APMO problem

\(\begin{array} {} P(x,y): & f(x^4+y) = x^3 f(x) + f(f(y)) \\ P(0,y): & f(y) = f(f(y)) & \small \color{blue} \text{Let }f(y) = a \\ & a = f(a) \implies f(x) = x & \small \color{blue} \text{Putting }f(x) = x \text{ in }P(x,y) \\ P(x,y): & f(x^4+y) = x^3 f(x) + f(f(y)) \\ & f(x^4+y) = x^4 + f(y) \\ & f(x^4+y) = x^4 + y & \small \color{blue} \text{Proving that }f(x) = x \end{array} \)

Therefore, $f(2016) = \boxed{2016}$ .

Putting $x = 0$ , we get $f(y) = f(f(y))$ for all $y$ ...........(1).

Putting $x = 1, y = 0$ , we get $f(0) = 0$ . Hence putting $y = 0$ , $f(x^4) = x^3f(x)$ ..........(2).

If $f(1) = 0$ , then $f(2) = f(1 + 1) = f(1) + f(1) = 0$ , so $f(3) = f(1 + 2) = f(1) + f(2) = 0$ and so on. Contradiction (only finitely many zeros). So $f(1) = k$ for some non-zero $k$ . By (1), $f(k) = k$ .

Suppose $f(h) = 0$ for some $h$ not $0$ or $1$ . Then $f(h^4) = h^3f(h) = 0$ , so $f(x) = 0$ for any of the distinct values $x = h, h^4, h^{16}, h^{64}$ , ... . Contradiction (only finitely many zeros). So $f(h)$ is not $0$ for any non-zero $h$ .

Given any $x$ , put $z = f(x^4) - x^4$ . Then $f(x^4) = f(f(x^4)) = f(x^4 + z) = x^3f(x) + f(z)$ . But $f(x^4) = x^3f(x)$ , so $f(z) = 0$ . Hence $z = 0$ . So for any positive x we have $f(x) = x$ . But $f(x^4) = f( (-x)^4 ) = - x^3 f(-x)$ . Hence $f(-x) = - f(x)$ . So $f(x) = x$ for all $x$ .

Put X=0 ,remaining expression is f(y)= f(f(y)) From above expression and observation we have f(y)=y