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Math Prize 2011 Problem 10

Source:

September 19, 2011
trigonometryfunction

Problem Statement

There are real numbers aa and bb such that for every positive number xx, we have the identity tan⁡−1(1x−x8)+tan⁡−1(ax)+tan⁡−1(bx)=π2 . \tan^{-1} \bigl( \frac{1}{x} - \frac{x}{8} \bigr) + \tan^{-1}(ax) + \tan^{-1}(bx) = \frac{\pi}{2} \, . (Throughout this equation, tan⁡−1\tan^{-1} means the inverse tangent function, sometimes written arctan⁡\arctan.) What is the value of a2+b2a^2 + b^2?
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