Q: Showing the irreducibility of a polynomial I am asked to show that $f(X)=X^5 - aX -1$ is irreducible over $\mathbf{Z[X]}$ unless $a=0,2,-1$. ($\mathbf{Z}$ being the ring of integers.)
I have been able to do this, but my solution is a lengthy one.
I used the fact that if $f$ is not irreducible, then we have $f(X)=g(X)h(X)$, where $g$ and $h$ are polynomials over $Z$. 
We can ignore the case where either one is constant, because the g.c.d. of the coefficients is 1. Hence,there are two possibilities - one where one of the polynomials is linear, and the second where one of them has degree 2 and the other has degree 3. For the first case, f must have a rational root, and it is easy to show that this is only possible if $a$ is either $0$ or $2$. 
In the latter case, we can assume $g(X)= AX^3 + BX^2 + CX + D$ and $H(X)= A'X^2 + B'X +C'$, after which a laborious calculation involving comparing coefficients shows that that only possibility is $a=-1$. 
However, I feel that there is a much more simple and obvious solution that I am missing. Since the coefficients of $f$ have g.c.d 1, f is irreducible over $Z$ iff it is irreducible over $Q$. I was wondering if there was any way in which we could perhaps use that to find a shorter solution. 
Thanks in advance! 
