Find all  prime number  $p$,  such that  there exist an infinite number of  positive integer $n$  satisfying the following condition:
$p|n^{ n+1}+(n+1)^n.$

(September 29, 2012, Hohhot)
1. **Step 1: Exclude the prime number 2**

   We need to check if \( p = 2 \) satisfies the condition \( p \mid n^{n+1} + (n+1)^n \). 

   For \( p = 2 \):
   \[
   2 \mid n^{n+1} + (n+1)^n
   \]
   Consider the parity of \( n \):
   - If \( n \) is even, \( n = 2k \):
     \[
     n^{n+1} = (2k)^{2k+1} \text{ is even}
     \]
     \[
     (n+1)^n = (2k+1)^{2k} \text{ is odd}
     \]
     Thus, \( n^{n+1} + (n+1)^n \) is odd, and 2 does not divide an odd number.
   - If \( n \) is odd, \( n = 2k+1 \):
     \[
     n^{n+1} = (2k+1)^{2k+2} \text{ is odd}
     \]
     \[
     (n+1)^n = (2k+2)^{2k+1} \text{ is even}
     \]
     Thus, \( n^{n+1} + (n+1)^n \) is odd, and 2 does not divide an odd number.

   Therefore, \( p = 2 \) does not satisfy the condition.

2. **Step 2: Consider the general case for odd primes \( p \)**

   If there exists an \( n \) such that \( p \mid n^{n+1} + (n+1)^n \), then there exist infinitely many such \( n \). Consider numbers in the form \( n + kp\varphi(p) \) for all natural numbers \( k \).

3. **Step 3: Prove the existence of such \( n \) for an arbitrary odd prime \( p \)**

   Assume the order of 2 modulo \( p \) is \( d \). Let \( n \) be in the form \( pk - 2 \). We want to define \( k \) such that:
   \[
   p \mid (pk-2)^{pk-1} + (pk-1)^{pk-2}
   \]
   Simplifying, we get:
   \[
   p \mid (-2)^{pk-1} + (-1)^{pk-2}
   \]
   Since \( pk-1 \) and \( pk-2 \) have different parities:
   \[
   p \mid 2^{pk-1} - 1
   \]
   We need to choose \( k \) such that \( d \mid pk-1 \). This \( k \) obviously exists because \( (d, p) = 1 \).

   Therefore, for any odd prime \( p \), there exists an \( n \) such that \( p \mid n^{n+1} + (n+1)^n \), and hence there are infinitely many such \( n \).

\(\blacksquare\)

The final answer is all primes except \( \boxed{ 2 } \).