Irreducibility of a particular polynomial I've got this problem for my homework: find out whether the polynomial
$$f(x)=x(x-1)(x-2)(x-3)(x-4) - a$$
is irreducible over the rationals, where $a$ is integer which is congruent to $3$ modulo $5$. 
It is easy to verify that $f(x)$ has no integer zeros (and no rational zeros too) and, because $f(x)$ is primitive, it is irreducible over $\mathbb{Z}$ iff it is irreducible over $\mathbb{Q}$. It is also clear to me that, since $f(x)$ has no integer zeros, only way to factorize $f(x)$ over $\mathbb{Z}$ is $(x^3 + bx^2 + cx +d)(x^2 + ex +f)$, i.e. as a product of one irreducible cubic polynomial and one irreducible quadratic polynomial, both primitive. I've got pretty ugly system of $5$ equations with $5$ variables. So I decided to put the whole story in $(\mathbb{Z}/5\mathbb{Z})[x]$. Then I've got the polynomial $g(x) = x^5 + 4x +2$. Since I firmly believe that $f(x)$ is irreducible over $\mathbb{Z}$, it means that I must check whether the $g(x)$ is irreducible over $\mathbb{Z}/5\mathbb{Z}$. Is there any other way than brute force?

Note that mod $5$ the polynomial can be written
$$f(x)=(x-m-2)(x-m-1)x(x-m+1)(x-m+2)-3$$
for any integer $m$.  Now if there is an irreducible quadratic factor, then $f(x)$ has a root that is a root of that factor.  But the roots of the irreducible quadratics mod $5$ are all of the form $m+n\sqrt3$.  Plugging this in gives
$$0=(n\sqrt3-2)(n\sqrt3-1)n\sqrt3(n\sqrt3+1)(n\sqrt3+2)-3$$
which implies 
$$\sqrt3=(3n^2-4)(3n^2-1)n$$
which is a contradiction.
Added later:  The same argument mod $p$ (where $p$ is a prime) shows that any polynomial of the form
$$f(x)=(x-r_1)(x-r_2)\cdots(x-r_p)-a$$
where $a$ is a quadratic non-residue mod $p$ and $r_k\equiv k$ mod $p$ for $1\le k\le p$, has no linear or quadratic factors.  For $p\ge7$, this argument leaves open the possibility that the polynomial might still be reducible.  Pablo Rotondo's answer, however, shows this is not the case, even when $a$ is a quadratic residue.
