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Find the sum of all real solutions for $x$ to the equation $(x^2 + 2x + 3)^{(x^2+2x+3)^{(x^2+2x+3)}} = 2012.$

I just know $x^{x^x}$ is increasing in $x$ and hence the equation has a unique solution, nut then I dont know how to move on, I also know viete' formula but I dont know if it helps here, thanks in advance.

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@FlybyNight: $a^{b^c}$ means $a^{(b^c)}$ by standard convention, which matches how it's entered into LaTeX. – Jonas Meyer Aug 22 '13 at 19:44
@JonasMeyer I can't be that standard or else I wouldn't have asked. $a^b^c$ – Fly by Night Aug 22 '13 at 19:51
It is standard. One justification for it is that even though exponentiation is not associative there is rarely any reason to write $(a^b)^c$ because we can just write $a^{b \cdot c}$ instead. So $a^{b^c}$ gets to have the more useful meaning. – Dan Brumleve Aug 22 '13 at 19:57
@FlybyNight Well, LaTeX allows for $a^{b^c}$ and ${a^b}^c$, so to talk about the standard indeed requires a good eye :) – Hagen von Eitzen Aug 22 '13 at 19:57
@FlybyNight: Has your question not been answered? – Jonas Meyer Aug 22 '13 at 20:14

The sum is $-2$. Can you see why? Hint: I have not computed the solutions.

Details: As you observed, there is a unique positive $b$ such that $b^{(b^b)}=2012$. Moreover, this $b$ is in the interval $(2,3)$, by the Intermediate Value Theorem, since $2^{(2^2)}$ is too small and $3^{(3^3)}$ is too big.

Note that $x^2+2x+3=(x+1)^2+2$, so $x^2+2x+3$ attains a minimum value of $2$. Thus the equation $x^2+2x+3-b=0$ has two real solutions. The sum of these is the negative of the coefficient of $x$, that is, $-2$.

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(+1) very clever. – achille hui Aug 22 '13 at 20:04
I will wait a while on the (very small) amount of explanation that would turn the answer into a full solution. I would like to give the OP an opportunity to use the information to discover the solution. – André Nicolas Aug 22 '13 at 20:28
the quadratic term is symmetric about x = -1. If there are 2 complex solutions and they are conjugates and their real parts are -1, the imaginary terms cancel when adding and the real parts sum up to -2. But that's not the main reason, I understand that...Just trying to get some clue – imranfat Aug 22 '13 at 20:31
@imranfat, more hints: $2^{2^2}<2012$ and V...a – njguliyev Aug 22 '13 at 22:54
@Prism: We have to be careful with longer towers. The min of $x^2+2x+3$ is $2$. So if the tower has height more than $3$, and we still use $2012$, there is no real root. – André Nicolas Aug 23 '13 at 0:29

Hint: Look at Vieta's Formula then compare your equation.

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Sketch: $y^{y^y}=2012$ must have only one real solution, call it $a$ (we don't need to compute it). Then we are looking for the roots of $x^2 +2 x+3 - a = (x- x_0)(x-x_1)=x^2 - (x_0+x_1)x +x_0 x_1$ Hence...

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@JoelReyesNoche : fixed. it really does not matter – leonbloy Aug 23 '13 at 3:31
So even if the given number is 2013, the answer still would have been -2? – imranfat Aug 23 '13 at 15:15
@imranfat Yes. The only important thing is that the cuadratic has two real solutions (positive discriminant), which here happens when $a>2$ – leonbloy Aug 23 '13 at 15:24
This is outright EVIL :) Sometimes I wonder, forget about the answer, how does anyone come up with such a question to begin with... – imranfat Aug 23 '13 at 15:39

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