the number of polynomial functions of degree satisfy
(A) 0
(B) 1
(C) 2
(D) infinetly many
I get answer (B).
f(x^2) = f(x)f(x) = f(f(x))
Consider just the part
f(x^2) = f(f(x))
So try x^2 = f(x).
Verify that this does not contradict the other equality given.
Given the constraint on the degree we cannot choose f(x) = 0 or f(x) = 1.
It also happens that 2 is the only nonzero real number satisfying n*n = n+n. (Look at the exponent to see why I say this.)