1. ## Question #2: Prime

Adapt the proof of Theorem 1.5.9 to prove:

Theorem 1.5.9 - There is an infinit number of primes.

(a) There is an infinite number of primes of the form 4n + 3.

(b) There is an infinite number of primes of the form 6n + 5.

2. ## Dirichlet's theorem on arithmetic progressions

Use Dirichlet's theorem on arithmetic progressions.
This website should tell you all you need to know

Dirichlet's theorem on arithmetic progressions - Wikipedia, the free encyclopedia

3. Thanks, that helped alot. Another question. Is there another way to show a number is prime other than just being divisible by another number.

4. yes there is, using modular arithmetic

Wilson's Theorem states that (p-1)! = -1 mod p if and only if p i prime

also using fermat's little theorem if a^(n-1) = 1 mod n, then n is said to have a strong probability of being prime.

5. Can I have an example using the number 301.

6. Using Dirichlet's theorem is complete overkill here.

Originally Posted by Mel
(a) There is an infinite number of primes of the form 4n + 3.
Assume there are fininitely many prime of that form $\displaystyle p_1,...,p_n$.
Let $\displaystyle N = 4(p_1\cdot .... \cdot p_n) - 1$.
This odd number can be factored into prime divisors.
It cannot be that each prime divisor has form $\displaystyle 4k+1$ because the product of numbers of the form $\displaystyle 4k+1$ still has the form $\displaystyle 4k+1$. Therefore, there must be a prime divisor of the form $\displaystyle 4k+3$. Thus, $\displaystyle N$ be divisible by one of $\displaystyle p_1,...,p_n$. But that implies $\displaystyle p_i | 1$ for $\displaystyle 1\leq i \leq n$. And this is impossible.

(b) There is an infinite number of primes of the form 6n + 5.
I think this is similar to the one above. Try modifing the proof.