Question

Let $p_1$ and $p_2$ denote two arbitrary prime numbers. Which one of the following statements is correct for \emph{all values of $p_1$ and $p_2$?}

• A. $p_1+p_2$ is not a prime number.
• B. $p_1p_2$ is not a prime number.
• C. $p_1+p_2+1$ is a prime number.
• D. $p_1p_2+1$ is a prime number.

Hint:

To check "for all" statements, try to \emph{disprove} options with quick counterexamples; if you can't, attempt a short proof (like "product of two primes is always composite").

Answer 1

The Correct Option is B

Step 1: Translate "for all primes."
The statement must hold for every choice of primes \(p_1, p_2\) (they could be equal, e.g., both \(2\), or distinct).

Step 2: Test universality of each option.
Option (A): If one prime is \(2\) and the other is \(3\), then \(p_1+p_2=5\) which is prime \(\Rightarrow\) (A) is false.
Option (C): With \(p_1=2, p_2=3\), we get \(2+3+1=6\), not prime \(\Rightarrow\) (C) is false.
Option (D): Take \(p_1=3, p_2=5\): \(p_1p_2+1=16\), not prime \(\Rightarrow\) (D) is false (a single counterexample kills universality).

Step 3: Prove (B) is always true.
For primes \(p_1,p_2 > 1\), the product \(p_1p_2\) has at least the two nontrivial divisors \(p_1\) and \(p_2\), with \(1 < p_1\leq p_1p_2\) and \(1 < p_2\leq p_1p_2\). Hence \(p_1p_2\) has more than two positive divisors \(\Rightarrow\) it is composite, never prime. Therefore, (B) holds for all primes.

\[ \boxed{\text{(B) is the only universally true statement.}} \]