If $p\in\mathbb{N}$ is a prime, is $x^n+px+p^2$ irreducible in $\mathbb{Z}[x]$?
I've proved that any non-unit factor in $\mathbb{Z}[x]$ must have degree at least 2.
Eisenstein's criterion doesn't hold, but perhaps we can make some minor change so that it does? (E.g. a linear substitution, as with cyclotomic polynomials??)
Or could we start from scratch, assuming $x^n+px+p^2 = (b_kx^k+\dots+b_0)(c_{n-k}x^{n-k}+\dots+c_0)$ and trying to get conditions on the $b$'s and $c$'s that give a contradiction. I've got that $b_0=\pm p$, $c_0=\pm p$, $b_1=0$, $c_1=\pm 1$, and $c_2=\pm b_2$, but can't see how this helps!
A hint given in the question says 'Consider powers of $p$ dividing coefficients.' Does this suggest some variant of Eisenstein, or an unusual way of implementing Eisenstein?
Thanks for any help with this!
As you write $$x^n+px+p^2=(b_kx^k+\dots+b_0)(c_{n-k}x^{n-k}+\dots+c_0). $$
First, we may assume $b_k=c_{n-k}=1$, and $k>0,n-k>0$. Then $\bmod p$, it will give $x^n=\text{polynomial}\times \text{polynomial }$ in $\mathbb{Z}/p[x]$. This forces that $p\mid b_i,p\mid c_j$ for $i\neq k,j\neq n-k$. Now consider the term of degree one of the original polynomial and obtain $px=(b_0c_1+c_0b_1)x$. This gives a contradiction if $n-k>1,k>1$.
Now in the case $$ x^n+px+p^2=(x^{n-1}+b_{n-2}x^{n-2}+\dots+b_0)(x+c_0) $$ we have $c_0=p,-p$, thus $p$ or $-p$ is a root of $x^n+px+p^2$, but this is not true.