$\left| (4 \mathbb{N} -1) \cap \mathbb{P} \right| \ = \ \infty$ where $\mathbb{P}$ is the set of prime numbers.

Question

I try to show that there are infinitely many prime numbers in the set $ \{ 4n-1 \ : \ n \in \mathbb{N} \}$. I've been told that I needed to adjust Euclid's proof a bit so that it would work for numbers of this kind, so I did, but without finding something useful.

My attempts

First, I supposed that there are there are only finitely many prime numbers of the shape $4n-1$, lets say $S \ = \ \{n_1,n_2, \cdots n_t \}$. Then I took a look at this, hoping that it would be prime. $$ \prod_{1 \leq j \leq t}(4n_j -1)+1 $$ But, if $n$ is odd, the ones cancel out and $2$ would divide everything if I'm not mistaking. If $n$ is even, I guess everything would be even as well.

It seems that I misunderstood the hint. Could you give me a sketch about how this really works? You can make use of rings and groups if you need to, and you can use easy analytic stuff, but please don't use very advanced knowledge.

Have a look at Help understand the proof of infinitely many primes of the form $4n+3$, which may well answer your Question. — hardmath, Sep 02 '14 at 06:58

score 4 · Answer 1 · answered Sep 02 '14 at 07:00

4

Suppose $p_1,\cdots,p_n$ are all of them.

Then consider $4p_1\cdots p_n-1$

answered Sep 02 '14 at 07:00

voldemort

13,182

$\left| (4 \mathbb{N} -1) \cap \mathbb{P} \right| \ = \ \infty$ where $\mathbb{P}$ is the set of prime numbers.

1 Answers1