Exercise 3.1

Proof. Let $n$ any integer such that $n\ge 3$, and $N=n!-1=2\times 3\times \cdots \times n-1>1$.

Then $N\equiv -1(\mathit{mod}6)$. As $6k+2,6k+3,6k+4$ are composite for all integers $k$, every prime factor of $N$ is congruent to $1$ or $-1$ modulo $6$. If every prime factor of $N$ was congruent to 1, then $N\equiv 1(\mathit{mod}6)$ : this is a contradiction because $-1\not\equiv 1(\mathit{mod}6)$. So there exists a prime factor $p$ of $N$ such that $p\equiv -1(\mathit{mod}6)$.

If $p\le n$, then $p\mid n!$, and $p\mid N=n!-1$, so $p\mid 1$. As $p$ is prime, this is a contradiction, so $p>n$.

Conclusion :

for any integer $n$, there exists a prime $p>n$ such that $p\equiv -1(\mathit{mod}6)$ : there are infinitely many primes congruent to $-1$ modulo $6$. □