Question 1 Is the following proof of the infinitude of primes $\equiv 1\mod 4$ okay?
Consider a prime divisor $p\mid (n!)^2+1$. Then $(n!)^2\equiv -1 \mod p$, hence $n!$ has multiplicative order $4$ in $\Bbb F_p^\times$ (question: is this conclusion true? what did I use here?).
Thus $n!\in \Bbb F_p^\times$ generates a subgroup of order $4$ and by Lagrange, $4\mid p-1$, i.e. $p\equiv 1\mod 4$.
Now we do the same thing, replacing $n$ by $p$. This will give a new prime $\equiv 1\mod 4$ and this prime is bigger than $p$ (otherwise it would divide $1$).
Question 2 My lecture notes say that from this one can conclude that there are infinitely many primes $p\equiv 3\mod 4$ by considering $n!-1$, which leaves remainder $3$ upon division by $4$. How does this imply infinitude of such primes?
Best Answer
The notes means that, for $n\geq 4$, $n!-1$ has remainder $3$ modulo $4$. It cannot have any prime factor $p\leq n$. It also must have a prime factor $p\equiv 3\pmod 4$ because if $N=p_1p_2\dots p_k$ and $p_i\equiv 1\pmod 4$ then $N\equiv 1\pmod 4$.
So given any $n$, there must be a $p$ not in $1,2,\dots,n$ such that $p\equiv 3\pmod 4$.