http://www.artofproblemsolving.com/Forum/viewtopic.php?f=36&t=395023&p=2195560&hilit=irreducible#p2195560

mavropnevma wrote: Let $p$ be a prime number,let $n_1, n_2, \ldots, n_p$ be positive integer numbers, and let $d$ be the greatest common divisor of the numbers $n_1, n_2, \ldots, n_p$. Prove that the polynomial \[\dfrac{X^{n_1} + X^{n_2} + \cdots + X^{n_p} - p}{X^d - 1}\]is irreducible in $\mathbb{Q}[X]$. Beniamin Bogosel Firstly, we recall a useful lemma. Lemma. Let $f$ be an integer polynomial with $|f(0)|$ a prime. Suppose for any root $\alpha \in \mathbb{C}$ of $f$ we have $|\alpha|>1$; then $f$ is irreducible over $\mathbb{Q}[X]$. (Proof) Assume to the contrary. By Gauss' lemma, we can pick $g, h \in \mathbb{Z}[X]$ with $\min(\text{deg} \, f, \text{deg} \, g)>0$ and $f=gh$. Now $|g(0)|>1$ and $|h(0)|>1$ are integers, but $|g(0)| \cdot |h(0)|$ is a prime; contradiction! $\blacksquare$ Now we claim that $$f(x)=\frac{\sum^p_{i=1} (x^{n_i}-1)}{x^d-1}$$does not share a common root with $x^d-1$. By L'hopital's rule, $$\lim_{x \rightarrow \alpha} f(x)=\frac{\sum^p_{i=1} n_i\alpha^{n_i}}{d \cdot \alpha^d}=\frac{n_1+\dots+n_p}{d}>0$$so that is true. Finally, if $\beta$ is a root of $f$ with $|\beta|<1$ then we have $$p=|\beta^{n_1}+\dots+\beta^{n_p}| \le |\beta|^{n_1}+\dots+|\beta|^{n_p}$$which can only occur if $\beta^d=1$. Hence the lemma applies and $f$ is irreducible.

Suppose that the numerator $f$ factors as $f = (X^d-1) \cdot g \cdot h$. Then one of $|g(0)|$ or $|h(0)|$ equals $1$, so there exists a root of $f$, say $r$, with magnitude at most $1$. On the other hand, we then have $$|r^{n_1}+r^{n_2}+\cdots+r^{n_p}| \leq p,$$with equality occurring if and only if $r^{n_i} = 1$ for each $i$, i.e. $r$ is a $d$th root of unity. On the other hand, the numerator cannot have any $\omega^d = 1$ as a double root, as its derivative $$f'(\omega) = \omega^{-1}(n_1+n_2+\cdots+n_p) \neq 0.$$Hence contradiction.

$Proof:$ $P(x) $ is primitive,so we only need to prove that $P(x)$ is irreducible in $\mathbb{Z}[X]$.Noting that any complex root $r$ we have $p=\left|r^{n_1}+r^{n^2}+…+r^{n_p}\right|\leq\left|r\right|^{n_1}+\left|r\right|^{n_2}+…+\left|r\right|^{n_p}$,which implies $\left|r\right|\geq 1$. The equal establishes which needs all the $r^{n_j}$ has the same argument.Take $r$ satisfies $\left|r\right|=1$. Noting that the sum of $r^{n_j}$ is $p\in \mathbb{R}$,so all the $r^{n_j}\in \mathbb{R}$,so $r$ is a unit root.Let $r=e^{\frac{2\pi ki}{n}}$ and $gcd(k,n)=1$,then $n|n_j$ for all $j$,so $r$ is a root of $x^d-1$.Because for $P(x)$ the multiplicity of $r$ is $1$,done.