Q. Alternate the proof for Euclid's infinite number of primes to show there are infinitely prime numbers of the form $6n-1$ where n is an integer.
suppose by contradiction there are finitely many primes, i.e. suppose $p_1 = 6(1) -1$, $p_2 = 6(2) - 1$...$p_n = 6n -1$ are all prime
If we take the product of all the primes $p_1...p_n$ we'll end up with $p_1...p_n = 6m - 1$ or $6n+1$, hence if the product is $6m - 1$ let $N_1 = p_1....p_n - 6 = 6(m-1) - 1 = 6k - 1$ By the prime factorisation theorem, there exists a prime factor q
$p_i$ does not divide $N_1$ and $q \not = p_i$ so q divides $N_1$ and is of the form $6k-1$
if $p_1...p_n = 6m + 1$ let $N_2 = p_1...p_n - 2$ and use a similar argument
is this the right approach?
edit: on second thoughts, wouldn't letting $N = 6p_1...p_n - 1$ work? As no $p_i$ divides N (as you have a remainder of -1) and any divisor of N (q), will be in the form of $6n-1$ as $p_1...p_n$ is just some integer.