Let $L = \mathbb{Q}(x_1,...,x_7)$ and $K = \mathbb{Q}(e_1,...,e_7)$ where the $e_i$ are the elementary symmetric polynomials. I am asked to determine whether or not there is an element $\alpha \in L$ such that $\alpha ^2\in K$ but $\alpha \notin K$.
My Idea was to use the fact that (unless I am mistaken) $Gal(L/K)=S_n$ and $A_n \le S_n$ of index two. Therefore we get a field $K\le F$ of degree two. Then take some $\alpha \in F-K$ so we must have $K \le F(\alpha)$ of degree two. So looking at minimal polynomial gives some $$a\cdot\alpha^2+b\cdot\alpha+c = 0$$ and I need to show that $b=0$. Not sure where to go from here or whether this is even the right approach.
As suggested, use the discriminant $\displaystyle\prod_{i<j} (x_i-x_j)$. It is not symmetric, but it’s square is.
For your solution, first, WLOG, say that $a=1$. Then use $\alpha+\frac{b}{2}$ regardless of the value of $b$ (found by completing the square).