Peter wrote: Show that a polynomial of odd degree $2m+1$ over $\mathbb{Z}$, \[f(x)=c_{2m+1}x^{2m+1}+\cdots+c_{1}x+c_{0},\] is irreducible if there exists a prime $p$ such that \[p \not\vert c_{2m+1}, p \vert c_{m+1}, c_{m+2}, \cdots, c_{2m}, p^{2}\vert c_{0}, c_{1}, \cdots, c_{m}, \; \text{and}\; p^{3}\not\vert c_{0}.\] We can find the theorem in vol 48(1897) of Mathematische Annalen. However, I remember that it is not written in English.

Sorry for the following, but I think it's the most natural way: The polynomial is even irreducible in $\mathbb{Q}_p$, the field of $p$-adic numbers. We denote the $p$-adic (additive) valuation by $v_p$ (restricted to $\mathbb{Z}$, this is how often the prime factor $p$ occurs in the given number). Now we have the tool of Newton polygons (we already have them in $\mathbb{Q}$, but it's more natural to consider it in $\mathbb{Q}_p$). Given a polynomial $f(x)=a_n x^n + ... + a_1 x+ a_0$, one considers the set $S$ of points $(k,v_p(a_k))$. For this, one takes the "lower convex hull" (don't know the correct term), that is the smallest set of segments joining $(0, v_p(a_0))$ with $(n, v_p(a_n))$ always $\leq$ the convex hull of $S$ (thus just the lower border). There are two things to know: - the valuation $v_p$ can be extended to any algebraic extension in an unique way - if $(k,v_p(a_k))$ to $(m, v_p(a_m))$ is a segment in the polygon with slope $s$, then there are exactly $m-k$ roots of $f$ having valuation $-s$ Back to the problem: we see that the Newton polygon is the line from $(0,2)$ to $(2m+1 , 0)$, thus all roots have valuation $\frac{2}{2m+1}$. But if $f=gh$, then $v_p(g(0))=\frac{2}{2m+1} \cdot \mathrm{deg}(g)$ since $g$ has only roots of $f$ as roots. But $v_p(g(0))$ has to be an integer, so $\mathrm{deg}(g)=0, 2m+1$, giving $f$ irreducible. It's probably possible to elementarize this proof (then it would be similar to that of Eisensteins criterium).