$$\sqrt[4]{161-72 \sqrt{5}}$$

I tried to solve this as follows:

the resultant will be in the form of $a+b\sqrt{5}$ since 5 is a prime and has no other factors other than 1 and itself. Taking this expression to the 4th power gives $a^4+4 \sqrt{5} a^3 b+30 a^2 b^2+20 \sqrt{5} a b^3+25 b^4$. The integer parts of this must be equal to $161$ and the coeffecients of the roots must add to $-72$. You thus get the simultaneous system:

$$a^4+30 a^2 b^2+25 b^4=161$$ $$4 a^3 b+20 a b^3=-72$$

In an attempt to solve this, I first tried to factor stuff and rewrite it as:

$$\left(a^2+5 b^2\right)^2+10 (a b)^2=161$$ $$4 a b \left(a^2+5 b^2\right)=-72$$

Then letting $p = a^2 + 5b^2$ and $q = ab$ you get

$$4 p q=-72$$ $$p^2+10 q^2=161$$

However, solving this yields messy roots. Am I going on the right path?

  • $\begingroup$ $161-72 \sqrt 5 = (2- \sqrt 5)^4$ $\endgroup$ – mercio May 31 '14 at 22:53
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    $\begingroup$ Since you are taking the $4$th root, it might be cleaner to just apply your process to square roots twice. $\endgroup$ – Peter Woolfitt May 31 '14 at 22:54

$$\sqrt[4]{161-72\sqrt5}=\sqrt[4]{81-72\sqrt5+80}=\sqrt[4]{(9-4\sqrt{5})^2}=\sqrt{9-4\sqrt{5}}=\sqrt{4-4\sqrt{5}+5}=\sqrt{(2-\sqrt{5})^2}=\sqrt5-2$$ The trick is to notice that $72$ factors into $2*9*4$ and since $9^2+(4\sqrt5)^2=161$ you get this

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    $\begingroup$ Notice you can perform the same on $\sqrt{9-4\sqrt{5}}$. $\endgroup$ – Sebastien May 31 '14 at 22:56
  • $\begingroup$ I guess my brain isn't working this late,fixed $\endgroup$ – kingW3 May 31 '14 at 23:03

Denesting $\sqrt w = \sqrt{a+b\sqrt{n}}\,$ can be done by a simple formula that I discovered in my youth. $ {\bf Simple\ Denesting\ Rule} \ \ \overbrace{\rm \color{#0a0}{subtract\ out}\ \sqrt{norm}^{\phantom .}}^{\textstyle\!\!\! w \to w - \sqrt{ww'} =:\, s\!\!\!\!\!\!}\!\!\!, \ {\rm then}\ \ \overbrace{\color{brown}{\rm divide\ out}\ \sqrt{{\rm trace}}^{\phantom .}}^{\textstyle s\,\to\, s/\sqrt{s+s'}\!\!\!\!\!\!\!}$ from that.

$\!\begin{align}{\rm Recall}\ \ w = a + b\sqrt{n}\rm \ \ has\ \ {\bf norm}\ &=\: w\:\cdot\: w' = (a + b\sqrt{n})\ \cdot\: (a - b\sqrt{n})\ =\: a^2\! - n\: b^2\\[4pt] {\rm and,\ furthermore,\ }w\rm \ \ has\ \ {\bf trace}\ &=\: w+w' = (a + b\sqrt{n}) + (a - b\sqrt{n})\: =\: 2\,a\end{align}$

In the norm/trace sqrts either sign works e.g. $\sqrt 1 = \pm1,\,$ so we choose what proves simplest.

Here $\:161-72\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{#0a0}{subtracting\ out}\ \sqrt{norm}\ = -1\ $ yields $\ 162-72\sqrt 5\:$

which has $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{324}\ =\ 18.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\ \,$ of the above yields $\ \ \ 9-4\sqrt 5$

Checking: $\ (9 - 4\sqrt 5)^2 = 9^2\!+\! 4^2(5)- 2(9)4\sqrt 5 = 161-72\sqrt 5 \ \ \large \color{#c00}\checkmark$

Next $\:9-4\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{#0a0}{subtracting\ out}\ \sqrt{norm}\ = 1\ $ yields $\ 8-4\sqrt 5\:$

with $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{16}\ =\ 4.\ \ \ \ \rm \color{brown}{Dividing\ it\ out}\,\ $ of the above yields $\,\ \ \ 2-\sqrt 5$

Checking: $\ (2 - \sqrt 5)^2 = 2^2\!+\! 5 - 2\cdot 2\sqrt 5 = 9 - 4\sqrt 5 \ \ \large \color{#c00}\checkmark$

Negating $\,2-\sqrt 5\,$ to get the positive square-root yields the sought result. We chose the signs in $\,\sqrt 1 = \pm 1$ so that arithmetic is simplest. Any choice will work as the proof below shows (e.g. we do both here). For many worked examples see prior posts on denesting. Below is a sketch of a proof.

