f(f(x)) = x² - x + 1
f(0) = ?
(@edsaperia inflicted this on me earlier :P )
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Yes. We know what f^2 is; it's some particular function g with 1 as unique fixed point and g^{-1}(1) = {0, 1}.
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This is all that's important; everything else follows from this.
End of conversation
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oh, duh, true. My solution got started by reasoning that f(x) = f(y) => x^2 - x + 1 = y^2 - y + 1, which then => either x = y or x = 1 - x
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and then applying that to f(f(1)) = f(f(0)), and considering cases f(1) = f(0) and f(1) = 1 - f(0), then algebra. Lovely puzzle in any case.
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