Irreducibility of $X^{p-1} + \cdots + X+1$ 
Can someone give me a hint how to the irreducibility of $X^{p-1} + \cdots + X+1$, where $p$ is a prime, in $\mathbb{Z}[X]$ ? 

Our professor gave us already one, namely to substitute $X$ with $X+1$, but I couldn't make much of that.

In the solution for this problem I will use the well-know result as $\color{blue}{\text{Eisenstein's Irreducibility Criterion}}$.

Eisenstein's Irreducibility Criterion: Let $$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+\cdots a_{0}$$ be a polynomial with coefficients $a_{n},a_{n-1},\ldots,a_{0}\in \mathbb{Z}$. Suppose that there exists a prime $p$, such that:

*

*$p\not|a_{n}$. In words: $p$ does not divide $a_{n}$.

*$p| a_{n-1},a_{n-2},\ldots,a_{1},a_{0}$. In words: $p$ divide each $a_{i}$ for $0\leq i <n$.

*$p^{2}\not| a_{0}$. In words: $p^{2}$ does not divide $a_{0}$.


Then $f(x)$ is is irreducible over the integers $\mathbb{Z}[x]$.


It is a corollary of $\color{blue}{\text{Gauss's lemma}}$ that $f(x)$ is also irreducible over the rational numbers .

Gauss's lemma: A non-constant polynomial in $\mathbb{Z}[x]$ is irreducible in $\mathbb{Z}[x]$ if and only if it is both irreducible in $\mathbb{Q}[x]$ and primitive in $\mathbb{Z}[x]$.

Now, your question is well-know, as that $\color{blue}{\text{Cyclotomic polynomials}}$ are irreducibles.

Problem: Let $p$ un prime numbe, so $$\Phi_{p}(x)=x^{p-1}+x^{p-2}+\cdots+1=\frac{x^{p}-1}{x-1}$$ is irreducible.

Proof: Note that $$\Phi_{p}(x)=\frac{x^{p}-1}{x-1}=x^{p-1}+x^{p-2}+\cdots+1$$
does not satisfy the conditions of Eisenstein's criterion, but we can see that $$\Phi_{p}(x+1)=\frac{(x+1)^{p}-1}{x}=x^{p-1}+\binom{p}{1}x^{p-2}+\binom{p}{2}x^{p-3}+\cdots+\binom{p}{p-2}x+\binom{p}{p-1}$$ does. Indeed, note that $\boxed{1}$ we can see that $p\not| 1$, in $\boxed{2}$ we can see that $p| \binom{p}{1}, \binom{p}{2},\ldots,\binom{p}{p-2},\binom{p}{p-1}$ and finally in $\boxed{3}$ we have that $p^{2}\not| \binom{p}{p-1}=p$. So, $\Phi_{p}(x+1)$ is irreducible. But, if $\Phi_{p}(x)$ factored as $f(x)g(x)$ so then $\Phi_{p}(x+1)=f(x+1)g(x+1)$. So, $\Phi_{p}(x)$ is irreducible.  $\boxed{}$


