Show that a polynomial is irreducible in $\mathbb{Q}$ Show that polynomial $f\left( x \right) = x\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right)\left( {x - 4} \right) - a$ is irreducible in $\mathbb{Q}$, where $a \in  - 2 + 5\mathbb{Z}$.
I have shown that $f$ doesn't have zeroes in $\mathbb{Q}$. We also know that since $f$ is primitive, by Gauss lemma, it is irreducible in $\mathbb{Q}$ iff it is irreducible in $\mathbb{Z}$.
I figured we could use $\pi :\mathbb{Z}\left[ X \right] \to \left( {\mathbb{Z}/5\mathbb{Z}} \right)\left[ X \right]$, show that it is a homomorphism and that irreducibility of $\pi \left( f \right)$ implies irreducibility of $f$.
But, assuming all these statements are true, which I haven't bothered yet to check, how can we see that $\left( {\pi \left( f \right)} \right)\left( x \right) = {x^5} + 4x + 2$ is irreducible in $\left( {\mathbb{Z}/5\mathbb{Z}} \right)\left[ X \right]$?
 A: In $\mathbb{F}_p[x]$, we have $\prod_{i=0}^{p-1} \left ( x-i \right )  = x^p - x$.
Now it is known that if $p \nmid a$ ,then $x^p - x  + a$ is irreducible over $\mathbb{F}_p[x]$.
Proof: Let $\alpha \not\in \mathbb{F}_p$ be a root of $x^p - x + a$ over $\mathbb{F}_p$. Then all the elements $\alpha, \alpha + 1 ,\ldots, \alpha + p-1$ are roots of $x^p - x + a$ over $\mathbb{F}_p$.So $\mathbb{F}_p (\alpha)$ is the splitting field of $x^p -x + a$ over $\mathbb{F}_p$.Now its easy to see that the degree of the minimal polynomial of $\alpha$ over $\mathbb{F}_p$ divides $p$.(*) Since $p$ is a prime, degree of the minimal polynomial of $\alpha$ over $\mathbb{F}_p$ is $p$. (It can't be $1$ since $\alpha \not\in \mathbb{F}_p$.) This proves that the minimal polynomial is actually $x^p - x + a$. So it must be irreducible and we are done.
An elementary proof of it for the case $p=5$, can be found in http://www.artofproblemsolving.com/Forum/viewtopic.php?p=2450995#p2450995. 
Edit: (*)
The degree of the minimal polynomial $g_0(x)$ of $\alpha$ over $\mathbb{F}_p[x] $ is equal to $[\mathbb{F_p}(\alpha) : \mathbb{F_p}] = n$ (let). Since $\mathbb{F_p}(\alpha) = \mathbb{F_p}(\alpha  + t)$,  minimal polynomial $g_k(x)$ of $\alpha + t$ also has degree $n$. Note that $g_k(x) | x^p - x + a$.So roots of $g_k(x) \in \left \{ \alpha, \ldots, \alpha + p-1 \right \} $. Also note that from the uniqueness of minimal polynomial $g_r(x)$ and $g_s(x)$ has no common root for $r \ne s$. So roots of $g_k(x)$ partitions $\left \{ \alpha, \ldots, \alpha + p-1 \right \} $ with $n$ elements in each class. So $n | p$.
A: An alternative, possible, but really terrible, approach, is to prove that there no polynomials of the form $q(x)=x^2-Ax-B$ that divide $p(x)=x(x^2-1)(x^2-4)-a$.
If we explicitly compute $p(x)\pmod{q(x)}$, we get:
$$ (B^2+(3A^2-5)B+(A^2-1)(A^2-4))\,x+AB(2B+A^2-5)-a, $$
and four times the coefficient of $x$ is:
$$ (2B+3A^2-5)^2-(5A^4-10A^2+9).$$
Since the last quantity must be zero, it is necessary that $(5A^4-10A^2+9)$ is a square, say:
$$(\clubsuit)\qquad 5(A^2-1)^2 + 4 = Q^2.$$
Using the theory of Pell's equations we can write down the entire family of integer solutions to 
$$ 5X^2-Y^2 = -4 $$
and look for the solutions $(X,Y)$ in which $X+1$ is a square, in order to prove that the only integer solutions to $(\clubsuit)$ occur for $A=0,\pm 1,\pm 2,\pm 3$. Do not try this at home. 
