Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$

Question

Let $x \in \mathbb{N}$ and let $p$ be a prime divisor of $x^4+x^3+x^2+x+1$, prove that $p = 5$ or $p \equiv 1 \mod{5}$

Note that $(x^4+x^3+x^2+x+1)(x-1) = x^5 - 1$.

I tried reducing the equation $\mod{5}$, which gave me some information on $x$, but I was not able to utilize this information.

Answer 1

If $r= ord _p(x)$ then by Fermat $r\mid p-1$ and by proposition $r\mid 5$.

If $r=5$ then $p-1 =5k$ so $p\equiv_5 1$

If $r=1$ then $x=1$ so $p\mid 5$ and thus $p=5$.

Answer 2

• Notice that $x^4+x^3+x^2+x+1 = \frac{x^5-1}{x-1}$. Now, $\frac{x^5-1}{x-1} \equiv 0 \pmod p \implies x^5\equiv 1 \pmod p$. It follows that the order of $x$, call it $r$, $\pmod p$ divides $5$. But $5$ is a prime number, hence $r=1,5$.

• We know that The order of any number $\pmod p$ divides $\phi(p)$. If the order is $5$, then $5|(\phi(p)=p-1) \implies p\equiv 1 \pmod 5$. If otherwise, i.e. the order is $1$, then $x \equiv 1 \pmod p$. Therefore, $(x^1)^5-1 \equiv 0 \pmod 5$ implying that $5$ divides $x^5-1$.