Answer. The only solutions are $f(x)=cx^d+e$ where $c \in \{-1, 1\}$ , $d \ge 0$ is an odd divisor of $n$ and $e \in \mathbb{Z}$ for all integers $x$ .

Solution. It is clear that these satisfy the conditions. We will show that no other function works. Indeed, suppose a function $f$ other than those specified works.

Injectivity of $f$ follows since $f(a)=f(b) \Longrightarrow a^n=b^n$ and as $n$ is odd, $a=b$ . Shifting by a constant; let $f(0)=0$ , and for all integers $a$ we get $f(a) \mid a^n$ . As $f(1) \mid 1$ , scaling by a $\pm 1$ factor, we let $f(1)=1$ . For all primes $p$ , $f(p) \mid p^n \Longrightarrow f(p)=\pm \, p^k$ for some $k >0$ .

Firstly, let $f(p)=-p^k$ and observe that $p^k+1 \mid p^n-1$ . Write $n=kl+r$ for $0 \le r<k$ and note that $p^n \equiv p^r\cdot (-1)^l \pmod {p^k+1} \Longrightarrow p^k+1 \le \left|p^r\cdot (-1)^l-1\right| \le p^r+1,$ which is clearly false. It follows that $f(p)=p^{k_p}$ for all primes $p$ with $1 \le k_p \le n$ . Evidently, there exists $1 \le d \le n$ such that $k_p=d$ holds for infinitely many primes $p$ .

Fix a prime $p_1$ with $k_{p_1}=d$ and vary $p$ with $k_p=d$ . Write $n=md+r$ for $0 \le r<d$ and note that $p^d-p_1^d \mid p^n-p_1^n \Longrightarrow p_1^{md}\cdot \left(p^r-p_1^r\right) \equiv 0 \pmod {p^d-p_1^d}.$ This is fails to occur unless $d \mid n$ .

Finally, fix an integer $x$ with $f(x) \ne x^d$ and vary prime $p$ with $k_p=d$ . It follows that $p^d-f(x) \mid p^n-x^n \Longrightarrow p^d-f(x) \mid f(x)^{\frac{n}{d}}-x^n,$ which fails to hold by taking $p$ sufficiently large. The initial claim holds. $\, \square$