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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.

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  • $\begingroup$ Notice that your last question is either (case p=2) equivalent to looking for primes of the form $n^2 -2$, see (math.stackexchange.com/questions/591333/…) or (case pq odd) related to Goldbach's conjecture. In particular Goldbach predicts that for every $n\geq 3$ there are diferent primes $p,q$ such that $4n^2 = p +q$. $\endgroup$ – eduard Jan 31 at 23:50
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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.

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  • $\begingroup$ Your remark is really useful, it gave me a good intuition for these extensions. Thanks! $\endgroup$ – DrinkingDonuts Feb 5 at 13:26
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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.

$\square$

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  • $\begingroup$ I'm bothered by your Q-linear relation, which involves only $\sqrt p$ and $\sqrt q$, whereas a basis of K/Q is {$1, \sqrt p, \sqrt q, \sqrt {pq}$}. $\endgroup$ – nguyen quang do Feb 1 at 11:42
  • $\begingroup$ yes, thanks, a good point! have to rearrange things! $\endgroup$ – dan_fulea Feb 1 at 14:45

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