So the prof said in class that the proof of this is hard, but we might want to attempt at home. I won't be able to see him again until Wednesday, but I'm guessing there is some hole in my proof, since it didn't take me very long. Tough I can't tell what the problem is.
I hope it's not sloppy to read. I've never written a long proof with the intent of it being read before.
THEOREM
For all integers $ a, b, m $ and $ GCD(a, b) = 1 $, there are infinitely many primes of the form $ am + b $
PROOF
(Let $a, b, c, n_i $ be integers)
By contradiction, let us assume our theorem to be false.
Then it follows that there exist two integers $ a, b $, where $GCD(a, b) = 1 $, such that, for all $ m $,
$ \displaystyle \frac{am + b}{c} = n_1 $
That is, $ c | am + b $.
For some $ c $ and some $ n_1 $. ( The value of $ c $ and $ n_1 $ changing as the value of $m$ changes.)
Since $a$ and $b$ have no factors in common, it follows that:
$ \displaystyle \frac{m}{c} + \frac{b}{c} = n_2 $
(That is, that $ GCD(m, b) > 1 $ in order for Equality 1 to hold, and thus, both are divisible by some $c$.)
But this contradicts (1.) in the following way:
It is clear that, for some set of primes $ \{ p_{b1}...p_{bk} \} $, $ b = p_{b1} p_{b2} ... c ... p_{bk} $.
But suppose an $m > b$ such that
$ m = p_{m1} p_{m2} ... p_{mq}$, and
$ \{p_{b1}...p_{bk}\} \cap \{p_{m1}...p_{mq}\} = \emptyset $
For example: $ m = p_{bk+1} $
Clearly, not all $ m $'s have a common factor with $b$.
From this it follows that there must be at least one integer $ m $ for which $ am + b $ is prime.
Proving there are infinitely many more is possible once this single case is found (this part we did in class, so I won't write it).