Assume $K:\mathbb Q$ is a field extension and $[K:\mathbb Q] = 2$. Show that there is a unique squarefree $d \in \mathbb Z$ such that $K = \mathbb Q(\sqrt d)$.
I know that $K$ is generated by say $\{1,a\}$ over $\mathbb Q$ for some $a \in K$ and $1 \in K$. Since the extension is finite we know that $a$ is algebraic over $\mathbb Q$ and thus we have a minimum polynomial (monic) $f \in \mathbb Q[X]$ with $f(a) = 0 \in K$. Then $\deg(f) = 2$ and since of the form $f(x) = x^2 +bx +c$ so $$ 0 = a^2 +ba+c $$
What now ? :D
As Ben commented:
$$D:=\Delta=b^2-4c$$
does the trick, but its uniqueness is only up to a rational square, meaning: for any
$$0\neq r\in\Bbb Q\;,\;\;\Bbb Q(\sqrt D)=\Bbb Q(\sqrt{r^2D})\,$$