Study irreducibility of $f(x) = x^{20} + 5x^{15}+25x^{10}+125x^5+625$ in $\mathbb Q[x]$

Question

I started with the idea of looking for a $b \in \mathbb Z$ where $f(x+b)$ is a polinomyal that I can use Einsestein for $p = 5$. According with $\dbinom{5}k$ are multiples of $5$ I just need the independent term dividing $5$ and not $5^2$. Lets write this term $i(b) = b^{20} +5b^{15}+25b^{10}+125b^5+625$ according to Fermats Theorem $b^5 \equiv b \ (mod 5)$ then $i(b) \equiv b^4 \ mod(5)$ but I can't find the b with this and I do not know how to continue, some advide? Thanks

Answer 1

Because $2$ is a primitive root modulo $25$, the cyclotomic polynomial $\Phi_{25}(x)$ is irreducible modulo two (or in the ring $\Bbb{Z}_2[x]$). Your polynomial reduces to $\Phi_{25}(x)$ modulo two, so it is irreducible in $\Bbb{Z}[x]$. Gauss's Lemma and friends then imply that it is irreducible in $\Bbb{Q}[x]$ as well.

Answer 2

To note, since there is a lot of material now in the comments.

If you set $5y=x^5$ you obtain $625(y^4+y^3+y^2+y+1)$ and you can proceed from there.

(Jyrki Lahtonen's approach is more general)

Answer 3

I'll present a self-contained solution without cyclotomic polynomial essentially following Jykri Lahtonen's answer. By Gauss's lemma, it suffices to prove this is irreducible over $\mathbb Z$ hence over $\mathbb F_2$.

Over $\mathbb F_2$, we have the polynomial is just $x^{20}+x^{15}+x^{10}+x^5+1=0$ which is a geometric progression. Note that this implies $\frac{1-\alpha^{25}}{1-\alpha^5}=0$ for any root $\alpha$, hence $\alpha^{25}=1$. Therefore the order of $\alpha$ in the cyclic group $(\mathbb F_2(\alpha))^{\times}$ can only be $1$ (impossible, as $1$ is not a solution to the polynomial), $5$ (impossible, as if $\alpha^5=1$, then $\alpha$ cannot satisfy the polynomial) and $25$ (the only possibility).

Therefore we have $25 \mid 2^n-1$ where $n:=[\mathbb F_2(\alpha):\mathbb F_2]$. Now we can easily check that none of $n<20$ satisfies this relation, hence $n=20$.