Lemma $\ \ \sqrt w\, =\, \dfrac{s}t,\ \ \ \begin{align}s &\,=\, w \pm \sqrt{ww'}\\[.1em] t &\,=\: \pm\sqrt{s+s'}\end{align}\ $ when $\ \ \color{#90f}{\sqrt{ww'}\in\Bbb Q}$

Proof $\quad\ s^2 =\, w (w+w' \pm 2\sqrt{ww'})\, =\, w\, t^2$

Necessarily $\ \color{#90f}{\sqrt{ww'}\in \Bbb Q}\,$ if a denesting $\sqrt w = v = c + d\sqrt n\,$ exists, since

$$w = v^2\,\Rightarrow\, w' = v'^2\Rightarrow\, ww' = (vv')^2\in\Bbb Q^2$$

  • $\begingroup$ how you obtain 162 ? and begining $9-4\sqrt{5}$ which have norm 1 you arrive at $8-4\sqrt{5}$ I want to understand this method, can you give me more details ? what role have norm ? $\endgroup$ – Lucas May 2 '15 at 17:35
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    $\begingroup$ @Lucas As said, the sign of square roots are chosen to simplify arithmetic. In the first case we chose $\,\sqrt{\rm norm} = \sqrt{1} = \color{#c00}{-1},\,$ so subtracting it from $\,{\rm w} =162\!-\!72\sqrt 5\,$ yields $\,w -(\color{#c00}{-1}) = w\!+\!1.\,$ In the second case we chose $\,\sqrt{1} = 1.\,$ Any choice of signs will work (but you may need to adjust the result for normalization to positive / principal branch to get analytic vs. algebraic roots). Follow the above link for many more examples. $\endgroup$ – Bill Dubuque May 2 '15 at 17:51
  • $\begingroup$ I can't understand how $\sqrt{1}=-1$ $\endgroup$ – Lucas May 2 '15 at 18:00
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    $\begingroup$ @Lucas It's just a notation that proves convenient here (and in other algebra) where the result doesn't depend on the branch of the root, i.e if the result only needs $\,x^2 = a\,$ then it is convenient to let $\, x = \sqrt a\,$ denote any such root. In general algebra there is not necessarily any way to distinguish the roots, e.g. mod $\,5\,$ the sqrts of $1$ are $1$ and $4\ (\equiv -1)$ and there is no notion of "positive" here. $\endgroup$ – Bill Dubuque Jun 21 '16 at 23:50
  • $\begingroup$ There are details in the method that became clear to me only after reading the proof. First, the rule applies only if $\sqrt{\rm norm}$ is rational. Then you can add $\pm\sqrt{\rm norm}$ to get $s$, take the trace of $s$, and divide $s$ by $\pm\sqrt{\rm trace}.$ (For complex denesting both signs of $\pm\sqrt{\rm trace}$ are the two complex square roots; for real denesting one gives you the positive square root and one gives you the negative root.) $\endgroup$ – David K Feb 1 at 14:45

Another approach. We can apply twice the following general algebraic identity involving nested radicals \begin{equation*} \sqrt{a-\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}-\sqrt{\frac{a-\sqrt{ a^{2}-b}}{2}}\tag{1} \end{equation*} to get \begin{equation*} \sqrt[4]{161-72\sqrt{5}}=\sqrt[4]{161- \sqrt{25\,920}}=\sqrt{5}-2. \end{equation*} The numerical computation can be carried out as follows:

\begin{eqnarray*} \sqrt[4]{161-72\sqrt{5}} &=&\left( \sqrt{\frac{161+\sqrt{161^{2}-25\,920}}{2} }-\sqrt{\frac{161-\sqrt{161^{2}-25\,920}}{2}}\right) ^{1/2} \\ &=&\left( \sqrt{\frac{161+1}{2}}-\sqrt{\frac{161-1}{2}}\right) ^{1/2} \\ &=&\sqrt{9-\sqrt{80}} \\ &=&\sqrt{\frac{9+\sqrt{9^{2}-80}}{2}}-\sqrt{\frac{9-\sqrt{9^{2}-80}}{2}} \\ &=&\sqrt{\frac{9+1}{2}}-\sqrt{\frac{9-1}{2}}\\ &=&\sqrt{5}-2. \end{eqnarray*}

ADDED. Note: If the radical were of the form $\sqrt{a+\sqrt{b}}$, then the applicable identity would be

\begin{equation*} \sqrt{a+\sqrt{b}}=\sqrt{\frac{a+\sqrt{a^{2}-b}}{2}}+\sqrt{\frac{a-\sqrt{ a^{2}-b}}{2}}.\tag{2} \end{equation*}

