If (x+\sqrt{1+x^2})(y+\sqrt{1+y^2})=1, where x and y are <strong>REAL</strong> numbers, then what is the value of (x+y)^2?

Explanation

Rationalizing gives (x+\sqrt{1+x^2})(\sqrt{1+x^2}-x) = 1. Comparing with the given equation, y+\sqrt{1+y^2} = \sqrt{1+x^2}-x. Doing the same for y yields x+y = 0. Therefore, (x+y)^2 = 0.

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