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

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.
