1987 OIM Problems/Problem 1

Problem

Find all $f(x)$ such that: $\left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x$ for $x \ne 0$ , $x \ne 1$ , $x \ne -1$ ,

~translated into English by Tomas Diaz. ~orders@tomasdiaz.com

Solution

We have the following equations: $(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right)=64x$ $(2) \left[ f\left(\frac{1-x}{1+x}\right) \right]^2f\left( x \right)=64 \cdot \frac{1-x}{1+x}$ Multiplying $(1)$ and $(2)$ , we have $(3) f(x)f\left(\frac{1-x}{1+x}\right) = 16 \cdot \sqrt[3]{x\cdot \frac{1-x}{1+x}}$ Dividing $(1)$ by $(3)$ gives $f(x) = 4 \cdot \sqrt[3]{x^2\cdot \frac{1+x}{1-x}}$ Checking to see if it works… $(1) \left[ f(x) \right]^2f\left( \frac{1-x}{1+x} \right) = 64 \cdot \sqrt[3]{x^4 \cdot \left(\frac{1+x}{1-x}\right)^2 \cdot \left(\frac{1-x}{1+x}\right)^2 \cdot \frac{1}{x}} = 64x$ ~Archieguan

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1987 OIM Problems/Problem 1

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Solution

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