Galois group of a certain degree 3 polynomial A degree 3 irreducible polynomial $f\in\mathbb{Q}[x]$ has Galois group $A_3$ if and only if its discriminant $\Delta(f)$ is a square in $\mathbb{Q}$ (otherwise the Galois group is $S_3$). 
Consider the polynomial $f=x^3-ax-1\in\mathbb{Q}[x]$ where $a\in\mathbb{Z}$, whose discriminant is $4a^3-27$. In Example 2.2 of this paper Keith Conrad says $f$ has Galois group $A_3$ iff $a=3$, but he says it is hard to prove it. So my question is of course, how can I prove it?
Note that the statement is equivalent to $4a^3-27$ is a perfect square iff $a=3$ (hence the tag numer theory).
Someone said to me the words "Branched Covers and Monodromy Groups", and it looks to me like the problem has to do with that (?).
 A: Asking when $4a^3 - 27$ is a square is equivalent to asking for integer solutions $x,y$ such that $y^2=4x^3-27$. This equation is known as an elliptic curve and the beauty of these is that they come with a group structure.
Making a change of variables, we can rewrite this as $E: y_1^2 = x_1^3 - 314928$, where $y_1=y/108, x_1=x/36$.
From the theory of elliptic curves, we find that this has exactly two solutions in the rationals which are $(108,972)$ and $(108,-972)$. Now in order to get integral points on our original curve, we require that $x_1$ is divisible by $36$ and $y_1$ is divisible by $108$. Fortunately this happens and gives us the integral points $(3,\pm 9)$.
This shows that $a=3$ is the only value for which the discriminant is square (in our case the square of 9)$.
NB: We didn't actually require the discriminant to be an integer, only a square, which means our life easier on the elliptic curve side of things as searching for rational solutions is easier than searching for integral ones. 
In our case they amounted to the same thing, but in general elliptic curves may have infinitely many rational solutions (each "one" corresponding to a value of $a$ that gives a square discriminant), but only finitely many integral solutions.
Edit
The theory of elliptic curves that I have used is:


*

*They have the structure of an abelian group

*This group is finitely generated (Mordell-Weil Theorem)

*The torsion subgroup is isomorphic to $C_3$ (Lutz-Nagell which gives the points)

*It has rank $0$ (performing a $2$-descent)


All of this can be found in Silverman's Arithmetic of Elliptic Curves in Chapters 3 and 8).
