Find the number of positive integers `n` such that 105 is a divisor of `n^(2)+n+1`

Explanation

To solve the problem of finding the number of positive integers \( n \) such that \( 105 \) is a divisor of \( n^2 + n + 1 \), we will follow these steps:

### Step 1: Factorize 105
First, we factor \( 105 \):
\[ 105 = 3 \times 5 \times 7 \]
This means that for \( 105 \) to divide \( n^2 + n + 1 \), \( n^2 + n + 1 \) must be divisible by \( 3 \), \( 5 \), and \( 7 \).

### Step 2: Check divisibility by 3
We will check the expression \( n^2 + n + 1 \) modulo \( 3 \):
- If \( n \equiv 0 \mod 3 \):
\[ n^2 + n + 1 \equiv 0^2 + 0 + 1 \equiv 1 \mod 3 \]
- If \( n \equiv 1 \mod 3 \):
\[ n^2 + n + 1 \equiv 1^2 + 1 + 1 \equiv 3 \equiv 0 \mod 3 \]
- If \( n \equiv 2 \mod 3 \):
\[ n^2 + n + 1 \equiv 2^2 + 2 + 1 \equiv 4 + 2 + 1 \equiv 7 \equiv 1 \mod 3 \]

Thus, \( n^2 + n + 1 \) is divisible by \( 3 \) if \( n \equiv 1 \mod 3 \).

### Step 3: Check divisibility by 5
Next, we check \( n^2 + n + 1 \) modulo \( 5 \):
- If \( n \equiv 0 \mod 5 \):
\[ n^2 + n + 1 \equiv 0^2 + 0 + 1 \equiv 1 \mod 5 \]
- If \( n \equiv 1 \mod 5 \):
\[ n^2 + n + 1 \equiv 1^2 + 1 + 1 \equiv 3 \mod 5 \]
- If \( n \equiv 2 \mod 5 \):
\[ n^2 + n + 1 \equiv 2^2 + 2 + 1 \equiv 7 \equiv 2 \mod 5 \]
- If \( n \equiv 3 \mod 5 \):
\[ n^2 + n + 1 \equiv 3^2 + 3 + 1 \equiv 13 \equiv 3 \mod 5 \]
- If \( n \equiv 4 \mod 5 \):
\[ n^2 + n + 1 \equiv 4^2 + 4 + 1 \equiv 21 \equiv 1 \mod 5 \]

Thus, \( n^2 + n + 1 \) is never divisible by \( 5 \).

### Step 4: Check divisibility by 7
Finally, we check \( n^2 + n + 1 \) modulo \( 7 \):
- If \( n \equiv 0 \mod 7 \):
\[ n^2 + n + 1 \equiv 1 \mod 7 \]
- If \( n \equiv 1 \mod 7 \):
\[ n^2 + n + 1 \equiv 3 \mod 7 \]
- If \( n \equiv 2 \mod 7 \):
\[ n^2 + n + 1 \equiv 7 \equiv 0 \mod 7 \]
- If \( n \equiv 3 \mod 7 \):
\[ n^2 + n + 1 \equiv 13 \equiv 6 \mod 7 \]
- If \( n \equiv 4 \mod 7 \):
\[ n^2 + n + 1 \equiv 21 \equiv 0 \mod 7 \]
- If \( n \equiv 5 \mod 7 \):
\[ n^2 + n + 1 \equiv 31 \equiv 3 \mod 7 \]
- If \( n \equiv 6 \mod 7 \):
\[ n^2 + n + 1 \equiv 43 \equiv 1 \mod 7 \]

Thus, \( n^2 + n + 1 \) is divisible by \( 7 \) if \( n \equiv 2 \mod 7 \) or \( n \equiv 4 \mod 7 \).

### Conclusion
Since \( n^2 + n + 1 \) is never divisible by \( 5 \), we conclude that there are no positive integers \( n \) such that \( 105 \) is a divisor of \( n^2 + n + 1 \).

### Final Answer
The number of positive integers \( n \) such that \( 105 \) is a divisor of \( n^2 + n + 1 \) is \( \boxed{0} \).