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

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)
