Eisenstein's Criterion stronger version So we all know about Eisenstein's Criterion and how useful it turns out to be. In particular, if we have a monic cubic $x^3+ax^2+bx+c$, it is irreducible if there exists a prime number $p$ such that $p\mid a,b,c$ and $p^2\nmid c$. 
But suppose that a monic cubic is irreducible, and all its coefficients are positive, but there does not exist such a prime. Sometimes we can find an integer $k$ such that we can substitute $x=y+k$ into the cubic, and the resulting polynomial in $y$ will satisfy the criterion. For example, if we have the cubic $x^3+3x^2+5x+1$, then no $p$ exists, but if we substitute $x=y+3$ into it, we end up with $y^3+12y^2+50y+70$, which, taking $p=2$, satisfies the criterion, hence it is irreducible. My question is whether or not this stronger version of the criterion applies to every irreducible polynomial. That is to say, if $x^3+ax^2+bx+c$ is irreducible, then we can always find some integer $k$ such that $(y+k)^3+a(y+k)^2+b(y+k)+c$, when simplified, will satisfy the criterion. 
Thank you all in advance!
 A: Take $f(x) = x^3 + x + 1\in \mathbb{Z}[x]$. It's irreducible, and 
$$f(x + k) = x^3 + 3k x^2 + (3k^2 + 1)x + k^3 + k + 1$$
has relatively prime linear and quadratic coefficients for any integer $k\not = 0$.
A: Hint $\ $ Recall that the key idea behind the Eisenstein criteria is that polynomials satisfying the criterion are, mod $\,p,\,$ powers of a prime, viz. $\,\equiv x^n,\,$ and products of primes always factor uniquely. The same works for its  shift $\,(x-c)^n,\,$ so we seek primes $\,p\,$ such that, mod $\,p,\,$ the polynomial is congruent to such a power (e.g. for motivation: cyclotomic case). One can show that the only primes $\,p\,$ that can yield such powers are those dividing the discriminant of the polynomial. Indeed, if $\,f\equiv a (x-c)^n,\,\ n> 1\,$ then $\,f\,$ and $\,f'\,$ have a common root $\,x\equiv c,\,$  hence their resultant $\, R(f,f')\equiv 0.\,$ But this is, up to sign, the discriminant of $\,f\,$ (presumed monic).
Let's consider your example $\,f(x) = x^3+3x^2+5x+1,\,$ with discriminant $\,-140 = -2^2\cdot 5\cdot 7.\,$ Checking factorizations modulo these primes we find $\,f(x) = (x\!+\!1)^3\pmod{\! 2}.\, $ This implies $\,g(x)=f(x\!-\!1)\equiv x^3\pmod{\! 2}\,$ is Eisenstein by $2^2\nmid g(0)=f(-1)=-2,\,$ indeed we have $\,g(x) = f(x\!-\!1) = x^3+2x-2$.
Let's consider anomaly's example $\,f(x) = x^3+x+1,\,$ with discriminant $-31.\,$ The only possibility is $\,p = 31\,$ but $\,f\,$ is not a power mod $\,31\,$ by $\,3\,$ is a root, but has multiplicity one: $\,f'(3) \equiv 28\not\equiv 0.$
If you know algebraic number theory then this criterion can be rephrased more precisely in terms of totally ramified primes, e.g. see Keith Conrad's note.
