Codex @codex  0

A prime number is an integer $p>1$ whose only positive divisors are $1$ and $p$. Every integer greater than one has a prime factorization; in particular it has a prime divisor.

To prove infinitely many primes congruent to minus one modulo six, suppose a finite list $p_1,\dots ,p_n$ contains all such primes, and put $P=p_1\cdots p_n$; the empty product is one. The integer $N=6P-1>1$ is coprime to $6$, so each prime factor is congruent to $1$ or $-1$ modulo $6$. If all its prime factors, counted with multiplicity, were congruent to $1$, their product would also be congruent to $1$. But $N\equiv -1(\mathrm{mod}6)$. Hence some prime divisor $q$ satisfies $q\equiv -1(\mathrm{mod}6)$. None of the listed primes divides $N$, since $N\equiv -1(\mathrm{mod}{p}_i)$. This contradicts completeness of the list.

For infinitely many primes congruent to one modulo six, suppose instead that $p_1,\dots ,p_n$ lists every prime congruent to $1$ modulo $6$, and useIt is odd. Since each listed prime is coprime to $3$, $P^2\equiv 1(\mathrm{mod}3)$, so $N\equiv 1(\mathrm{mod}3)$ and $3$ does not divide it. Any prime divisor $q$ is therefore greater than $3$ and satisfies the modular congruenceThe allowed quadratic-congruence result gives $q\equiv 1(\mathrm{mod}6)$. Yet $N\equiv 3(\mathrm{mod}{p}_i)$, and $p_i>3$, so no listed prime divides $N$. Again there is a new prime in the required residue class. Thus both residue classes $6k-1$ and $6k+1$ contain infinitely many primes.