# Let $\alpha = \sqrt{2}+\sqrt{3}$. Can we make polynomial $r(\alpha) = \sqrt{n}$?

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 $$n\neq 2,3$$?

• $x^2-2,x^2-3,x^2-6$ stay irreducible over $\Bbb{Q}(\sqrt{n})$ whenever $n \ne d^2m$ with $m\in 2,3,6$. Thus $\sqrt{n} \not \in \Bbb{Q} \implies \sqrt{n} \not \in \Bbb{Q}(\sqrt{2},\sqrt{3})$ Sep 3 '19 at 15:25
• This is a duplicate of math.stackexchange.com/questions/3343122/… Sep 3 '19 at 15:38
• 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.. Sep 3 '19 at 16:01
• @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. Sep 3 '19 at 16:06

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 $$\{1,2,3,6\}$$.