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Let $p$ be prime, $n \in \mathbb{Z}$. Show that if $\Phi_{5}(n) = n^4 + n^3 + n^2 + n + 1 \equiv 0 \pmod{p}$, then $p = 5$ or $p \equiv 1 \pmod{5}$.

If $p = 5$, then we are done. If $p \neq 5$, then I believe I may follow an argument here by Thomas Andrews, whose answer I do not fully understand. We notice that $n \neq 1 \pmod{p}$ (because, otherwise, $n^4 + n^3 + n^2 + n + 1 \equiv 5 \not \equiv 0 \pmod{p}$). How does one then conclude that $n^5 \equiv 1 \pmod{p}$? Also, how does one know that $5$ is the order of $n$-- and not some power between $1$ and $5$?

I can piece together the remainder of the argument: In noticing that $n \neq 0 \pmod{p}$, one can use Fermat's Little Theorem to conclude $n^{p-1} \equiv 1 \pmod{p}$. In establishing that $n$ has order $5$, we may conclude that $5|p-1$, hence that $p \equiv 1 \pmod{5}$, as desired.

J.Dmaths
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V. Elizabeth
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Note that $(n-1)(n^4+n^3+n^2+n+1) = n^5-1$. If the product is $0$ Modulo $p$ (which it will be if $n^4+n^3+n^2+n+1\equiv 0\pmod{p}$, regardless of whether $n-1\equiv 0\pmod{p}$ or not), then $n^5-1\equiv 0\pmod{p}$, so $n^5\equiv 1\pmod{p}$.

In general, if $a^k\equiv 1\pmod{p}$, then the multiplicative order of $a$ modulo $p$ divides $k$. You can verify this using the division algorithm, and remembering that the multiplicative order is the least positive integer $m$ such that $a^m\equiv 1\pmod{p}$. So if $n^5\equiv 1\pmod{p}$, then the multiplicative order of $n$ modulo $p$ divides $5$. Hence it is either $1$ or $5$, and cannot be anything in between.

But if you are also assuming that $n\not\equiv 1\pmod{p}$, then the order is not $1$, and hence must be equal to $5$.

Arturo Magidin
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