# Is $x^n+px+p^2$ irreducible in $\mathbb{Z}[x]$?

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 $x^n+px+p^2 = (b_kx^k+...+b_0)(c_{n-k}x^{n-k}+...+c_0)$ 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!

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HINT: Consider that is factorizable as $fg$, the reduce mod p, and you will have $\bar{f}\bar{g}=x^n$, which by unique factorization in $\mathbb{F}_p[x]$ gives that all coefficients except the leading ones are multiple of $p$. Use that fact for obtaining a contradiction when considereng the terms of dregrees zero and one of those polynomials. –  Josué Tonelli-Cueto Apr 28 '12 at 10:55
Example 2.2.4 from the book Prasolov: Polynomials gives a solution of this problem using Newton diagram. (However, I do not know this technique and, based on the hint you wrote, this is obviously not the solution you're after.) –  Martin Sleziak Apr 28 '12 at 11:13

As you write $$x^n+px+p^2=(b_kx^k+\ldots+b_0)(c_{n-k}x^{n-k}+\ldots+c_0)$$
First, we may assume $b_k=c_{n-k}=1$, and $k>0,n-k>0$. Then $\mod p$, it will give $x^n=polynomial\times polyonoimal$ 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 the degree one of the original polynomial, we obtain $px=(b_0c_1+c_0b_1)x$, this gives a contradiction if $n-k>1,k>1$.
Now if in the case $$x^n+px+p^2=(x^{n-1}+b_{n-2}x^{n-2}+\ldots+b_0)(x+c_0)$$ We have $c_0=p,-p$, thus $p$ or $-p$ is a root of $x^n+px+p^2$, but this is not true.
A little plug for newton polygons: Looking at the newton polygon of the polynomial, we see that over $\mathbb{Q}_p,$ the polynomial is either irreducible or factors as a product of a degree $n-1$ eisenstien polynomial times a linear factor with a root of valuation 1. So if the polynomial factors over $\mathbb{Z}$ a linear factor with a root of valuation 1 must appear. Now evaluate at $p$ and $-p$ to obtain the result. –  jspecter Apr 28 '12 at 11:59