pq =[3(2p 2)+ 1][3(2q 2+ 1)]=(6p 2+ 1)(6q 2+2)
= 36p2q 2+ 12p 2+6q 2+2 = 3( 12p2q 2+4p 2+2q 2)+2
And 2n-3=3k+2-3=3k- 1, which is possible.
PQ = (3p1+1) (3q1+1) Two prime numbers in this form, p 1 and p2 can only be even numbers, so there are.
pq =[3(2p 2)+ 1][3(2q 2)+]=(6p 2+ 1)(6q 2+ 1)
= 36p2q 2+6p 2+6q 2+ 1 = 3( 12p2p2q 2+2 p2+2q 2)+ 1
And 2n-3=3k+2-3=3k- 1, which is impossible.
Pq=(3p 1+2)(3q 1+2) Two prime numbers in this form, p 1 and p2 can only be odd numbers, so there are
pq =[3(2p 2+ 1)+2][3(2p 2+ 1)+2]=(6p 2+5)(6q 2+5)
= 36p2q 2+30p 2+30q 2+25 = 3( 12p2q 2+ 10p 2+ 10q 2+8)+ 1
And 2n-3=3k+2-3=3k- 1 is also impossible.
Judging from the above speculation, your guess is correct, which can be strengthened as follows: when any positive even number n is divided by 3, the remainder is 2. When n-3 can be decomposed into the product of two prime numbers (prime numbers) P and Q, pq can only be in this form: after P and Q are divided by 3, one of the residuals is 1 and the other is 2.