This is a particularly fun problem! Its on a homework that I already turned in. I used the proof by contradiction method. I just need a clarification point. I started by assuming finite number of primes of form 12k-1. suppose N = (6*P1*P2...*Pn)^2 - 3 and set the congruence (6*P1*P2..*Pn)^2 congruent to 3 (mod p) Then N = 36k-3 N must have a q such that q | N and q | (6*P1*P2..*Pn) leaving q|3, but q is of the form 12k-1, and cannot divide 3. is this correct? could someone straighten this out a bit more for me? make it more simple/concise?