &= \frac{(x^2+y^2)^2 + (x^2-y^2)^2}{x^4-y^4} \\ &= \frac{2x^4 + 2y^4}{x^4 - y^4}. \end{align*} Dividing both sides by two yields $$\frac{k}{2} = \frac{x^4 + y^4}{x^4 - y^4}.$$ That means \begin{align*} \frac{x^4 + y^4}{x^4 - y^4} + \frac{x^4 - y^4}{x^4 + y^4} &= \frac{k}{2} + \frac{2}{k} \\ \frac{(x^4 + y^4)^2 + (x^4 - y^4)^2}{x^8 - y^8} &= \frac{k^2 + 4}{2k} \\ \frac{2x^8 + 2y^8}{x^8 - y^8} &= \frac{k^2 + 4}{2k}. \end{align*} Dividing both sides by two yields $$\frac{x^8 + y^8}{x^8 - y^8} = \frac{k^2 + 4}{4k}.$$ That means \begin{align*} \frac{x^8 + y^8}{x^8 - y^8} - \frac{x^8 - y^8}{x^8 + y^8} &= \frac{k^2 + 4}{4k} - \frac{4k}{k^2 + 4} \\ &= \frac{k^4 + 8k^2 + 16 - 16k^2}{4k(k^2 + 4)} \\ &= \frac{k^4 - 8k^2 + 16}{4k(k^2 + 4)} \\ &= \boxed{\frac{(k^2 - 4)^2}{4k(k^2 + 4)}}. \end{align*}