China Team Selection Test

We are asked to find all positive integers $m$ such that there exists a prime number $p$ for which $n^m - m$ is not divisible by $p$ for any integer $n$.

We claim that the answer is all $m \neq 1$.

First, consider $m = 1$. In this case, the expression becomes $n - 1$, which can clearly be a multiple of any prime $p$ by choosing $n \equiv 1 \pmod{p}$.

Now, consider $m > 1$. Let $p$ be an arbitrary prime factor of $m$. Write $m = p^k l$, where $\gcd(l, p) = 1$. Assume that no prime $q$ exists such that $n^m - m \equiv 0 \pmod{q}$ has no solution for $n$.

Consider the expression $(p^k l)^{p-1} + (p^k l)^{p-2} + \cdots + p^k l + 1$. Since the left-hand side is not congruent to $1 \pmod{p^{k+1}}$, we can choose $q$ such that $q \not\equiv 1 \pmod{p^{k+1}}$. We will show that this $q$ leads to a contradiction.

First, note that the remainder of $m^{p-1} + m^{p-2} + \cdots + 1$ when divided by $m - 1$ is $p$, which is relatively prime to $m - 1 = p^k l - 1$. Thus, $\gcd(q, m - 1) = 1$, so $m \not\equiv 1 \pmod{q}$.

Since $n^m \equiv m \pmod{q}$, we have $n^{p^k l} \equiv p^k l \pmod{q}$, so $n^{p^{k+1} l} \equiv (p^k l)^p \equiv 1 \pmod{q}$. Let the order of $n \pmod{q}$ be $x$. This means $x \mid p^{k+1} l$. However, since $n^{p^k l} \equiv p^k l \pmod{q}$, which is not congruent to $1 \pmod{q}$, we have that $x$ is not a factor of $p^k l$, so $p^{k+1} \mid x \mid (q - 1)$, implying $q \equiv 1 \pmod{p^{k+1}}$, which is a contradiction.

Thus, there exists a prime $q$ such that $q$ is not a factor of $n^m - m$ for all integers $n$.

The answer is: $\boxed{m \neq}$ 1}.