Prove that $\sqrt{3\pm\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3\mp\sqrt{7}})$. I'm currently solving a fairly long exercise related to Galois theory in which I've come across having to prove that $\sqrt{3+\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3-\sqrt{7}})$ and $\sqrt{3-\sqrt{7}} \not\in \mathbb{Q}(\sqrt{3+\sqrt{7}})$. So far I haven't been able to find an "easy" or simple and understandable way to do so given that this isn't the main part of the problem. 
Any help is appreciated!
 A: Let $x_{\pm}=\sqrt{3 \pm \sqrt{7}}$. 
It is easy to see that we have the following quadratic extensions:
$\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}) \subset \mathbb{Q}(x_+)$, 
$\mathbb{Q} \subset \mathbb{Q}(\sqrt{7}) \subset \mathbb{Q}(x_-)$.
Assume that $x_+ \in K=\mathbb{Q}(x_-)$. Then $\sqrt{2} = x_+x_- \in K$, thus $L=\mathbb{Q}(\sqrt{2},\sqrt{7}) \subset K$. 
Since these fields have the same degree over $\mathbb{Q}$, $K \subset L$, ie $x_+=a+b\sqrt{2}+c\sqrt{7}+d\sqrt{14}$ for rationals $a,b,c,d$. 
Taking squares, we get $3+\sqrt{7}=(a^2+2b^2+7c^2+14d^2) + (2ab+14cd)\sqrt{2} + (2da+2bc)\sqrt{14} + (2ca+4bd)\sqrt{7}$. 
Thus $ab=-7cd$, $ad=-bc$, $2ca+4bd=1$, $a^2+2b^2+7c^2+14d^2=3$. 
Assume $a=0$: then $d \neq 0$ thus $c=0$ and $bd=1/4$, $2b^2+14d^2=3$. Usual quadratic theory yields then a contradiction. 
Thus $b=-7cd/a$, and $ad=7c^2d/a$ thus $a^2=7c^2$ hence $a=0$. A contradiction, hence the result.
A: HINT: Show that both have minimal polynomial $f:=X^4-6X^2+2$ over $\Bbb{Q}$, and hence that $[\Bbb{Q}(\sqrt{3\pm\sqrt{7}}):\Bbb{Q}]=4$, but that the splitting field of $f$  over $\Bbb{Q}$ has degree greater than $4$.
A: In general, in this kind of problem, it is better not to mix the two operations + and $\times$. Let me give an illustration here, using only $\times$. Introduce the quadratic field $k=\mathbf Q(\sqrt 7)$. Adopting the notation $x_{\pm}=\sqrt {3 \pm \sqrt 7}$ suggested by @Mindlack, let us write $K_{\pm}=k(x_{\pm})$. These are two extensions of $k$ of degree at most $2$ :


*

*if  $K_{+}$ or $K_{-} =k$, i.e. $(3 \pm\sqrt 7)\in {k^*}^2$, norming down to $\mathbf Q$ shows that $N(3\pm\sqrt 7)=2$ is a square in $\mathbf Q^*$: impossible

*if both degrees are 2, $K_{\pm}\subset K_{\mp}$ iff $K_{\pm}= K_{\mp}$, iff $k(x_{+})= k(x_{-})$, iff $2=(3+\sqrt 7)(3-\sqrt 7)\in {k^*}^2$  (no specific calculation, this is rudimentary Kummer theory over $k$), iff $\mathbf Q(\sqrt 2)=\mathbf Q(\sqrt 7)$, iff $2.7$ is a square in $\mathbf Q^*$(again by Kummer): impossible because $\mathbf Z$ is a UFD  ./.
