There are infinite primes of the form 3k+2

Let's assume there are finite primes of the form \(3k+2\). Let \(p_0 = 2\) and let \(N\) be:

$$N = 3(p_1 p_2 p_3 \ldots p_r) + 2$$

The set of primes \(\{ p_1, p_2, p_3, \ldots, p_r \}\) are all the finite primes of the form \(3k+2\) (other than 2). \(N\) itself cannot be prime because it's of the form \(3k+2\) and is greater than \(p_r\). Now let's think about the prime factors of \(N\). Since multiplying two numbers of the form \(3k+1\) gives another number of the form \(3k+1\), then all prime factors of \(N\) cannot be of the form \(3k+1\), at least one has to be of the form \(3k+2\).

Let \(p\) be a prime factor of \(N\) of the form \(3k+2\). If \(p=2\), then:

$$2|N \implies 2|3(p_1 p_2 p_3 \ldots p_r) $$

But 2 cannot divide 3 or any prime \(p_i\), so \(p\) cannot be 2. If instead \(p\) is one of \(\{ p_1, p_2, p_3, \ldots, p_r \}\), then \(p|3(p_1 p_2 p_3 \ldots p_r)\), but if \(p\) divides both \(N\) and \(3(p_1 p_2 p_3 \ldots p_r)\), then:

$$p|N-3(p_1 p_2 p_3 \ldots p_r) \implies p|2 $$

This is a contradiction, so \(p\) can't be one of \(\{ p_1, p_2, p_3, \ldots, p_r \}\), which mean's there exist primes (like \(p\)) of the form \(3k+2\) that don't belong to \(\{ p_1, p_2, p_3, \ldots, p_r \}\). This contradicts our assumption that there are finite primes of the form \(3k+2\).

Let's assume there are finite primes of the form \(3k+2\). Let \(p_0 = 2\) and let \(N\) be:

$$N = 3(p_1 p_2 p_3 \ldots p_r) + 2$$

The set of primes \(\{ p_1, p_2, p_3, \ldots, p_r \}\) are all the finite primes of the form \(3k+2\) (other than 2). \(N\) itself cannot be prime because it's of the form \(3k+2\) and is greater than \(p_r\). Now let's think about the prime factors of \(N\). Since multiplying two numbers of the form \(3k+1\) gives another number of the form \(3k+1\), then all prime factors of \(N\) cannot be of the form \(3k+1\), at least one has to be of the form \(3k+2\).

Let \(p\) be a prime factor of \(N\) of the form \(3k+2\). If \(p=2\), then:

$$2|N \implies 2|3(p_1 p_2 p_3 \ldots p_r) $$

But 2 cannot divide 3 or any prime \(p_i\), so \(p\) cannot be 2. If instead \(p\) is one of \(\{ p_1, p_2, p_3, \ldots, p_r \}\), then \(p|3(p_1 p_2 p_3 \ldots p_r)\), but if \(p\) divides both \(N\) and \(3(p_1 p_2 p_3 \ldots p_r)\), then:

$$p|N-3(p_1 p_2 p_3 \ldots p_r) \implies p|2 $$

This is a contradiction, so \(p\) can't be one of \(\{ p_1, p_2, p_3, \ldots, p_r \}\), which mean's there exist primes (like \(p\)) of the form \(3k+2\) that don't belong to \(\{ p_1, p_2, p_3, \ldots, p_r \}\). This contradicts our assumption that there are finite primes of the form \(3k+2\).

There Are Infinite Primes Of The Form 3k+2