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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    $\begingroup$ Denesting radicals is hard, but there's an algorithm for it. For the case you're interested in, Wikipedia describes a simple procedure. $\endgroup$
    – Zhen Lin
    Commented Sep 15, 2012 at 14:14
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    $\begingroup$ You may want to look at: cybertester.com/data/denest.pdf searching this site reveals related questions such as: math.stackexchange.com/questions/194030/… $\endgroup$
    – NoChance
    Commented Sep 15, 2012 at 14:15
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    $\begingroup$ Can you post your sources of these CRAZY nested radicals from Ramanujan's works? I'd like to read some more about his ideas. $\endgroup$
    – SasQ
    Commented Dec 14, 2014 at 13:25
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    $\begingroup$ isibang.ac.in/~sury/ramanujanday.pdf $\endgroup$
    – JSCB
    Commented Dec 15, 2014 at 9:29
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    $\begingroup$ "The basic method I learned is to set this equal to $\sqrt{x-y\sqrt{c}}$" and do what with it? "but I found out that this doesn't work with $\sqrt{5+2\sqrt{6}}$" why not? What was supposed to happen? $\endgroup$
    – fleablood
    Commented Feb 20, 2019 at 1:27

6 Answers 6


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

$$\bbox[1px,border:1px solid #0a0]{\bbox[8px,border:1px solid #0a0]{\rm {\bf Simple\ Denesting\ Rule}\!:\ \ \color{blue}{subtract\ out}\ \sqrt{norm},\, \ then\ \ \color{brown}{divide\ out}\ \sqrt{trace}\ }}\qquad\ \ $$

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

Here $\:61-24\sqrt{5}\:$ has norm $= 29^2.\:$ $\rm \color{blue}{Subtracting\ out}\ \sqrt{norm}\ = 29\ $ yields $\ \color{#0a0}{32\:\!-2\:\!4\sqrt{5}}\:$

and $\rm\color{#0a0}{this}$ has $\rm\ \sqrt{trace}\: =\: 8,\ \ thus, \ \ \color{brown}{dividing \ it \ out}\, $ of $\rm\color{#0a0}{this}$ yields the sqrt: $\,\pm( 4\,-\,3\sqrt{5}).$

See here for a simple proof of the rule, and see here for many examples of its use.

  • $\begingroup$ What about when $a^2 < nb^2$ ? $\endgroup$
    – DanielV
    Commented Dec 19, 2014 at 7:36
  • $\begingroup$ @DanielV It will always yield the denesting (if it exists), see the linked proof. $\endgroup$ Commented Oct 18, 2020 at 6:45
  • $\begingroup$ I'm familiar with norm and trace of matrices, but not numbers: How do you define norm and trace of individual numbers? And what is $w'$? $\endgroup$ Commented Jun 29, 2023 at 12:22
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    $\begingroup$ @SRobertJames $\rm\,w'$ is the conjugate of $\rm\,w,\,$ i.e. $\rm\,a-b\sqrt n.\,$ The norm and trace are defined above, i.e. $\,\rm ww'$ and $\,\rm w+w'\,$ resp. These are (quadratic) special cases of general notions from algebraic field extensions, see Field Norm and Field Trace, which can be viewed concretely in matrix terms. $\endgroup$ Commented Jun 29, 2023 at 13:09
  • $\begingroup$ To be clear: the green box means to instruct us to divide out the $\rm {\sqrt{trace} }$ of that intermediate result, rather than to to divide out the $\rm {\sqrt{trace} }$ of the expression inside the given surd. $\endgroup$
    – ryang
    Commented Aug 24, 2023 at 2:12

There are the following identities. $$\sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}$$ and $$\sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^2-b}}{2}}-\sqrt{\frac{a-\sqrt{a^2-b}}{2}},$$ where all numbers under radicals are non-negatives.

For example: $$\sqrt{5+2\sqrt6}=\sqrt{5+\sqrt{24}}=\sqrt{\frac{5+\sqrt{5^2-24}}{2}}+\sqrt{\frac{5-\sqrt{5^2-24}}{2}}=\sqrt3+\sqrt2.$$ This is interesting, when $a$ and $b$ are rationals and $a^2-b$ is a square of a rational number.

The first identity is true because $$\left(\sqrt{\frac{a+\sqrt{a^2-b}}{2}}+\sqrt{\frac{a-\sqrt{a^2-b}}{2}}\right)^2=$$ $$=\frac{a+\sqrt{a^2-b}}{2}+\frac{a-\sqrt{a^2-b}}{2}+2\sqrt{\frac{a+\sqrt{a^2-b}}{2}}\cdot\sqrt{\frac{a-\sqrt{a^2-b}}{2}}=a+\sqrt{b}.$$

  • 3
    $\begingroup$ This is interesting. I had never seen the identities you begin with. $\endgroup$
    – Lubin
    Commented Feb 20, 2019 at 2:19
  • $\begingroup$ @Lubin same with me too. $\endgroup$
    – Max0815
    Commented Feb 20, 2019 at 2:20
  • $\begingroup$ We can prove it. It's not hard. $\endgroup$ Commented Feb 20, 2019 at 2:34
  • $\begingroup$ How would you prove the first one @MichaelRozenberg? I can get the second on I think because it is conjugate of first, which should be easy. $\endgroup$
    – Max0815
    Commented Feb 20, 2019 at 3:08
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    $\begingroup$ To get some more intuition how these identities were derived in the first place. To that end, you've got to view the polynomial $x^4-2ax^2+(a^2-b)$ for which $\sqrt{a+\sqrt{b}}$ is a zero, as the product of $x^2\pm c x + \sqrt{a^2-b}$. It turns out $c$ needs to satisfy the equation $-2a = -c^2 + 2\sqrt{a^2-b}$ and you just need to find the roots of this simple polynomial, $x^2- c x + \sqrt{a^2-b}$ :) $\endgroup$ Commented Feb 20, 2019 at 10:03

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 $d=\frac {a+\sqrt{a^{2}-b^{2}c}}{2}.\,$ Thus

$$\sqrt{a+b\sqrt{c}}\,=\, \sqrt{\frac {a+\sqrt{a^{2}-b^{2}c}}{2}} +\sqrt{\frac {a-\sqrt{a^{2}-b^{2}c}}{2}}$$


(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.


