Let $t = m^2+m+7$ such that $m \in \mathbb{Z}$. Show that the Galois group of $f(x) \in \mathbb{Q}[x]$ denoted by $G(f(x),\mathbb{Q})$ is $A_3$ with $f(x) = x^3-tx-t$.
Proof: Note that $f(x)$ is a polynomial of degree $3$ , then $G(f(x),\mathbb{Q}) \leq S_3$. Recall that $G(f(x),\mathbb{Q}) = A_3$ if and only if $f(x)$ is irreducible over $\mathbb{Q}$ and the discriminant of $f(x)$ denoted by $D$ is an square in $\mathbb{Q}$, i.e. $\sqrt{D} \in \mathbb{Q}$. After a while doing the calculations we get the following discriminant:
$$D = t^2(4t-27) = t^2(4m^2+4m+1) = t^2(2m+1)^2 = (t(2m+1))^2.$$
Then, $D$ is an square in $\mathbb{Q}$. Now, I am stucked trying to proof the irreducibility of $f(x)$. As $f(x)$ is a polynomial of degree $3$ it is enough to show that $f(x)$ has no roots in $\mathbb{Q}$ and by some irreducibility tests, I can conclude that the only candidates to roots of $f(x)$ are the divisors of $t$ but this set is a little bit chaotic in the sense that the set of divisors of a number changes a lot with the number, could anyone tell me a hint?
Note that $m(m+1)$ is always even because if $m$ is odd, then $m+1$ is even. By definition of $t$, we can conclude that $t$ is always odd or $t = 1$ (mod $2$). Now, we see that $\overline{f(x)} = x^3-x-1 \in \mathbb{Z}_2[x]$ and has no roots in this field, then it is irreducible. Finally, by the modular criteria, $f(x) \in \mathbb{Q}[x]$ is irreducible.
With this and the result about the discriminant, we conclude that $G(f(x),\mathbb{Q}) = A_3$.
Thanks to reuns for the hint :)