= Solution
A <prime number> is an integer greater than one whose only positive divisors are one and itself. In particular, one is not a <prime number>. Every integer greater than one has a <prime number> divisor, by taking its smallest divisor greater than one.
Suppose there were only finitely many <prime numbers> congruent to $-1$ modulo six, and let their product be $P$, with empty product one. The integer $N=6P-1>1$ has no divisor two or three and is divisible by none of the listed <prime numbers>, since $N\equiv-1\pmod{p_j}$. Every <prime number> divisor of $N$ is therefore $1$ or $-1$ modulo six. If all these divisors were $1$ modulo six, their product, with multiplicities, would also be $1$ modulo six. But $N\equiv-1\pmod6$. Hence at least one divisor is an unlisted <prime number> congruent to $-1$ modulo six, a contradiction. \b[There are infinitely many primes of the form $6k-1$.]
For the other class, suppose its <prime numbers> were a finite list with product $P$. None is divisible by three. The integer $M=4P^2+3$ is odd and is $1$ modulo three, since $P^2\equiv1\pmod3$. Thus a <prime number> divisor $q$ of $M$ satisfies $q>3$. The residue $x=2P$ solves $x^2\equiv-3\pmod q$, so the allowed quadratic-congruence fact gives $q\equiv1\pmod6$. However, for every listed divisor $p_j$, $M\equiv3\pmod{p_j}$, and $p_j>3$, so none divides $M$. We have again found an unlisted <prime number> in the supposedly complete list. \b[There are infinitely many primes of the form $6k+1$.] Both arguments remain valid for an empty starting list.
Back to article page
