Proving that $[\mathbb{Q}(\sqrt{\sqrt{p+q}+\sqrt{q}},\sqrt{\sqrt{p+q}-\sqrt{q}}):\mathbb{Q}]=8$. Some days ago I posted a question in MSE in order to correct a solution to the problem of Prove that $[\mathbb{Q}(\sqrt{4+\sqrt{5}},\sqrt{4-\sqrt{5}}):\mathbb{Q}]=8$.
After posting this another question, I found a general argument for this type of extensions. I think that the ideas at the solution of Bill Dubuque in this question could be used to solve the following problem:

Let $p$ and $q$ be distinct positive prime numbers such that $p+q$ is a perfect square. Then $[\mathbb{Q}(\sqrt{\sqrt{p+q}+\sqrt{q}},\sqrt{\sqrt{p+q}-\sqrt{q}}):\mathbb{Q}]=8.$

My attempt of solution:
Let $\alpha_1 = \sqrt{\sqrt{p+q}+\sqrt{q}}$ and $\alpha_2=\sqrt{\sqrt{p+q}-\sqrt{q}}).$ Let $\mathbb{K}=\mathbb{Q}(\alpha_1,\alpha_2)$.
First observe that $$\alpha_1^2 = \sqrt{p+q}+\sqrt{q},$$ and $$\alpha_1 \alpha_2 = \sqrt{p}.$$
Let $\mathbb{L}=\mathbb{Q}(\alpha_1^2,\alpha_1 \alpha_2)=\mathbb{Q}(\sqrt{q},\sqrt{p}).$ We have that $[\mathbb{L}:\mathbb{Q}]=4,$ hence $\mathbb{L}$ is a 2-dimensional vector space over $\mathbb{Q}(\sqrt{q}),$ with basis $\{1,\sqrt{p}\}$. We will prove now that $\alpha_1 \not\in \mathbb{L}:$
Suppose that $\alpha_1 \in \mathbb{L}$ (this imply directly that $\alpha_2 \in \mathbb{L}$ too), then exists unique $a,b \in \mathbb{Q}(\sqrt{q})$ with $$\alpha_1 = a + b\sqrt{p}.$$ Hence, $$\sqrt{p+q}+\sqrt{q} = a^2 + p b^2 + 2ab\sqrt{p},$$ or equivalently: $$2ab\sqrt{p} = \sqrt{p+q}+\sqrt{q} - a^2 - p a^2.$$
Since the right member of the equality is in $\mathbb{Q}(\sqrt{q}),$ must be $a=0$ or $b=0$.


*

*If $a=0$ then $\alpha_1 = b\sqrt{p}=b\alpha_1 \alpha_2,$ hence $1=b\alpha_2$ and we conclude that $\alpha_2^{-1}=b \in \mathbb{Q}(\sqrt{q}).$

*If $b=0$ then $\alpha_1=a \in \mathbb{Q}(\sqrt{q}).$
Both cases gets a contradiction since $\sqrt{\sqrt{p+q}\pm\sqrt{q}}\not\in\mathbb{Q}(\sqrt{q}).$ If we suppose that $$\sqrt{\sqrt{p+q}\pm\sqrt{q}}\in\mathbb{Q}(\sqrt{q}),$$ then exists unique $a,b \in \mathbb{Q}$ such that $$\sqrt{\sqrt{p+q}\pm\sqrt{q}}=a+b\sqrt{q}.$$ Hence
$$\sqrt{p+q}\pm\sqrt{q} = a^2 + qb^2+2ab\sqrt{q},$$ and must be $ab=\pm1/2$ and $\sqrt{p+q} = a^2 + qb^2.$ Solving for $a$ we get that $a$ is a root of the polynomial $$4x^4-4\sqrt{p+q}x^2+q.$$ Hence $a$ have one of the following four values: $$\pm\sqrt{\frac{\sqrt{p+q}}{2}\pm\frac{\sqrt{p}}{2}},$$ but any of these values is a rational, if not, $$\bigg(\pm\sqrt{\frac{\sqrt{p+q}}{2}\pm\frac{\sqrt{p}}{2}}\bigg)^2=\frac{\sqrt{p+q}}{2}\pm\frac{\sqrt{p}}{2} \in \mathbb{Q}.$$
With this we conclude the proof and get the original claim.
End.
The problem I posted some days ago is a special case with $p = 11$ and $q = 5$.
Is this approach correct? I'm interested in reading Galois-type solutions since I think they are more "beautiful". Which are the pair of distinct positive primes whose sum is a perfect square? I see the pairs $(11,5)$, $(23,2)$ and $(31,5)$ for example.
Thaks to everyone.
 A: We have $\sqrt {p+q}=n\in\Bbb N$, so i will use this $n$ below.
Let us consider the tower of fields:
$\require{AMScd}$
\begin{CD}
{} @. L=\Bbb Q\left(\ \sqrt {n\pm\sqrt q}\ \right)\\
@.    @AAA\\
{} @. K=\Bbb Q(\ \sqrt p, \sqrt q\ )\\
@.\nearrow @.\nwarrow\\
\Bbb Q(\sqrt p) @. {} @. \Bbb Q(\sqrt q )\\
@.\nwarrow @.\nearrow\\
{} @. \Bbb Q  @.{}
\end{CD}
Some remarks first:


*

*The extension $K=\Bbb Q(\sqrt p,\sqrt q):\Bbb Q$ has degree four, else $\sqrt p$, $\sqrt q$ would differ by a rational factor, but $p\ne q$.

