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Wacky Polynomial

Let f(x)=x(4x23)(64x696x4+36x23)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 xx satisfying f(2014)(x)=xf^{(2014)}(x) = x.

Details and Assumptions

  • f(2014)(x)f^{(2014)}(x) denotes the function ff applied 20142014 times successively on xx, i.e. f(2014)(x)=f(f((f(x))))2014 timesf^{(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) x514x4+86x3298x2+573x468=0x^5-14 x^4 + 86 x^3 - 298 x^2 + 573 x - 468=0 are {3,3,4,23i,2+3i} \left \{ 3, 3, 4, 2-3i, 2+3i \right \} , but its number of distinct real solutions is 22.

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

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