Question

Let \( p_1 \) and \( p_2 \) denote two arbitrary prime numbers. Which one of the following statements is correct for all values of \( p_1 \) and \( p_2 \)?

• A. \( p_1 + p_2 \) is not a prime number.
• B. \( p_1 p_2 \) is not a prime number.
• C. \( p_1 + p_2 + 1 \) is a prime number.
• D. \( p_1 p_2 + 1 \) is a prime number.

Hint:

Product of two primes is always composite because it has at least three divisors: 1, each prime, and their product.

Answer 1

The Correct Option is B

Step 1: Understanding the question.
We are given two arbitrary prime numbers \( p_1 \) and \( p_2 \). We must find a statement that is true for all possible choices of primes.
Step 2: Recall definition of prime numbers.
A prime number has exactly two divisors: 1 and itself. Any number that has more than two divisors is composite.
Step 3: Analyze each option.
(A) \( p_1 + p_2 \) is not a prime number.
Counterexample: If \( p_1 = 2 \), \( p_2 = 3 \), then \[ p_1 + p_2 = 5 \] which is prime. Hence, this statement is not always true.
(B) \( p_1 p_2 \) is not a prime number.
Since both \( p_1 \) and \( p_2 \) are primes and greater than 1, their product has at least three divisors: \[ 1,\; p_1,\; p_2,\; p_1 p_2 \] Therefore, \( p_1 p_2 \) is always composite, never prime. This statement is true for all primes.
(C) \( p_1 + p_2 + 1 \) is a prime number.
Counterexample: If \( p_1 = 2 \), \( p_2 = 3 \): \[ 2 + 3 + 1 = 6 \] which is composite. Hence, this is false.
(D) \( p_1 p_2 + 1 \) is a prime number.
Counterexample: If \( p_1 = 3 \), \( p_2 = 5 \): \[ 3 \times 5 + 1 = 16 \] which is composite. Hence, this is false.
Step 4: Conclusion.
The only statement that is always correct for any two primes is: \[ \boxed{p_1 p_2 \text{ is not a prime number}} \]