If p be a prime number, then `p^2 + 1` can not have its unit digit is :

Options

A. 3
B. 9
C. 7
D. all of these

Answer

To solve the problem, we need to determine the possible unit digits of \( p^2 + 1 \) where \( p \) is a prime number.

### Step-by-Step Solution:

1. **Identify Possible Unit Digits of Prime Numbers**:
The prime numbers less than 10 are 2, 3, 5, and 7. The unit digits of these prime numbers are 1, 2, 3, 5, 7, and 9.

2. **Calculate the Square of Each Prime Number**:
- For \( p = 2 \):
\[ p^2 = 2^2 = 4 \quad \text{(unit digit is 4)} \]
- For \( p = 3 \):
\[ p^2 = 3^2 = 9 \quad \text{(unit digit is 9)} \]
- For \( p = 5 \):
\[ p^2 = 5^2 = 25 \quad \text{(unit digit is 5)} \]
- For \( p = 7 \):
\[ p^2 = 7^2 = 49 \quad \text{(unit digit is 9)} \]
- For \( p = 11 \):
\[ p^2 = 11^2 = 121 \quad \text{(unit digit is 1)} \]
- For \( p = 13 \):
\[ p^2 = 13^2 = 169 \quad \text{(unit digit is 9)} \]
- For \( p = 17 \):
\[ p^2 = 17^2 = 289 \quad \text{(unit digit is 9)} \]
- For \( p = 19 \):
\[ p^2 = 19^2 = 361 \quad \text{(unit digit is 1)} \]

3. **List the Possible Unit Digits of \( p^2 \)**:
From the calculations above, the possible unit digits of \( p^2 \) are:
- 0 (from 10)
- 1 (from 1, 11, 19)
- 4 (from 2)
- 5 (from 5)
- 9 (from 3, 7, 13, 17)

4. **Calculate \( p^2 + 1 \)**:
Now we add 1 to each of the possible unit digits of \( p^2 \):
- If the unit digit of \( p^2 \) is 0, then \( p^2 + 1 \) has a unit digit of 1.
- If the unit digit of \( p^2 \) is 1, then \( p^2 + 1 \) has a unit digit of 2.
- If the unit digit of \( p^2 \) is 4, then \( p^2 + 1 \) has a unit digit of 5.
- If the unit digit of \( p^2 \) is 5, then \( p^2 + 1 \) has a unit digit of 6.
- If the unit digit of \( p^2 \) is 9, then \( p^2 + 1 \) has a unit digit of 0.

5. **Determine Impossible Unit Digits**:
The unit digits that can result from \( p^2 + 1 \) are:
- 0
- 1
- 2
- 5
- 6

The unit digits that cannot be obtained are:
- 3
- 7
- 8
- 9

### Conclusion:
Thus, the unit digits that \( p^2 + 1 \) cannot have are 3, 7, and 9. Therefore, the answer is that \( p^2 + 1 \) cannot have its unit digit as 3, 7, or 9.