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Let $N = 5(n!)^2 -1$ for some $n \in \mathbb{N}$. Prove there is a prime $p$ such that $p$ divides $N$ and $p \equiv -1$ mod $5$.

My work so far is to first take a prime $p$ that divides $N$. Then $5(n!)^2 \equiv 1$ mod $p$ and hence $5 \equiv ((n!)^{-1})^2$ mod $p$. So $\left( \frac{5}{p}\right) = 1$. But $\left( \frac{5}{p}\right) = \left( \frac{p}{5}\right)$ so $p \equiv 1$ mod 5 or $p \equiv 4 \equiv -1$ mod 5.

Not sure where to go from here. My goal ultimately is to show there are infinitely many primes that satisfy $p \equiv -1$ mod 5, using this $N$ as a helper. I don't know how to do this.

Thank you.

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    $\begingroup$ Hint: $N\equiv-1\pmod 5$, so not all prime factors of $N$ can be $\equiv1\pmod 5$. $\endgroup$ – Jyrki Lahtonen Apr 4 at 14:23
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    $\begingroup$ Just a nitpick, but if n=1, then N=4 which only has 2 as a divisor. Presumably for all n>2? $\endgroup$ – Chris Moorhead Apr 4 at 14:27

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