Let f(x)=x(4x2−3)(64x6−96x4+36x2−3). Find the last three digits of the number of distinct positive reals x satisfying f(2014)(x)=x.
Details and Assumptions
f(2014)(x) denotes the function f applied 2014 times successively on x, i.e. f(2014)(x)=2014 timesf(f(⋯(f(x))⋯)).
Any non-real / repeating solutions of the equation should be ignored. For example, the roots of the equation (counted with multiplicity) x5−14x4+86x3−298x2+573x−468=0 are {3,3,4,2−3i,2+3i}, but its number of distinct real solutions is 2.
This problem appeared in the Proofathon Algebra contest, and was posed by me.
Your answer seems reasonable.
Find out if you're right!