TITLE: Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$
QUESTION [6 upvotes]: Is there any way to determine the Galois group of $X^{5}-2X+7$ over $\mathbb{Q}$ not using the discriminant? Thanks!

REPLY [3 votes]: Following Nishant's suggestion the calculation might go as follows.
Modulo $p=2$ our polynomial $f(x)=x^5-2x+7$ factors as
$$
f(x)\equiv x^5+1=(x+1)(x^4-x^3+x^2-x+1).
$$
This tells us many things. For example the only possible non-trivial factorization over $\Bbb{Q}$ must have a linear factor. But the rational root test quickly tells us that those don't exist. Thus $p(x)$ is irreducible in $\Bbb{Q}[x]$, and the Galois group $G\le S_5$ thus acts transitively on the set of five roots of $f(x)$. 
By Dedekind's theorem we also see that $G$ contains a 4-cycle, so we can tell that $G$ is doubly transitive, and that $|G|\ge 5\cdot4=20$. Note that the modulo $2$ factors above were distinct. This was crucial, as otherwise we could not have made this deduction. This also implies that $p=2$ does not divide the discriminant.
What other primes to try? Somehow $p=7$ suggests itself. We immediately get the factorization
$$
f(x)=x^5-2x=x(x^4-2).
$$
Here we get a bit of luck as modulo $7$ we have $2\equiv9=3^2$. Therefore we can factor further:
$$
f(x)=x(x^4-9)=x(x^2-3)(x^2+3)=x(x^2-3)(x^2-4)=x(x^2-3)(x-2)(x+2).
$$
It is, indeed, our lucky day, as $3$ is a quadratic non-residue modulo $7$, so we have a single irreducible quadratic factor and three distinct linear factors. Dedekind's theorem thus tells us that $G$ contains a 2-cycle.
It is a known fact that when a subgroup of $S_5$ contains a 5-cycle and a 2-cycle, it must be all of $S_5$. Alternatively we could deduce that $G$ must have at least $40$ elements, and, because it contains a 2-cycle, is not a subgroup of $A_5$.
Anyway: $G=S_5$ is the only remaining option.