Determine all primes $p$ such that there exists an integer $x$ satisfying $x^{2010}+x^{2009}+$ $\cdots+1 \equiv p^{2010}\left(\bmod p^{2011}\right)$.
The answer is all $p \equiv 1(\bmod 2011)$.

First, note that if $x$ satisfies the problem condition, then we in particular have $\Phi_{2011}(x) \equiv 0(\bmod p)$ which implies $p \equiv 1(\bmod 2011)$.

Conversely, suppose $p \equiv 1(\bmod 2011)$ and fix an $a(\bmod p)$ with order 2011 (hence $a \not \equiv 1(\bmod p))$. Let $A=\left\{x\left(\bmod p^{2011}\right) \mid x \equiv a(\bmod p)\right\}$ and $B=\left\{x\left(\bmod p^{2011}\right) \mid x \equiv 0(\bmod p)\right\}$ We claim $\Phi_{2011}: A \rightarrow B$ is injective: indeed if $\Phi_{2011}(x) \equiv \Phi_{2011}(y)\left(\bmod p^{2011}\right)$ then

$$
\left(x^{2011}-1\right)(y-1) \equiv\left(y^{2011}-1\right)(x-1) \quad\left(\bmod p^{2011}\right)
$$

hence

$$
(x-y)\left(x y \sum_{k} x^{2010-k} y^{k}-\sum_{k} x^{2011-k} y^{k}+1\right) \equiv 0 \quad\left(\bmod p^{2011}\right) .
$$

The sum in parentheses is nonzero modulo $p$ because it equals

$$
\begin{aligned}
a^{2} \cdot 2011 a^{2010}-2012 a^{2011}+1 & =2011 a^{2012}-2012 a^{2011}+1 \\
& \equiv 2011 a-2012+1 \quad(\bmod p) \\
& =2011(a-1) \\
& \not \equiv 0 \quad(\bmod p) .
\end{aligned}
$$

This proves $\Phi_{2011}: A \rightarrow B$ is injective; but now $|A|=|B|$ implies surjectivity too. Since $p^{2010} \in B$, this completes the proof.