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 prime number is an integer greater than $1$ that has exactly two positive divisors: $1$ and itself.

Let $p_1$ and $p_2$ be any two prime numbers ($p_1 \ge 2$ and $p_2 \ge 2$).

• Analysis of (B): The product $p_1 p_2$ is an integer $\ge 4$.

  • If $p_1 \neq p_2$, the positive divisors of $p_1 p_2$ are $1, p_1, p_2,$ and $p_1 p_2$ (at least four distinct divisors).

  • If $p_1 = p_2 = p$, the positive divisors of $p^2$ are $1, p,$ and $p^2$ (three distinct divisors).

  In both cases, $p_1 p_2$ has more than two distinct positive divisors, so $p_1 p_2$ is composite and never a prime number.

• Counterexamples for other options:

  • (A) If $p_1 = 2$ and $p_2 = 3$, then $p_1 + p_2 = 5$, which is prime.

  • (C) If $p_1 = 3$ and $p_2 = 5$, then $p_1 + p_2 + 1 = 9 = 3^2$, which is not prime.

  • (D) If $p_1 = 3$ and $p_2 = 5$, then $p_1 p_2 + 1 = 16 = 2^4$, which is not prime.

Therefore, statement (B) is correct for all values of $p_1$ and $p_2$.