Q17. Let both p and k be prime numbers such that (p² + k) is also a prime number less than 30. What is the number of possible values of k?

We are given that p and k are prime numbers, and (p² + k) is also a prime number less than 30. The prime numbers less than 30 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. Since p² + k < 30, p² must be less than 30.

Possible prime values for p are 2, 3, and 5. Case 1: p = 2 If p = 2, then p² = 4. We need 4 + k < 30, so k < 26. Possible prime values for k (less than 26) are 2, 3, 5, 7, 11, 13, 17, 19, 23.

Let's check (4 + k) for primality: - 4 + 2 = 6 (not prime) - 4 + 3 = 7 (prime) -> k = 3 is a solution - 4 + 5 = 9 (not prime) - 4 + 7 = 11 (prime) -> k = 7 is a solution - 4 + 11 = 15 (not prime) - 4 + 13 = 17 (prime) -> k = 13 is a solution - 4 + 17 = 21 (not prime) - 4 + 19 = 23 (prime) -> k = 19 is a solution - 4 + 23 = 27 (not prime) For p = 2, there are 4 possible values for k: {3, 7, 13, 19}. Case 2: p = 3 If p = 3, then p² = 9.

We need 9 + k < 30, so k < 21. Possible prime values for k (less than 21) are 2, 3, 5, 7, 11, 13, 17, 19. Let's check (9 + k) for primality: - If k = 2, 9 + 2 = 11 (prime) -> k = 2 is a solution - If k is any other prime (which would be an odd number), then 9 + k will be an even number greater than 2 (odd + odd = even), and thus not prime.

So, for p = 3, k = 2 is the only solution. For p = 3, there is 1 possible value for k: {2}. Case 3: p = 5 If p = 5, then p² = 25. We need 25 + k < 30, so k < 5. Possible prime values for k (less than 5) are 2, 3.

Let's check (25 + k) for primality: - 25 + 2 = 27 (not prime) - 25 + 3 = 28 (not prime) For p = 5, there are no possible values for k. If p is any prime greater than or equal to 7, p² would be 49 or greater, making p² + k > 30.

So, no further cases are possible. Total number of possible values for k = 4 (from p=2) + 1 (from p=3) = 5. This question tests the understanding of prime numbers and systematic case analysis, which are important for number theory problems in UPSC CSAT.