The Answer to the Question
is below this banner.
Here's the Solution to this Question
Note that if we multiply 2 numbers of the form 6k + 1 together, we get another number of the same form
If there were only finitely many primes of the form 6k + 5, say
p0 = 5 < p1 < p2 < · · · < pn.
Consider the integer
N = 6p1p2 · · · pn + 5.
Clearly N > 1 is not divisible by 2 and the note with which we began this solution implies that N has at least one prime divisor p of the form 6k + 5. If p = 5, we get that
5|N − 5 = 6p1p2 · · · pn,
while, if p > 5, we have that
5|N − 6p1p2 · · · pn = 5.
In either case, we have a contradiction.