I was working on Dirichelet's Theorem, but I can't quite get it. Perhaps I am missing something.
Dirichelet's Theorem says that for positive integers a and b, if gcd(a, b) = 1 then the airthmetic progression: a, a + b, a + 2b, a + 3b, ... will have an infinite number of primes.
I can come up with a proof using a special case; that is:
Let a = 3, and b = 4
The arithmetic progression then is 3, 3 + 4, 3 + 2*4, 3 + 3*4, ...
Proof (by contradiction)
The above implies that there exists an infinite number of primes of the form 4n + 3, say p_0 = 3, p_1, p_2, ..., p_r in increasing order.
Consider Q = 4*p_1*p_2*...*p_4 + 3
Certainly, Q > p_1 and Q = 4*integer + 3 says that Q is composite. Therefore, some prime < Q must divide Q.
Q is odd, and thus prime divisors of Q must be modd. An odd prime is of the form 4n + 1 or 4n + 3
Claim: There exists a prime of the form 4n + 3 that divides Q
**Proof (by contradiction):
Assume all primes dividing Q are of the form 4*integer + 1
Look at (4m + 1)*(4s + 1) = 4(4ms + s + m) + 1
(4m + 1)*(4s + 1) = 16ms + 4s + 4m + 1
Q has to be 4 int + 1
Going back to the other proof;
Q has to be a prime divisor, say q, of the form 4*int + 3
So q must come from the list 3 = p_0 = p_1, ..., p_r
Case 1: q = 3
3|Q and 3|3 so 3|(Q - 3)
And (Q - 3) = 4*p_1*p_2*...*p_r
Case 2: q = p_i for some i w/ 1 <= i <= r
q|Q and q|4*p_1*p_2*...*p_r so q|(Q - 4*p_1*p_2*...*p_r)
(Q - 4*p_1*p_2*...*p_r) = 3
Thus, the proof is complete.
Obviously this is much easier than the general theorem proof, which is what I have been trying to work on.