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I have recently read some passage about nested radicals, I'm deeply impressed by them. Simple nested radicals $\sqrt{2+\sqrt{2}}$,$\sqrt{3-2\sqrt{2}}$ which the later can be denested into $1-\sqrt{2}$. This may be able to see through easily, but how can we denest such a complicated one $\sqrt{61-24\sqrt{5}}(=4-3\sqrt{5})$? And Is there any ways to judge if a radical in $\sqrt{a+b\sqrt{c}}$ form can be denested?

Mr. Srinivasa Ramanujan even suggested some CRAZY nested radicals such as: $$\sqrt[3]{\sqrt{2}-1},\sqrt{\sqrt[3]{28}-\sqrt[3]{27}},\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}, \sqrt[3]{\cos{\frac{2\pi}{7}}}+\sqrt[3]{\cos{\frac{4\pi}{7}}}+\sqrt[3]{\cos{\frac{8\pi}{7}}},\sqrt[6]{7\sqrt[3]{20}-19},...$$ Amazing, these all can be denested. I believe there must be some strategies to denest them, but I don't know how.

I'm a just a beginner, can anyone give me some ideas? Thank you.

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Denesting radicals is hard, but there's an algorithm for it. For the case you're interested in, Wikipedia describes a simple procedure. – Zhen Lin Sep 15 '12 at 14:14
You may want to look at: searching this site reveals related questions such as:… – NoChance Sep 15 '12 at 14:15
Can you post your sources of these CRAZY nested radicals from Ramanujan's works? I'd like to read some more about his ideas. – SasQ Dec 14 '14 at 13:25
up vote 20 down vote accepted

There do exist general denesting algorithms employing Galois theory, but for the simple case of quadratic algebraic numbers one can employ a simple rule that I discovered as a teenager.

Simple Denesting Rule $\rm\ \ \ \color{blue}{subtract\ out}\ \sqrt{norm}\:,\ \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace} $

Recall $\rm\: w = a + b\sqrt{n}\: $ has norm $\rm =\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2 - n\: b^2 $

and, furthermore, $\rm\:w\:$ has trace $\rm\: =\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\:a$

Here $\:61-24\sqrt{5}\:$ has norm $= 29^2.\:$ $\rm\ \: \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 29\ $ yields $\ 32-24\sqrt{5}\:$

and this has $\rm\ \sqrt{trace}\: =\: 8,\ \ thus,\ \ \ \color{brown}{dividing\ it\ out}\ $ of this yields the sqrt: $\:\pm (4-3\sqrt{5}).$

For many further examples see my prior posts on denesting.

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Thanks~ ${}{}{}$ – ᴊ ᴀ s ᴏ ɴ Sep 17 '12 at 12:05
What about when $a^2 < nb^2$ ? – DanielV Dec 19 '14 at 7:36
@DanielV It will always yield the denesting (if it one exists). – Bill Dubuque Dec 19 '14 at 14:11

(I will assume $b$ is not a square, since otherwise it would not be a nested radical.)

A nested radical can be denested if and only if there exist $u,v\in\mathbb{N}$ such that the nested radical is of the form $\sqrt{u^2+v\pm2u\sqrt{v}}$ in which case it is also equal to $|u\pm\sqrt{v}|$.

It's not hard to show that those expressions are equal, which means that all nested radicals of that form can indeed be denested.

For the other way, let's consider the following equality where $a,b,c,d,e\in\mathbb{N}$: $$\sqrt{a\pm\sqrt{b}}=c\pm d\sqrt{e}$$ (Note that we can also write e.g. $\sqrt{3-2\sqrt{2}}$ in that form as $\sqrt{3-\sqrt{8}}$) If we square both sides, we get: $$a\pm\sqrt{b}=c^2+ed^2\pm2cd\sqrt{e}$$ This suggests we pick $u=c$ and $v=ed^2$. Then $a\pm\sqrt{b}=u^2+v\pm2u\sqrt{v}$ as claimed.

This isn't quite the end of the story since preferably we'd also like to know that $a$ actually corresponds to $u^2+v$ and $b$ to $4u^2v$, i.e. that you can't have two equal nested radicals of different forms. Depending on how deep you want to go, you can either take that for granted for now or observe that $a\pm\sqrt{b}$ satisfies the polynomial relations $(x-a)^2-b=0$ and $(x-u^2-v)^2-4u^2v=0$ and it follows from that that the two polynomials must be equal (using e.g. the concept of minimal polynomials or explicitly dividing the one by the other with remainder taking into account that $a\pm\sqrt{b}$ will satisfy no linear relation) which gives us our correspondence.

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You can derive a formula for $\sqrt{a+b\sqrt{c}}$. You will have to assume that $\sqrt{a+b\sqrt{c}}$ can be rewritten as the sum of two surds (radicands). So $$\sqrt{a+b\sqrt{c}}=\sqrt{d}+\sqrt{e}$$

Squaring both sides yields $$a+b\sqrt{c}=d+e+2\sqrt{de}$$

From that, we can see that $a=d+e$ so $e=a-d$ and $b\sqrt{c}=2\sqrt{de}\rightarrow b^{2}c=4de$.

Substituting $e$ with $a-d$ gives $b^{2}c=4d(a-d)$. So $b^{2}c=4ad-4d^{2}$. Rearranging the terms gives us $4d^{2}-4ad+b^{2}c=0$

Using the Quadratic Equation, we have $$d=\frac {a\pm\sqrt{a^{2}-b^{2}c}}{2}$$

And since $a=d+e$, $e$ is the conjugate of $d$. So $e=\frac {a-\sqrt{a^{2}-b^{2}c}}{2}$ and $d=\frac {a+\sqrt{a^{2}-b^{2}c}}{2}$.

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