How many natural numbers n are there such that `n! + 10` is a perfect square?

Options

A. 1
B. 2
C. 4
D. infinitely many

Answer

To solve the problem of how many natural numbers \( n \) exist such that \( n! + 10 \) is a perfect square, we can break it down into a systematic approach.

### Step 1: Understand the Problem
We need to find natural numbers \( n \) such that \( n! + 10 = k^2 \) for some integer \( k \). This means \( n! + 10 \) must be a perfect square.

### Step 2: Analyze Small Values of \( n \)
Let's check the values of \( n \) starting from 1 up to 5, since factorials grow rapidly and will likely yield non-square results for larger \( n \).

- **For \( n = 1 \)**:
\[ 1! + 10 = 1 + 10 = 11 \quad (\text{not a perfect square}) \]

- **For \( n = 2 \)**:
\[ 2! + 10 = 2 + 10 = 12 \quad (\text{not a perfect square}) \]

- **For \( n = 3 \)**:
\[ 3! + 10 = 6 + 10 = 16 \quad (4^2 \text{ is a perfect square}) \]

- **For \( n = 4 \)**:
\[ 4! + 10 = 24 + 10 = 34 \quad (\text{not a perfect square}) \]

- **For \( n = 5 \)**:
\[ 5! + 10 = 120 + 10 = 130 \quad (\text{not a perfect square}) \]

### Step 3: Analyze Larger Values of \( n \)
Now we need to consider \( n \geq 6 \).

For \( n \geq 6 \), we note that \( n! \) is even (since \( n! \) includes the factor 2). Therefore, \( n! + 10 \) is also even.

However, we can analyze the parity of \( n! + 10 \):
- \( n! \) for \( n \geq 2 \) is even.
- Adding 10 (which is also even) keeps the sum even.

### Step 4: Check the Form of Perfect Squares
Perfect squares can be either:
- Even: \( k^2 \) where \( k \) is even.
- Odd: \( k^2 \) where \( k \) is odd.

Since \( n! + 10 \) is even for \( n \geq 6 \), we focus on even perfect squares.

### Step 5: Factorial Growth
For \( n \geq 6 \):
- \( n! \) grows very quickly, and we can observe that \( n! \) will be significantly larger than 10.
- The difference \( n! + 10 \) will not yield a perfect square because the gap between consecutive perfect squares increases as \( n \) increases.

### Conclusion
After checking all values up to 5 and analyzing values greater than 5, we find that the only solution is for \( n = 3 \).

Thus, the total number of natural numbers \( n \) such that \( n! + 10 \) is a perfect square is:

\[ \boxed{1} \]