# Verify that there are not infintely many primes of the form 5k + 4 [closed]

How does one verify that there are not infinitely many primes of the form 5k + 4 where k $\in \mathbb{Z}$ without using Dirichlet's theorem? The question is, this congruence class isn't in the set of arithmetic-progressions, so how does one verify that this doesn't hold.

## closed as off-topic by Jack D'Aurizio, Lord Shark the Unknown, Xander Henderson, I am Back, Matthew ConroyOct 14 '17 at 17:59

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• There are. ${}$ – punctured dusk Oct 14 '17 at 14:15

To prove that there are infinitely many primes of the form $5n+4$, note that each number of the form $20N^2-1$, with $N\in\Bbb N$ has a prime factor of that form.

• How do you get $20N^{2} - 1$ from $5n+4?$ – Kagamine Len Oct 14 '17 at 14:46
• @KagamineLen Quadratic reciprocity. – Lord Shark the Unknown Oct 14 '17 at 14:49
• How do I get that with quadratic reciprocity? – Kagamine Len Oct 14 '17 at 14:54
• @KagamineLen For any prime factor $p$ of $20N^2-1$ we get that $5$ is a quadratic residue modulo $p$ (the multiplicative inverse of $2N$ is a square root), hence, by QR, $p\equiv 1,4\pmod 5$. Now note that some prime factor of this number must not be congruent to $1\pmod 5$, since $20N^2-1\not\equiv 1\pmod 5$. Now take $N$ to be the product of all primes $4\pmod 5$ after supposing there are finitely many of those. – Wojowu Oct 14 '17 at 15:13

We may write:

$$p= 5 k +4 = 5( k +1) - 1$$ It can be seen that if (k+1) is even then p may be a prime. The even values of n+1 make an arithmetic progression with 4 as first term and common differences 2 , 4 and 6 alternatively as follows: $$4, 6, 12, 16, 22, 26, 30 . . .$$ This is an infinite series of numbers, hence we can conclude that there can be infinitely many primes of form 5 k + 4