Proof (from Sebastião e Silva, Silva Paulo, Compêndio de Álgebra II, 1963). To find two rational numbers $x,y$ such that

\begin{equation*} \sqrt{a+\sqrt{b}}=\sqrt{x}+\sqrt{y},\text{ with }a,b\in \mathbb{Q}, \end{equation*}

we square both sides and rearrange the terms

\begin{equation*} 2\sqrt{xy}=a-x-y+\sqrt{b}. \end{equation*}

Squaring again yields \begin{equation*} 4xy=\left( a-x-y\right) ^{2}+2\left( a-x-y\right) \sqrt{b}+b. \end{equation*} Since $x,y\in \mathbb{Q}$, $a-x-y=0$, which means that $x,y$ satisfy the system of equations

\begin{equation*} x+y=a,\qquad xy=\frac{b}{4}. \end{equation*}

Consequently they are the roots of \begin{equation*} X^{2}-aX+\frac{b}{4}=0, \end{equation*}


\begin{eqnarray*} x &=&X_{1}=\frac{a+\sqrt{a^{2}-b}}{2} \\ y &=&X_{2}=\frac{a-\sqrt{a^{2}-b}}{2}. \end{eqnarray*}


A (more complicated) approach that works on any nested radical, would be to use the Zippel Denesting Theorem.

$\sqrt[4]{161-72\sqrt{5}}$ is a fourth power exponent in $\mathbb{Q}(\sqrt{5})$ so setting the radical equal to its primitive root of unity and finding its roots gives us the simplification.

So we have: $\sqrt[4]{161-72\sqrt{5}}=x\iff x^4+72\sqrt{5}-161=0\iff (\sqrt{5}-2-x)(x+\sqrt{5}-2)(4\sqrt{5}-9-x^2)=0$ with the first one giving the correct denesting of $\sqrt{5}-2$.

  • $\begingroup$ I mention some of the general denesting algorithms and structure theorems in some of my older posts, e.g. here. $\endgroup$ – Bill Dubuque Jul 31 '16 at 1:38
  • $\begingroup$ @BillDubuque Pardon for my stupidity (sometimes) but isn't the Denesting Structure Theorem something only Math Majors learn in college? $\endgroup$ – Frank Jul 31 '16 at 1:46
  • $\begingroup$ No, it is not something usually taught (although the Kummer theory needed to derive the results would be known to most algebraic number theorists). Only with relatively recent emphasis on computation (CAS) has the problem received more intensive interest (circa 1980). What is your reference for Zippel's results? $\endgroup$ – Bill Dubuque Jul 31 '16 at 1:49
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    $\begingroup$ @BillDubuque: Like, where I got it? From Susan Landau's How to Tangle with a Nested Radical document. researchgate.net/publication/… $\endgroup$ – Frank Jul 31 '16 at 3:08

I think this problem is a lot easier than the other answers would have one believe. Since $$161^2-5\cdot72^2=(161+72\sqrt5)(161-72\sqrt5)=1$$ We can see that $161+72\sqrt5$ is a unit in the ring of integers of $\mathbb{Q}(\sqrt5)$ Thus it must be $(\pm1)$ times a power of the fundamental unit, $\phi=\frac{1+\sqrt5}2$: $$161+72\sqrt5=\phi^n$$ Solving for $n$ we have $$n=\frac{\ln(161+72\sqrt5)}{\ln\left(\frac{1+\sqrt5}2\right)}=12$$ Thus $$(161+72\sqrt5)^{1/4}=\phi^3=2+\sqrt5$$


Number's Simple Denesting Rule (https://math.stackexchange.com/q/816527) is incorrect. According to it,

$ \sqrt{a + b \sqrt{n}} = \frac{a + b \sqrt{n} - \sqrt{(a + b \sqrt{n})(a - b \sqrt{n})}}{\sqrt{2 a}}.$

The latter formula is equivalent to

$\sqrt{a + b \sqrt{n}} - \sqrt{a - b \sqrt{n}} = \sqrt{2a},$

which is not an identity.

Moreover, according to Number's definition of the trace of an expression $a + b \sqrt{n}$, the trace for $161 - 72 \sqrt{5}$ is $322$ (not $324$, as Number wrote) and the trace for $9 - 4 \sqrt{5}$ is $18$ (not $16$). However, these miscalculations together with the incorrect Simple Denesting Rule led to the correct results

$\sqrt{161 - 72 \sqrt{5}} = 9 - 4 \sqrt{5}$, $\qquad \sqrt{9 - 4 \sqrt{5}} = 2- \sqrt{5}$.

  • $\begingroup$ Can you complete your objection? $\endgroup$ – max_zorn Jan 19 '18 at 8:29
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    $\begingroup$ $2-\sqrt{5}<0$. Need the nonnegative root by definition of $\sqrt{}$. $\endgroup$ – Oscar Lanzi Feb 28 '19 at 2:32
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    $\begingroup$ You misread the rule. $t := \sqrt{\rm trace}$ is divided out of the intermediate expression $s$ that we obtain after subtracting out the norm, i.e. the result is $\, s/t, \ $ for $\, t = \sqrt{ {\rm tr} s} = \sqrt{s + s'}.\ $ See the proof. $\endgroup$ – Bill Dubuque Oct 28 '19 at 18:58

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