On the minimal polynomial of constructible numbers Let us say we would like to compute the minimal polynomial of $\sqrt{3}+\sqrt{5}$ over $\mathbb Q$.
Following this idea, we pick the minimal polynomials $x^2-3$ and $x^2-5$ of $\sqrt{3}$ and $\sqrt{5}$, look at the roots of each polynomial $\{-\sqrt{3},\sqrt{3}\}$ and $\{-\sqrt{5},\sqrt{5}\}$ and compute the following product
$$
p(x)=\prod_{a\in\{-\sqrt{3},\sqrt{3}\}}\prod_{b\in\{-\sqrt{5},\sqrt{5}\}}(x-a-b),
$$
which is guaranteed to be a polynomial with rational coeficients, precisely,
$p(x)=x^4-16x^2+4$. Once we have this polynomial, we factorize it and find out the minimal polynomial of $\sqrt{3}+\sqrt{5}$.
Well, that surely made me question: 

Are the roots of the minimal polynomial of a constructible number precisely those obtained by changing the sign of each occurence of $\sqrt{ \ \ \ \ }$?

I ran some tests using sympy and this statement was verified to be true in all of them. I was not able to find a counterexample.
So I was wondering: is it really true? How to prove it? Any ideas?
 A: I'm going to describe an algorithm that answers my question, then discuss whether or not this algorithm gives the minimal polynomial of the constructible number.
Define the following notation:
1) For any number of the form $a=b+c\sqrt{u}\in F(\sqrt{u})$, 
denote $\overline a=b-c\sqrt{u}$.
2) For any polynomial $p(x)\in F(\sqrt{u})[x]$, denote by $\overline p(x)$ the polynomial obtained by applying the operation $a \mapsto \overline a$ to all the coeficients of $p(x)$.
It's not hard to prove the following:
Claim: For any polynomial $p(x)\in F(\sqrt{u})[x]$, we have that $p(x)\overline p(x)\in F[x]$.
Pick some constructible number $a$. Then $a\in Q(\sqrt{s_1})(\sqrt{s_2})...(\sqrt{s_n})$ for some chain of the form
$$
\mathbb Q \subseteq \mathbb Q(\sqrt{s_1})\subseteq \mathbb Q(\sqrt{s_1})(\sqrt{s_2}) \subseteq...\subseteq\mathbb Q(\sqrt{s_1})(\sqrt{s_2})...(\sqrt{s_n}),
$$
with $s_1\in\mathbb Q$, and $s_i \in \mathbb Q(\sqrt{s_1})...(\sqrt{s_{i-1}})$ for any $2\leq i\leq n$.
The algorithm goes as follows:
Let $p_0(x) = (x-a)$ and, for each $1\leq i \leq n$, $p_i(x)=p_{i-1}(x)\overline{p_{i-1}}(x)$.
So $p_0(x) \in \mathbb Q(\sqrt{s_1})(\sqrt{s_2})...(\sqrt{s_n})[x]$ and using the claim, $p_1(x) \in \mathbb Q(\sqrt{s_1})(\sqrt{s_2})...(\sqrt{s_{n-1}})[x]$, and using the claim again, $p_2(x) \in \mathbb Q(\sqrt{s_1})(\sqrt{s_2})...(\sqrt{s_{n-2}})[x]$, and so on.
So, apart from some boring proof by induction, we get that $p_n(x)\in \mathbb Q[x]$ and since $(x-a)=p_0(x)\ |\ p_1(x)\ |\ ...\ |\ p_n(x)$, we have that $p_n(a)=0$.
Therefore, $p_n(x)$ is a candidate to be the minimal polynomial of $a$.
The second part of my question was if $p_n(x)$ is always the minimal polynomial of $a$. And it turns out that it may not be.
If we consider the number $\sqrt{3+2\sqrt{2}}$, the polynomial obtained by th algorithm described above is $x^4-6x^2+1$, but
$$
x^4-6x^2+1 = (x^2-2x-1)(x^2+2x-1).
$$
But note that $3+2\sqrt{2}=(1+\sqrt{2})^2$, so $\sqrt{3+2\sqrt{2}}=1+\sqrt{2}\in\mathbb Q(\sqrt{2})$, so actually $Q(\sqrt{2})(\sqrt{3+2\sqrt{2}})=Q(\sqrt{2})$.
If we assume that the chain we picked initially is minimal, in the sense that it also satisfies that $\sqrt{s_1}\notin\mathbb Q$, $\sqrt{s_i} \notin \mathbb Q(\sqrt{s_1})...(\sqrt{s_{i-1}})$ for any $2\leq i\leq n$, and $a\notin \mathbb Q(\sqrt{s_1})...(\sqrt{s_{n-1}})$, then this algorithm certainly gives the minimal polynomial, since, in this case, $[a:\mathbb Q]=2^n$ and the degree of $p_n(x)$ is $2^n$.
