We know sometimes if we cannot use Eisenstein criterion directly to prove a polynomial $f(x)\in \mathbb{Z}[x]$ is irreducible,we can take $f(x+1)$ to use the criterion.
So I wonder can all irreducible $f[x]\in \mathbb{Z}[x]$ be judged with Eisenstein criterion at least after such a kind of translation?
You can take any irreducible polynomial of the form $f_{a,b}=X^4-2(a+b)X^2+(a-b)^2$, where $a,b\in\mathbb{Z}\setminus\{0\}$.
Such polynomial is proven to be irreducible if and only if three integers $a,b$ and $ab$ are not squares.
The underlying explanation is that , under this condition, $f_{a,b} $ is the minimal polynomial of $\sqrt{a}+\sqrt{b}$, which is a primitive element for $\mathbb{Q}(\sqrt{a},\sqrt{b})/\mathbb{Q}$. (It is easy to see it is reducible if the condition is not satisfied)
A bit of arithmetic shows that for any prime $p$, and $u,v\in\mathbb{Z}$ such that $p\nmid u$, the polynomial $f(uX+v)$ does not satisfy Eiseinstein criterion for $p$.
What is even funnier with this polynomial is that for any prime $p$, and $u,v\in\mathbb{Z}$ such that $p\nmid u$, the polynomial $f(uX+v)$ is reducible mod $p$.
So both standard irreducibility criteria do not work for $f$, even after translations.
