Is $x^{20} +x^{15}+ x^{10}+x^5+1$ irreducible in $\mathbb{Q}[x]$?
I think if $y=x^5$, then $P(y)=y^{4} +y^{3}+ y^{2}+y+1 =\displaystyle \frac{y^5-1}{y-1}$, $5$ is prime, then $P(y)$ is irreducible in $\mathbb{Q}[y]$ or $\mathbb{Q}[x^5]$ , how do I show that it is irreducible in $\mathbb{Q}[x]$?
Let $f(x) = x^{20}+x^{15}+x^{10}+x^5+1 \in \mathbb{Z}[x]$, we have $$\begin{array}{rrcl} & (x^5-1)f(x) &=& x^{25} - 1\\ \implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1\\ \implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1 \pmod 5\\ \iff & ((x^5+1)-1)f(x+1) &=& (x^5+1)^5 - 1 \pmod 5\\ \iff & x^5 f(x+1) &=& (x^{25}+1) - 1 \pmod 5\\ \iff & f(x+1) &=& x^{20} \pmod 5 \end{array}$$ This implies aside from the leading term $x^{20}$, the coefficients of $x^{k}$ in $f(x+1)$ where $0 \le k < 20$ are all divisible by $5$. Notice the constant term in $f(x+1)$ is $f(1) = 5$ which is not divisible by $5^2$. By Eisenstein's criterion, $f(x+1)$ is irreducible over $\mathbb{Q}[x]$. As a corollary, so does $f(x)$.