HMMT-Nov 2012 thm, Problem 4

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

Solution: When $y=x^{2}+2 x+3$, note that there is a unique real number $y$ such that $y^{y^{y}}=2012$ because $y^{y^{y}}$ is increasing in $y$. The sum of the real distinct solutions of the equation $x^{2}+2 x+3=y$ is -2 by Vieta's Formulae as long as $2^{2}+4(y-3)>0$, which is equivalent to $y>2$. This is easily seen to be the case; therefore, our answer is -2 . $\fbox{-2}$.

Answer: -2