One way of approaching this problem is by viewing it as a zero of an equation. Let me explain. Let's say you want to compute $\sqrt{x_0}$ where $x_0$ is a zero of some quadratic polynomial of the form $x^2-bx+1$. Now, one way to go is to note that if you have a zero of $x^2+ax+1$, then it will still be a zero if you multiply it with $x^2-ax+1$ which equals $$x^4 + (2-a^2) x^2 + 1$$ Now the idea is to work backwards. So, in particular, if you can find you can find an $a$ such that $b=a^2-2$, then you can conclude that the square root of you polynomial is equal to one of the zeros of the polynomials $x^2-ax+1$ or $x^2+ax+1$. It is usually not too hard to find out which. If you found out which, you can rewrite your square root accordingly to the desired form :)

To conclude, one of the tricks is to find the right form of your polynomials such that you end up with something useful. This method will however require some puzzling.

Edit applying this method to the example $\sqrt{5+2\sqrt{6}}$, you will find that the polynomial you need (thus the one for which you want to calculate the square root of a zero) is $x^2-10x+1$. Then according to the above method (which you derive on the go), your $a=\sqrt{12}$ and then you just need to solve $x^2-ax+1=0$ which is the only possibility since for the other one, filling in a positive number will yield a positive number. Solving this equation by completing the square is not too difficult. It turns out that the zeros lie at around 0.5 and 3. Hence, it is not difficult to note you need the larger zero which turns out to be exactly given by $\sqrt{2}+\sqrt{3}$. Does that make sense?

  • $\begingroup$ Yes. thanx!!!!! $\endgroup$
    – Max0815
    Commented Feb 20, 2019 at 1:43
  • $\begingroup$ If your polynomial ends with $+b$ instead of $1$, I think you need to work with $+\sqrt{b}$ in the polynomials with the $a$s. (Did not check this but I am sure this will work). $\endgroup$ Commented Feb 20, 2019 at 1:48
  • $\begingroup$ yes I believe so too. $\endgroup$
    – Max0815
    Commented Feb 20, 2019 at 2:19

There is also another radical identity, which generalises $\sqrt{a+\sqrt b}$ even further: \begin{align*}\sqrt{a\pm \sqrt{b+\sqrt{c}}}&= \frac{1}{\sqrt{2}}\left\{\sqrt{a^2+\sqrt{\frac{a^2-b+\sqrt{(a^2-b)^2-c}}{2}}-\sqrt{\frac{a^2-b-\sqrt{(a^2-b)^2-c}}{2}}}\right. \nonumber\\&\quad\left.\pm\,\sqrt{a^2-\sqrt{\frac{a^2-b+\sqrt{(a^2-b)^2-c}}{2}}+\sqrt{\frac{a^2-b-\sqrt{(a^2-b)^2-c}}{2}}}\right\}\end{align*} such that the $\pm$ signs are not independent of each other.

We also have:

If $a^2+b^2=c^2$, then $$(a+b-c)^2=2(c-a)(c-b).$$ Therefore, by letting $a=\sqrt x$ and $b=\sqrt{c^2-x}$, it follows for all $c$ and $x$ such that $c\in \mathbb{R}\setminus \big({-\sqrt x}, \sqrt x\big)$, $$\sqrt{2\big(c-\sqrt x\big)\big(c-\sqrt{c^2-x}\big)}=\sqrt x + \sqrt{c^2-x} -c.$$ Interestingly, given that $$(a+x)^2+(b+y)^2-(a+y)^2-(b+x)^2=2(a-b)(x-y)$$ then if $a^2+b^2=c^2$, we also have $$(a+b-c)^2=(2c)^2+(a+b)^2-(b+c)^2-(a+c)^2.$$ Through one of Ramanujan's classic identities, $$\big\{\sqrt [3]{(a+b)^2}-\sqrt [3]{a^2-ab+b^2}\big\}^3=3\big(\sqrt [3]{a^3+b^3}-a\big)\big(\sqrt [3]{a^3+b^3}-b\big)$$ it can be found that if $a^3+b^3=c^3$, then $$(a+b-c)^3=3(a+b)(c-a)(c-b)$$ Therefore, again, for all $c$ and $x$ such that $c\in\mathbb{R}\setminus ({-\sqrt [3] x}, \sqrt [3] x)$, it follows $$\sqrt [3]{3\big(\sqrt [3] x+\sqrt [3]{c^3-x}\big)\big(c-\sqrt [3] x\big)\big(c-\sqrt [3]{c^3-x}\big)}=\sqrt [3] x + \sqrt [3]{c^3-x}-c.$$ Unfortunately, due to Fermat's Last Theorem, there are no positive integer solutions for unique $a$, $b$ and $c$.

Try such substitutions yourself on the following one! :)

If $a^4+b^4=c^4$, then $$(a+b-c)^4=6(ab-ac-bc)^2-4\big\{a^3(c-b)+b^3(c-a)+c^3(a+b)\big\}$$


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