Suppose x=x+1. For what value of x is f(f(...(f(x))...)=0? with 2015 f's.
I don't think that there is a solution, beacuse if x=x+1 then there is no number that would satisfy x. Mabey you didn't correctly copy the questtion, if that is so, can you please re-write it? Sorry if this post isn't helpful.