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This is the extend version of my previous post Let $\alpha = \sqrt{2}+\sqrt{3}$. Find polynomial $r(x)$ such that $r(\alpha)=\sqrt{2}$. which I post due to find some standard process of obtaining polynomials.

And from @Hagen von Eitzen, I see the standard technique for this type of problem is compute the powers of $\alpha$ and linear combinations.

Based on this setup can we construct, $r(\alpha) = \sqrt{n}$, where $n\neq 2,3$?

It seems for $r(x) \in \mathbb{Q}[x]$, at least in the case of $n=p$, for prime $p$. this seems impossible. but I have no idea of writing formal mathematical proof.

For $r(\alpha) = \sqrt{3}$, from $\alpha^3 - 11 \alpha = -2\sqrt{3}$. I can deduce $r(x) = -\frac{1}{2} x^3 + \frac{11}{2} x$.

and for $r(\alpha) = \sqrt{6}$, from $\alpha^4 - \frac{49}{5} \alpha^2 = \frac{2}{5} \sqrt{6}$, I have $r(x) = \frac{5}{2} x^4 - 98x^2 $.. [In this case I guess we have to extend the degree of $r(x)$. ]

Is trial and error is the best ways to solving this kinds of problem?

I mean, Is there any formal way to find polynomial $r(x) \in \mathbb{Q}[x]$, with no degree assumption such that $r(\alpha) = \sqrt{n}$, where $n\neq 2,3$?

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    $\begingroup$ $x^2-2,x^2-3,x^2-6$ stay irreducible over $\Bbb{Q}(\sqrt{n})$ whenever $n \ne d^2m$ with $m\in 2,3,6$. Thus $\sqrt{n} \not \in \Bbb{Q} \implies \sqrt{n} \not \in \Bbb{Q}(\sqrt{2},\sqrt{3})$ $\endgroup$
    – reuns
    Sep 3 '19 at 15:25
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    $\begingroup$ This is a duplicate of math.stackexchange.com/questions/3343122/… $\endgroup$
    – Who am I
    Sep 3 '19 at 15:38
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    $\begingroup$ Well, every integer $n$ can be written as $C^2 *m$ where $C^2$ is a perfect square and $m$ is square free. So $\sqrt n = C\sqrt m$. If $m$ has a prime factor other than $2,3$ this is clearly impossible. So this is only possible if $m = 1, 2,3,6$ and those are all possibleby your method described.. $\endgroup$
    – fleablood
    Sep 3 '19 at 16:01
  • $\begingroup$ @HarshitGupta No it isn't. That question, linked to in the OP, is aboupt $p(\alpha) = \sqrt 2$. This this is for $p(\alpha)=\sqrt n; n\ne 2,3$. But reuns comment is really enough of an answer to this question. $\endgroup$
    – fleablood
    Sep 3 '19 at 16:06
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Here is a systematic method. It's just a change of basis. Write $$ \pmatrix{ \alpha^0 \\ \alpha^1 \\ \alpha^2 \\ \alpha^3 \\ } = \pmatrix{ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 5 & 0 & 0 & 2 \\ 0 & 11 & 9 & 0 \\ } \pmatrix{ 1 \\ \sqrt 2 \\ \sqrt 3 \\ \sqrt 6 \\ } $$ as $A=ME$. Then each element $\beta \in \mathbb Q[\sqrt 2,\sqrt 3]= \mathbb Q[\alpha]$ can be written as $\beta=b E = b M^{-1} A$.

For $\beta=\sqrt 2$, we have $b=(0 , 1 , 0 , 0)$ and so $b M^{-1}=(0, -9/2, 0, 1/2)$.

Try it now for $\beta=\sqrt 3$ and $\beta=\sqrt 6$.

This works for $\beta=\sqrt n$, provided $\sqrt n \in \mathbb Q[\sqrt 2,\sqrt 3]$. These $n$ are exactly those whose square-free part is in $\{1,2,3,6\}$.

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