Infinitely many primes of the form

Number theory
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tziaxri
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Joined: Mon Nov 09, 2015 4:32 pm

Infinitely many primes of the form

#1

Post by tziaxri »

Prove that there are infinitely many primes of the form: \[3k + 2\].
Tsakanikas Nickos
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Joined: Tue Nov 10, 2015 8:25 pm

Re: Infinitely many primes of the form

#2

Post by Tsakanikas Nickos »

- We will first show that every natural number of the form \( \displaystyle 3k+2 \, , \, k \in \mathbb{N}\cup\{0\} \) has a prime divisor of the same form.

Let \( \alpha \) be a natural number of the form \( \displaystyle 3k+2 \) for some \( k \in \mathbb{N}\cup\{0\} \), and let \( \displaystyle p \) be a prime divisor of \( \alpha \) (which exists,since \( \alpha > 1 \)). Then \( \displaystyle p \) has one of the following forms:
\(\text{(1) } \displaystyle p=3m \)
\(\text{(2) } \displaystyle p=3m+1 \)
\(\text{(3) } \displaystyle p=3m+2 \)
Since \( \displaystyle p \) is prime, we can discard case (1). Now, let us assume that all prime divisors of \( \alpha \) are of the form (2). If \( \displaystyle \alpha = p_{1}^{\alpha_{1}} \dots p_{r}^{\alpha_{r}} \) is the canonical representation of \( \alpha \), then
\begin{align*}
\displaystyle
\alpha &= (3m_{1}+1)^{\alpha_{1}} \dots (3m_{r}+1)^{\alpha_{r}} \\
&= (3l_{1}+1) \dots (3l_{r}+1) \\
&= 3M+1 \, ,
\end{align*}
which is a contradiction because of the form of \( \alpha \). Therefore, \( \alpha \) has a prime divisor of the form (3).

- We can now prove the assertion.

Suppose that there exist only finitely many primes of the given form and denote them by \( \displaystyle p_{1}, \dots , p_{n} \text{, where } n \in \mathbb{N} \). Set \( \displaystyle \Xi=3 p_{1} \dots p_{n}-1 \) and observe that
\[ \displaystyle \Xi=3 p_{1} \dots p_{n}-1 = 3 p_{1} \dots p_{n} -3 +2 = 3( p_{1} \dots p_{n} - 1) +2 = 3Q+2 . \]
Hence, \( \displaystyle \Xi \) has a prime divisor \( \displaystyle q \) of the same form. By hyphothesis, \( \displaystyle q \in \{ p_{1}, \dots, p_{n} \} \), so \( \displaystyle \exists s \in \{ 1,\dots,n \} : q=p_{s}. \) Obviously, \( \displaystyle q | 3 p_{1} \dots p_{n} = ( \Xi +1). \text{ Since } q|\Xi \, , \) we conclude that \( \displaystyle q|1 \) , which is a contradiction. Therefore, the primes of the form \( \displaystyle 3k+2 \) are infinite.
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