Show that there are infinitely many primes which are $\pm 1 \mod 5$ 
Show that there are infinitely many primes which are $\pm 1 \mod 5$.
HINT: Suppose that there are finitely many such primes, and let these primes be $q_i$ for $1 \leq i \leq n$. Let
$$N = \prod_{i = 1}^nq_i$$
and let $p$ be any primes dividing $N^2 - 5$. Show that $p \equiv \pm 1 \mod 5$, using question $3$ or otherwise, and that $p \neq q_i$ for any $1 \leq i \leq n$.

Question $3$ was show that (Legendre Symbol) $\left( \frac{5}{p} \right)= 1$ if and only if $p \equiv \pm 1 \pmod 5$ (which I have done).
What I have done for this question is that I said
$$p \mid N^2 - 5 \implies N^2 - 5 \equiv 0 \pmod p \implies N^2 \equiv 5 \pmod p.$$
Now I'm stuck on what to do as $5$ isn't a square number and so $N^2 \equiv 5 \mod p$ isn't a quadratic residue and so I can't use question $3$ to help.
Can someone help show that there are infintely many primes which are $\pm 1 \mod 5$ please.

Let $p$ be a prime dividing $N^{2} - 5$. Thus $N^{2} \equiv 5 \pmod{p}$, so by the very definition $5$ is a quadratic residue modulo $p$, so $p \equiv \pm 1 \pmod{5}$.
If $p = q_{i}$ for some $i$, then $p$ divides $N$, and since $p$ divides $N^2 - 5$, we have that $p$ divides $5$, a contradiction.
