More polynomials!

First, consider constant polynomials. $f(x)=c$ for $c=0$ or $c=1$ satisfies that $f(f(x))={f(x)}^{k}$ , but the condition $f(1)=1$ makes it so that only $f(x)=1$ is valid.

For non-constant polynomials, let $f(x)=z$ . Then we have that $\displaystyle f(z)={z}^{k}$ which implies $f(x)={x}^{k}$ . This also satisfies the condition that $f(1)=1$ . With this part out of the way, we know that $\displaystyle p(x)=1+x+{x}^{2}...+{x}^{2015}$ then if we multiply by $x-1$ we get: $\displaystyle (x-1)p(x)={x}^{2016}-1$ By definition, a root of the polynomial to the 2016th power must be 1 and there are 2016 roots. Therefore, by removing ${1}^{2016}$ : $\sum _{ p(x)=0 }^{ \quad }{ { x }^{ 2016 } } = 2015$

Let deg(f(x))=degree of f(x) f(f(x))=(f(x))^I, let deg(f(x))=n Therefore, deg(f(f(x)))=deg((f(x))^k) => n²=nk => n=0 or n=k. If n=0, f(x)=1 since f(1)=1. If n=k, f(f(x))=(f(x))ⁿ. Let 'a' be a root of f(x). Let f(x)=c(x-a1)(x-a2)...(x-an) Put x=a1, f(0)=(0)ⁿ=0, 0 is a root of f(x). Let an=0(we can choose any arbitrary ai=0). Therefore, cf(x)(f(x)-a1)...(f(x)-a(n-1))=cⁿxⁿ(x-a1)ⁿ.... Note xⁿ divides RHS, it must also divide LHS. But if all air's are nonzero then we cannot have linear factors of x. Thus we must make all roots of f(x) zero. Thus f(x)=cxⁿ. Putting this expression of f(x) in original equation we will get c=1. Thus f(x)=1,x,x²...x²⁰¹⁵. p(x)=1+x+x²+...+x²⁰¹⁵. Roots of p(x) are 2016th roots of unity except 1 itself, thus our final answer will be 2015