Show that any prime divisor of $x^4+x^3+x^2+x+1$, with $x\in\mathbb{N}$, is $5$ or $1$ mod $5$

Question

We can write the "polynomial" as follows: $$x^4+x^3+x^2+x+1=\frac{x^5-1}{x-1}.$$ For even $x=2y$, we have that $x^5-1=(2y)^5-1=32y^5-1\equiv1$ mod $5$.

For odd $x=2y+1$, we have that $(2y+1)^5-1\equiv_532y^5\equiv2$ mod $5$.

I find it hard to believe my computations above. I guess this makes me rather clueless. I was hoping there was some way out with Legendre symbols. Any help is appreciated.

Answer 1

Hint: Suppose that $$p\mid x^5-1$$ meaning $$x^5\equiv 1\pmod p.$$ This gives us $\text{ord}_p(x)\mid 5$. Suppose $x\not\equiv 1\pmod p$, meaning $\text{ord}_p(x)=5$. We know that for every $n$ and $x$ coprime to $n$ that $\text{ord}_n(x)\mid \phi(n)$. What does this mean in our scenario? How do we eliminate the case $x\equiv 1\pmod p$?

Answer 2

Suppose that $x^4+x^3+x^2+x+1\equiv0\bmod p$. Multiplying by $x-1$ and adding $1$ gives

$x^5\equiv1\bmod p$ Eq. 1

As $x$ is clearly nonzero $\bmod p$, Fermat's Little Theorem gives

$x^{p-1}\equiv1\bmod p$ Eq. 2

Both of the latter equations are true only if

$x^d\equiv1\bmod p; d=gcd(5,p-1)$ Eq. 3

If $p\equiv1\bmod5$ then $d=5$ and Eq. 3 is consistent with Eq. 1. Thus $p$ may he any prime $\equiv1\bmod 5$.

If $p\not\equiv1\bmod5$ then $d=1$ in Eq. 3, whereupon

$x^4+x^3+x^2+x+1\equiv5\bmod p,$

but by hypothesis this residue is also $0\bmod p$. This forces $p=5$.