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 $\begin{align*} \frac{x^8 + y^8}{x^8 - y^8} - \frac{x^8 - y^8}{x^8 + y^8} &= \frac{k^2 + 4}{4k} - \frac{4k}{k^2 + 4} \\ &= \frac{k^4 + 8k^2 + 16 - 16k^2}{4k(k^2 + 4)} \\ &= \frac{k^4 - 8k^2 + 16}{4k(k^2 + 4)} \\ &= \boxed{\frac{(k^2 - 4)^2}{4k(k^2 + 4)}}. \end{align*}$

1997 JBMO (Problems • Resources)
Preceded by Problem 1	Followed by Problem 3
1 • 2 • 3 • 4 • 5
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1997 JBMO Problems/Problem 2

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