Wacky Polynomial

Let \(f(x) = x\left( 4x^2-3 \right) \left( 64x^6-96x^4+36x^2-3 \right) \). 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. \(f^{(2014)}(x) = \underbrace{f ( f ( \cdots ( f(x) )\cdots ))}_{2014 \text{ times}}\).

• Any non-real / repeating solutions of the equation should be ignored. For example, the roots of the equation (counted with multiplicity) \(x^5-14 x^4 + 86 x^3 - 298 x^2 + 573 x - 468=0 \) are \( \left \{ 3, 3, 4, 2-3i, 2+3i \right \} \), but its number of distinct real solutions is \(2\).

This problem appeared in the Proofathon Algebra contest, and was posed by me.