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I'm aware there's certain expressions, such as $\sqrt{2 + \sqrt{3}}$, that can be expressed simply as a sum of two individual square roots, such as $\frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}$. This can be done by creating a system of equations as so:

$$ \sqrt{2 + \sqrt{3}} = \sqrt{a} + \sqrt{b} \\ 2 + \sqrt{3} = (\sqrt{a} + \sqrt{b})^2 $$ then solving for $a$ and $b$.

I was testing this method out on seperate expressions, specifically $\sqrt{5 + \sqrt{5}}$. However, this method doesn't hold for ${a, b} \in \mathbb{Q}$. My next idea was to attempt solving for so:

$$ \sqrt{5 + \sqrt{5}} = \sqrt{a} + \sqrt{b} + \sqrt{c} $$

although working backwards shows that:

$$ \sqrt{a} + \sqrt{b} + \sqrt{c} = \sqrt{(a + b + c) + 2\sqrt{ab} + 2\sqrt{ac} + 2\sqrt{bc}} $$

I'm a fresh mathematician so I'm not sure how to approach this properly, how can I properly express, or show that you can't express,

$$ \sqrt{5 + \sqrt{5}} = \pm \sqrt{x_1} \pm \sqrt{x_2} \pm \dots \pm \sqrt{x_n}. $$

My intuition: there could be the right combination of rational square roots that could equate $\sqrt{5 + \sqrt{5}}$ to a certain degree.

$$ 5 + \sqrt{5} = x_1 + x_2 + \dots + x_{n} \pm \sqrt{x_{1}x_{2}} \pm \sqrt{x_{1}x_{3}} \pm \dots \pm \sqrt{x_{1}x_{n}} \pm \dots \pm \sqrt{x_{n}x_{1}} \pm \sqrt{x_{n}x_{2}} \pm \dots \pm \sqrt{x_{n}x_{n-1}} $$

But I clearly bit off more than I could chew...

I wasn't sure how to search for this problem, sorry if this question has already been asked in some different way. Any answer, comment, or resources would be greatly appreciated thank you.

FishDrowned
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    There is some discussion here on wikipedia. – Jakobian Feb 13 '25 at 17:45
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    Are you familiar with Galois theory? I think that gives the most straightforward explanation for why the answer is no: $\mathrm{Gal}(\mathbb Q(\sqrt{5+\sqrt5})/\mathbb Q)\simeq C_4$, while the Galois group of $\mathbb Q(\sqrt{x_1},\ldots,\sqrt{x_n})$ is a subgroup of $C_2^n$, which cannot contain $C_4$. – Mathmo123 Feb 13 '25 at 17:47
  • @Mathmo123 Ah, this seems to have answered my question. I am rather unfamiliar with Galois theory but am absolutely willing to learn. I'll try scouting out some resources. If you'd like to write an answer with a small explanation that'd be much appreciated. Thank you. – FishDrowned Feb 13 '25 at 18:01
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    @FishDrowned I like this book if you want to learn it (although I haven't learned Galois theory from it, I was interested in transcendental bases and so on). – Jakobian Feb 13 '25 at 18:07
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    There's probably a more elementary method by computing the minimal polynomial of $a\sqrt{x}+b\sqrt{y}$ and showing that it cannot equal $X^4- 10X^2 + 20$, the minimal polynomial of $\sqrt{5+\sqrt 5}$. Your case reduces to this case by considering degrees of minimal polynomials. – Mathmo123 Feb 13 '25 at 18:25
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    See this answer for links to general radical denesting algorithms. The Wikipedia formula linked by @Jakobian above can be presented more simply using my Simple Denesting Rule. – Bill Dubuque Feb 25 '25 at 07:02
  • Hi @BillDubuque! I've seen your answers already, those only work for certain cases correct? I was looking for specifically $\sqrt{5 + \sqrt{5}}$ being represented as any $n$ sum of $\sqrt{x_i}$. Through Mathmo's comment, it seems it isn't possible, but how about $n$ sum of $\sqrt[k]{x_i}$? – FishDrowned Feb 25 '25 at 07:34
  • @FishDrowned The general denesting algorithm works generally - see the linked post. It is a constructive formulation of the Galois theoretic arguments mention above. – Bill Dubuque Feb 25 '25 at 07:55

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