Let $p$ be a prime number and $n$ an integer $\geq 2$. We set $P(X) = X^n + X + p.$
Problem
a) We assume $p \neq 2$ or $n$ even. Show that $P$ is irreducible over $\mathbb{Q}$.
b) We assume $p = 2$ and $n$ odd. Show that we have $P (X) = (X + 1) Q (X)$ with $Q$ irreducible on $\mathbb{Q}$.
My effort:
a) $P\in\mathbb{Z}[X]$ is primitive then $P$ irreducible in $\mathbb{Z}[X] \Leftrightarrow$ P irreducible in $\mathbb{Q}[X]$.
If $p \neq 2$
If we had a decomposition in $\mathbb Z[X]$ $P(X)=P_1(X)\cdot P_2(X)=(X^{n_1}+\dots+a)\cdot(X^{n_2}+\dots+b)$ we would deduce $a\cdot b=p$ so without losing generality, we can suppose that $a=\pm 1$. This means that for the complex zeros $z_i$ of $P_{1}(x)$ we would have $\prod_{i=1}^{n_1}|z_i|=1$, so that at least one of those zeros, say $z$, must satisfy $|z|\leq 1$.
But then, since $z$ is also a zero of polynomial $P$: $z^n+z+p=0$. This is clearly impossible since $p=|z^n+z|\leq |z|^n+|z|\leq 2$. This contradiction shows that $X^n + X + p$ is actually irreducible.
If $n$ even
I don't have an idea!!
b) $P (X) = X^n + X+2 $ Then $-1$ is zeros of $P$, so we have $P (X) = (X + 1) Q (X)$ with $Q=\sum_{k=1}^{n-1}{(-1)^kX^k +2}$. But I can't show that $Q$ is irreducible on $\mathbb{Q}$.
An idea please.