Let p, q be prime numbers such that n3pq – n is a multiple of 3pq fo

1.	Let p, q be prime numbers such that n3pq – n is a multiple of 3pq for all positive integers n. Find the least possible value of p + q.
Answer» n^3pq – n = 0 (mod 3) n^3pq – n = 0 (mod p) n^3pq – n = 0 (mod q) ∴ We need to fulfill following conditions : (i) (3 – 1) \| (pq – 1) ⇒ pq is odd (ii) (p – 1) \| (3q – 1) Now 3 must not divide (p – 1) as it doesn’t divide (3q – 1) ∴ p – 1 = 3k + 1 or 3k + 2, for some integer k. ∴ ⇒ p = 3k + 2 or 3k + 3 But p ≠ 3k + 3 (as its prime) ∴ p = 3k + 2 clearly p > 3 and k = odd = 2λ + 1 (say) ⇒ p = 6λ + 5 (ii) (q – 1) \| (3p – 1) so q will also be 5 (mod 6) by trial least values for p & q are 17 and 11. ∴ p + q = 28

Let p, q be prime numbers such that n3pq – n is a multiple of 3pq for all positive integers n. Find the least possible value of p + q.

Answer»

n^3pq – n = 0 (mod 3)

n^3pq – n = 0 (mod p)

n^3pq – n = 0 (mod q)

∴ We need to fulfill following conditions :

(i) (3 – 1) | (pq – 1) ⇒ pq is odd

(ii) (p – 1) | (3q – 1)

Now 3 must not divide (p – 1) as it doesn’t divide (3q – 1)

∴ p – 1 = 3k + 1 or 3k + 2, for some integer k.

∴ ⇒ p = 3k + 2 or 3k + 3

But p ≠ 3k + 3 (as its prime)

∴ p = 3k + 2 clearly p > 3 and k = odd = 2λ +

1 (say)

⇒ p = 6λ + 5

(ii) (q – 1) | (3p – 1) so q will also be 5 (mod 6)

by trial least values for p & q are 17 and 11.

∴ p + q = 28