October 18th
Today I learned how to generate an irreducible polynomial which can't be proven with the standard tricks behind Eisenstein's Criterion. We begin by creating a field extension $K/\QQ$ such that $K$ has no totally ramified primes.
We claim that $K=\QQ(\sqrt2,\sqrt5)$ does the trick. Indeed, fix a rational prime $p\in\ZZ,$ and we'll show that $p$ is not totally ramified. We show this by considering the subfields $\QQ(\sqrt2)$ and $\QQ(\sqrt5).$ On one hand, note that $\QQ(\sqrt2)\cap\overline\ZZ=\ZZ[\sqrt2],$ and\[\disc\left(\ZZ[\sqrt2]\right)=\det\begin{bmatrix} 1 & \phantom-\sqrt2 \\ 1 & -\sqrt2 \\\end{bmatrix}^2=(2\sqrt2)^2=8.\]It follows that if $p\ne2,$ then $p$ is not ramified in $\QQ(\sqrt2).$ In particular, for any $\mf P$ of $\QQ(\sqrt2,\sqrt5)$ over $\mf p$ of $\QQ(\sqrt2)$ over $p$ of $\ZZ,$ we have that\[e(\mf P/p)=e(\mf P/\mf p)e(\mf p/p)\le\left[\QQ(\sqrt2,\sqrt5):\QQ(\sqrt2)\right]\cdot1=2 \lt 4,\]so $p$ is not totally ramified in $\QQ(\sqrt2,\sqrt5).$ Now, on the other hand, note that $\QQ(\sqrt5)\cap\overline\ZZ=\ZZ\left[\frac{1+\sqrt5}2\right],$ and\[\disc\left(\ZZ\left[\textstyle\frac{1+\sqrt5}2\right]\right)=\det\begin{bmatrix} 1 & \frac{1+\sqrt5}2 \\ 1 & \frac{1-\sqrt5}2 \\\end{bmatrix}^2=(\sqrt5)^2=5.\]It follows that if $p\ne5,$ then $p$ is not ramified in $\QQ(\sqrt5).$ Using the same logic as before, we fix any prime $\mf P$ of $\QQ(\sqrt2,\sqrt5)$ over $\mf p$ of $\QQ(\sqrt5)$ over $p$ of $\ZZ$ so that\[e(\mf P/p)=e(\mf P/\mf p)e(\mf p/p)\le\left[\QQ(\sqrt2,\sqrt5):\QQ(\sqrt5)\right]\cdot1=2 \lt 4,\]so $p$ is not totally ramified in $\QQ(\sqrt2,\sqrt5)$ again. It follows that no prime other than $2$ can be totally ramified in $\QQ(\sqrt2,\sqrt5)$ from $\QQ(\sqrt2),$ and no prime other than $5$ can be totally ramified in $\QQ(\sqrt2,\sqrt5)$ from $\QQ(\sqrt5),$ so no prime at all can be totally ramified in $\QQ(\sqrt2,\sqrt5).$
To finish, we claim that $P(x)=\boxed{x^4-14x^2+9}=\left(x^2-7\right)^2-40$ is not $p$-Eisenstein for any prime $p,$ no matter how we transform it like $P(ax+b).$ Indeed, suppose for the sake of contradiction that $P(ax+b)$ is indeed $p$-Eisenstein, for some integers $a$ and $b$ with $a\ne0.$ Note that $P(x)$ is the minimal polynomial for $\sqrt2+\sqrt5$ (it's a quartic with $\sqrt2+\sqrt5$ as a root), so it follows $P(ax+b)$ is the minimal polynomial for\[\alpha=a\sqrt2+a\sqrt5+b.\]However, by our discussion yesterday, it follows that $p$ is totally ramified in $\QQ(\alpha)=\QQ(\sqrt2+\sqrt5).$ But we just showed that $\QQ(\sqrt2+\sqrt5)=\QQ(\sqrt2,\sqrt5)$ has no totally ramified primes! So we have our contradiction, and we're done here.
I guess I might want to show that $\QQ(\alpha)=\QQ(\sqrt2+\sqrt5)$ and $\QQ(\sqrt2+\sqrt5)=\QQ(\sqrt2,\sqrt5).$ Well, $\sqrt2+\sqrt5=\frac1a(\alpha-b)$ shows the first equality, and the second one holds because\[\sqrt2=\frac{(\sqrt2+\sqrt5)^3-11\sqrt2}6,\]and we can get $\sqrt5$ by writing $\sqrt2+\sqrt5-\sqrt2.$ The finer details are really not terribly interesting.
As an aside, I think my $P(x)$ is (accidentally) also reducible$\pmod p$ for every rational prime $p.$ Indeed, we can check that it's $(x+1)^4\pmod2,$ and then for all other primes $p,$ I think (not super sure)\[\left[\mathcal O_K:\ZZ[\sqrt2+\sqrt5]\right]=2,\]so the Dedekind-Kummer factorization algorithm for ideals says that the splitting of $p$ can be determined by factoring $P(x)\pmod p.$ But the Galois group of $K=\QQ(\sqrt2,\sqrt5)$ is $\ZZ/2\ZZ\times\ZZ/2\ZZ,$ which isn't cyclic, so no prime $p$ may remain inert in $\mathcal O_K.$ (An inert prime forces $e(\mf p/p)=r=1,$ so the decomposition field is $\QQ$ and the inertial field is $K,$ so we must have $G\cong D\cong D/E$ is cyclic.) It follows that $P(x)\pmod p$ can never remain irreducible$\pmod p.$