Show the progression (6k +1) (k is an integer) is closed under multiplication: Firstly I should check that I remember what this means... If it is closed when you multiply any 2 elements together you get an element that is in the set? So for this I thought just show (6k+1)(6n+1), where k and n are integers, is of the form (6k+1) but I don't think that works and can't think of any other ways to do it. Show that the progression (6k+5) contains a prime. Then show it contains infinitely many primes: Show that it contains a prime 6(1)+5=11 (is that part really that simple :S?) As for showing it has infinitely many.... Any help with this would be greatly appreciated. I'm actually on exchange at Lund Uni in Sweden (I come from Melbourne Uni, Australia) and I'll be stuffed if I can understand my lecturer (on Number Theory) so until I can get my hands on a copy of the text book this stuff is doing my head in!