Consider the statement : `" P(n) : n^(2)-n+41` is prime." Then, which one of the following is true?

Options

A. P(5) is false but P(3) is true
B. Both P(3) and P(5) are false
C. P(3) is false but P(5) is true
D. Both P(3) and P(5) are true

Answer

To determine the truth of the statement \( P(n) : n^2 - n + 41 \) being prime, we will evaluate this expression for specific values of \( n \).

### Step 1: Evaluate \( P(3) \)

We start by substituting \( n = 3 \) into the expression:

\[ P(3) = 3^2 - 3 + 41 \]

Calculating this step-by-step:

1. Calculate \( 3^2 \):
\[ 3^2 = 9 \]

2. Substitute back into the equation:
\[ P(3) = 9 - 3 + 41 \]

3. Simplify:
\[ 9 - 3 = 6 \]
\[ 6 + 41 = 47 \]

4. Check if 47 is prime:
- 47 has no divisors other than 1 and itself, so it is prime.

### Step 2: Evaluate \( P(5) \)

Next, we substitute \( n = 5 \) into the expression:

\[ P(5) = 5^2 - 5 + 41 \]

Calculating this step-by-step:

1. Calculate \( 5^2 \):
\[ 5^2 = 25 \]

2. Substitute back into the equation:
\[ P(5) = 25 - 5 + 41 \]

3. Simplify:
\[ 25 - 5 = 20 \]
\[ 20 + 41 = 61 \]

4. Check if 61 is prime:
- 61 has no divisors other than 1 and itself, so it is also prime.

### Conclusion

Both \( P(3) \) and \( P(5) \) yield prime numbers (47 and 61 respectively). Therefore, the statement \( P(n) : n^2 - n + 41 \) is true for \( n = 3 \) and \( n = 5 \).

### Final Answer

The correct conclusion is that both \( P(3) \) and \( P(5) \) are true.

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