I claim that if $ n $ is odd, $ f(x) = -x $ is the only solution, and if $ n $ is even, then there is no solution for $ f. $ Lemma 1: There exists a nonnegative integer $ k < n $ such that $ \frac{f(x)}{x^k} $ is a polynomial in $ x^n. $ Proof: Note that since $ f(x)^n = f(x^n + 1) + 1 $ is a polynomial in $ x^n $ by writing out the terms of $ f $ and expanding we can bash out the result. Now, if $ k = 0 $ we can write $ f(x) = g(x^n) $ and $ g $ will satisfy the same condition as $ f $ so by infinite descent we find that $ f(x) = cx^n $ for some constant $ c \in \mathbb{C}. $ This is clearly impossible so we must have $ k > 0. $ In other words, we have that $ f(0) = 0. $ We now proceed with casework on the parity of $ n. $ Case 1: $ n $ is even. In this case by letting $ x = 0 $ in the condition we find that $ f(1) = -1. $ Then by letting $ x = 1 $ in the condition we find that $ f(2) = 0. $ Proceeding in this fashion, we find that $ f(a_k) = 0 $ for all positive integers $ k $ where $ a_1 = 0 $ and $ a_k = (a_{k - 1}^n + 1)^n + 1. $ This means that $ f(x) $ has infinitely many roots, contradiction. Case 2: $ n $ is odd. In this case by letting $ x = 0 $ in the condition we find that $ f(1) = -1. $ Then by letting $ x = 1 $ in the condition we find that $ f(2) = -2. $ Proceeding in this fashion, we find that $ f(b_k) = -b_k $ for all positive integers $ k $ where $ a_1 = 0 $ and $ a_k = a_{k - 1}^n + 1. $ Therefore the polynomial $ f(x) + x $ has infinitely many roots, so it must be identically $ 0, $ as desired.

this solution is wrong. there are actually infinetely many solutions for odd n. i'll write a solution if anyones interested

giglio wrote: this solution is wrong. there are actually infinetely many solutions for odd n. i'll write a solution if anyones interested Yeah pleass type yourr solution i was try my best but too hard