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A friend came up with this problem, and we and a few others tried to solve it. It turned out to be really hard, so one of us asked his professor. I came with him, and it took me, him and the professor about an hour to say anything interesting about it.

We figured out that for positive $x$, assuming $f$ exists and is differentiable, $f$ is monotonically increasing. (Differentiating both sides gives $f'(x)*[\text{positive stuff}]=2x$). So $f$ is invertible there. We also figured out that f becomes arbitrarily large, and we guessed that it grows faster than any linear function. Plugging in $f{-1}(x)$ for $x$ gives $x+f(x)=[f^{-1}(x)]^2$. Since $f(x)$ grows faster than $x$, $f^{-1}$ grows slower and therefore $f(x)=[f^{-1}(x)]^2-x\le x^2$.

Unfortunately, that's about all we know... No one knew how to deal with the $f(f(x))$ term. We don't even know if the equation has a solution. How can you solve this problem, and how do you deal with repeated function applications in general?

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    It is trivial to come up with functional equations which are extraordinarily hard and probably beyond the abilities of all civilizations in this galaxy and a few others... Does your friend have some reason to be interested in this particular one?2011-11-24
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    Differentiating gives $f'(x)\left(1+f'(f(x))\right)=2x$... why is $1+f'(f(x))$ positive for positive $x$?2011-11-24
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    @mariano It's continuous, and it is obviously positive somewhere (since $f(x)$ becomes arbitrarily large). If it were negative anywhere, it would have to be zero somewhere, and that would make $f'$ blow up, contradicting the assumption that $f$ is differentiable. Of course, we don't know whether it's differentiable, or even continuous anywhere, but if there is a solution that is differentiable, the derivative is always positive for large enough $x$.2011-11-24
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    @MarianoSuárez-Alvarez My friend was messing around with function composition and came up with this problem; I don't really know why he chose this one. He noticed that if the right hand side was simply $x$, this would be easy, and then he tried solving it with the rhs equal to $x^2$, and got stuck.2011-11-24
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    @MarianoSuárez-Alvarez (Out of curiosity, why are so many simple-looking functional equations so hard?)2011-11-24
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    Lots of simple looking things are hard, the fact that this is a functional equation is irrelevant :)2011-11-24
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    See: http://math.stackexchange.com/questions/5303/why-do-statements-which-appear-elementary-have-complicated-proofs/5446#54462011-11-24
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    This one is easier: solve $f(f(x)) = \sin x.$ The method for doing this is due to J. Ecalle. The graph of the answer resembles a sine wave, slightly larger amplitude. Also $C^\omega$ except at multiples of $\pi.$ Despite appearances, this comes under the general heading of complex dynamics.2011-11-24
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    It might be worth looking at $g(g(x))=x^2$ as $f(x)$ may be close to $g(x)$ for large $x$. Clearly a solution is $g(x)=x^{\sqrt{2}}$2011-11-24
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    Regarding the simpler equation f(x)+f(f(x)) that Eric mentions, it's easy to find two solutions, but is there any easy way to show these are all? If they even are?2011-11-24

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