*The vertical arrow is an extension of fields.
First $\sqrt q\in L$, since $n\pm\sqrt q\in L$.
Also, because the product of the two numbers $\sqrt{n\pm\sqrt q}$
is $\sqrt{n^2-q}=\sqrt{(p+q)-q}=\sqrt p$, we also have $\sqrt p\in L$.

*For short, $L=
\Bbb Q\left(\ \sqrt {n+\sqrt q},\ \sqrt p,\ \sqrt q\ \right)
=
\Bbb Q\left(\ \sqrt {n-\sqrt q},\ \sqrt p,\ \sqrt q\ \right)
=K(\sqrt {n+\sqrt q})
=K(\sqrt {n-\sqrt q})
$.
It remains to show that the extension $L:K$ has degree two. 
If not, then we would have a linear relation over $\Bbb Q$ of the shape:
$$
\sqrt{n+\sqrt q}=A+B\sqrt q+\sqrt p(C+D\sqrt q)\in K\ .
$$
Apply now the Galois morphism $\sqrt p\to -\sqrt p$, $\sqrt q\to+\sqrt q$ of $K=\Bbb Q(\sqrt p,\sqrt q)$, to get parallely
$$
\begin{aligned}
    \sqrt{n+\sqrt q} &=A+B\sqrt q+\sqrt p(C+D\sqrt q)\in K\ ,\\
\pm \sqrt{n+\sqrt q} &=A+B\sqrt q-\sqrt p(C+D\sqrt q)\in K\ .
\end{aligned}
$$
(To be pedant and avoid any questions that i may put myself, i added that $\pm$ in the last relation, imposed by the minimal polynomial condition over $K$, the L.H.S being a root of $X^2 -n-\sqrt q\in \Bbb Q(\sqrt q)\ [X]$.)
The representation is unique, so we have either
$$
\begin{aligned}
\sqrt{n+\sqrt q} &=A+B\sqrt q\ ,\text{ or}\\
\sqrt{n+\sqrt q} &=\sqrt p(C+D\sqrt q)\ .
\end{aligned}
$$
We use now the other Galois morphism, $\sqrt p\to \sqrt p$, $\sqrt q\to-\sqrt q$, getting either 
$$
\begin{aligned}
\pm\sqrt{n-\sqrt q} &=A-B\sqrt q\ ,\text{ or}\\
\pm\sqrt{n-\sqrt q} &=\sqrt p(C-D\sqrt q)\ .
\end{aligned}
$$
We multiply, so $\pm \sqrt{n+\sqrt q} \cdot \sqrt{n-\sqrt q}=\pm \sqrt{n^2-q}=\pm\sqrt p$ is either $A^2-qB^2\in\Bbb Q$ or $p(C^2-qD^2)\in \Bbb Q$, thus a contradiction.
The linear relation cannot hold. So the degree of the vertial field extension is two.
$\square$ 
A: But this is again a quick application of the "kummerian argument" which I used in my answer to your question of a few days ago. Introduce $k=\mathbf Q(\sqrt p, \sqrt q)$, which is a biquadratic field because $pq$ cannot be a square in $\mathbf Q$ (by unique factorization in $\mathbf Z$)
. Consider then the extensions $k(\sqrt {\sqrt {p+q} \pm \sqrt q})$, where $p+q$ is a perfect square. Since $\sqrt {\sqrt {p+q}+\sqrt q} .\sqrt {\sqrt {p+q}-\sqrt q}=p$ is a square in $k^*$, the kummerian argument above $k$  shows that the extensions $k(\sqrt {\sqrt {p+q} \pm \sqrt q})$ are the same field, say $K$. Applying again Kummer over $\mathbf Q(\sqrt q)$ as base field, we see that $K=k=\mathbf Q(\sqrt q)(\sqrt p)$ iff $p(\sqrt {p+q}\pm \sqrt q)$ are squares in $\mathbf Q(\sqrt q)$; multiplying the two relations, we get that $p^3$ is a square in $\mathbf Q(\sqrt q)$ : impossible. Hence $[K:k]=2$ and $[K:\mathbf Q]=8$.
Remark:  In the kind of questions you are dealing with, the kummerian approach is more natural in the sense that it appeals only to the multiplicative structure of the fields involved, whereas a blunt direct approach mixes the multiplicative and additive structures.
