GATE 2025 IN GA Q.5: solution and answer

GATE 2025 · INGA Q.5 · 1 mark · MCQ

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.

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Answer and solution

ANSWER

(B): $p_{1} p_{2}$ is not a prime number. The product of two primes always has $p_{1}$ as a divisor strictly between $1$ and itself.

Official answer: BSource: Official answer key, GATE 2025 IN (IIT Roorkee)

The statement has to hold for every pair of primes, so one counterexample is enough to rule an option out.
(A) Take $p_{1} = 2, p_{2} = 3$ . Then $p_{1} + p_{2} = 5$ , which is prime. So (A) fails.
(C) Take $p_{1} = 2, p_{2} = 3$ . Then $p_{1} + p_{2} + 1 = 6 = 2 \times 3$ , which is not prime. So (C) fails.
(D) Take $p_{1} = 3, p_{2} = 5$ . Then $p_{1} p_{2} + 1 = 16 = 2^{4}$ , which is not prime. So (D) fails.
(B) A prime has no positive divisors other than $1$ and itself. $p_{1} p_{2}$ is divisible by $p_{1}$ , and $1 < p_{1} < p_{1} p_{2}$ : the first inequality holds because $p_{1} \geq 2$ , and the second because $p_{2} \geq 2$ gives $p_{1} p_{2} \geq 2 p_{1} > p_{1}$ . So $p_{1} p_{2}$ has a divisor other than $1$ and itself, which makes it composite. This covers $p_{1} = p_{2}$ as well, since $p^{2}$ has the divisor $p$ .
So $p_{1} p_{2}$ is never prime, and (B) holds for every pair of primes.

KEY IDEA

"For all values" means one counterexample kills an option. Test small primes, including $2$ , the only even prime. Prove the option that survives from the definition: a product of two numbers each at least $2$ is always composite.

WATCH OUT FOR

Two traps here. For (A), students often think "the sum of two primes is even, so it is not prime", which forgets the prime $2$ ( $2 + 3 = 5$ ). For (D), students often pick it because $2 \times 3 + 1 = 7$ is prime, which half-remembers Euclid's argument. Euclid's argument only shows that $p_{1} p_{2} + 1$ has a prime factor outside ${p_{1}, p_{2}}$ , not that it is prime itself, and $3 \times 5 + 1 = 16$ is a counterexample.

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GATE 2025 IN GA Q.5: solution and answer

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