n3pq – n = 0 (mod 3)
n3pq – n = 0 (mod p)
n3pq – 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