Answer : $f(n) = n.$ Now, to the proof (by induction) : At first, taking $n = 1$ in the given relation yields that $f(f(f(1))) = f(f(1)) = f(1) = 1.$ Then, let us assume that, $\forall i \leq n, \ \ f(i) = i.$ Thus $f$ is injective, so that $f(m) > n$ for $m > n.$ So we get that $f(f(f(n+1))) + f(f(n+1)) + f(n+1) \geq 3(n+1).$ But we know that $f(f(f(n+1))) + f(f(n+1)) + f(n+1) = 3(n+1),$ which yields $f(f(f(n+1))) = f(f(n+1)) = f(n+1) = n+1.$ And we're done. EDIT : I just see that it's also here, and we have the same solution..

Just to give another (and typical with those problems involving the iterates of a function - but (imho) less elegant) solution : for some integer $ x,$ we consider $ a_{n}(x) = f \circ f \ldots \circ f (x).$ (n times) Then we have : $ a_{n+3}+a_{n+2}+a_{n+1}= 3a_{n},$ whose characteristic polynomial is $ r^{3}+r^{2}+r-3 = (r-1)(r+1+i\sqrt{2})(r+1-i\sqrt{2}),$ hence $ a_{n}(x) = F(x)+(-1)^{n}(G(x)(1+i\sqrt{2})^{n}+H(x)(1-i\sqrt{2})^{n})$ where $ F(x), G(x)$ and $ H(x)$ are real constants. Now, letting successively $ n = 0, 1, 2,$ we get that : $ f(x) = a_{1}(x) = a_{0}(x) = x,$ which is indeed a solution of our functional equation, we're done.

mathmanman wrote: Just to give another (and typical with those problems involving the iterates of a function - but (imho) less elegant) solution : for some integer $ x,$ we consider $ a_{n}(x) = f \circ f \ldots \circ f (x).$ (n times) Then we have : $ a_{n + 3} + a_{n + 2} + a_{n + 1} = 3a_{n},$ whose characteristic polynomial is $ r^{3} + r^{2} + r - 3 = (r - 1)(r + 1 + i\sqrt {2})(r + 1 - i\sqrt {2}),$ hence $ a_{n}(x) = F(x) + ( - 1)^{n}(G(x)(1 + i\sqrt {2})^{n} + H(x)(1 - i\sqrt {2})^{n})$ where $ F(x), G(x)$ and $ H(x)$ are real constants. Now, letting successively $ n = 0, 1, 2,$ we get that : $ f(x) = a_{1}(x) = a_{0}(x) = x,$ which is indeed a solution of our functional equation, we're done. I understand, that $ a_0 (x) = x,$ but what is the value of $ a_1 (x)$ and $ a_2 (x)$??? We need them to define $ F(x), G(x)$ and $ H(x)$. Maybe you compare the imaginary part with $ 0$. But then we have: $ a_0(x) = F(x) + G(x) + H(x) = x$ $ a_1(x) = F(x) - G(x) - H(x) + i\sqrt {2}(H(x) - G(x)) = f(x)$ From second we get $ G(x) = H(x)$. Now: $ a_2(x) = F(x) + G(x)(1 + i\sqrt {2} - 2 + 1 - i\sqrt {2} - 2) = f(f(x))$ or $ a_2(x) = F(x) - 2G(x) = f(f(x))$ We have: $ F(x) + 2G(x) = x$ $ F(x) - 2G(x) = f(x)$ $ F(x) - 2G(x) = f(f(x))$ What now? And could we use this method for other equations, for example, for $ f(f(x)) = x$. In this case we have: $ r^2 - 1 = (r - 1)(r + 1)$ and $ a_{n}(x) = F(x) + ( - 1)^n G(x)$ $ a_0 (x) = F(x) + G(x) = x$ $ a_1 (x) = F(x) - G(x) = f(x)$ and what we should do next? And the last, is $ x$ should be integer (natural) or it may be real?

$f(f(f(1))) \ge 1 ,f(f(1)) \ge 1,f(1) \ge 1$ and $f(f(f(1)))+f(f(1))+f(1)=3 \implies f(1)=1$ and we can easily see that $f$ is one-to-one. now we use induction. statement: $\forall i<n, f(i)=i$ the induction step for $n=2$ is trivial so assume it's true for $n-1 \ge 2$ then we shall prove it for $n$, because $f$ in injective then $f(f(f(n))) \ge n ,f(f(n)) \ge n,f(n) \ge n$ which implies $f(n)=n$ so $\forall n \in N:f(n)=n$

Claim: $f$ is the identity. Proof. We prove by induction. Then the base case follows from $n=1.$ So if $f(k)=k ~\forall n\geq k,$ then $f(n+1)\geq n+1, f(f(n+1))\geq n+1$ and $f(f(f(n+1)))\geq n+1.$ We finish the induction considering $n=n+1$ in the original FE.