# 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.

• 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 $4n^2 = p +q$. – eduard Jan 31 '19 at 23:50

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.

• Your remark is really useful, it gave me a good intuition for these extensions. Thanks! – DrinkingDonuts Feb 5 '19 at 13:26

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.

$$\square$$

• I'm bothered by your Q-linear relation, which involves only $\sqrt p$ and $\sqrt q$, whereas a basis of K/Q is {$1, \sqrt p, \sqrt q, \sqrt {pq}$}. – nguyen quang do Feb 1 '19 at 11:42
• yes, thanks, a good point! have to rearrange things! – dan_fulea Feb 1 '19 at 14:45