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

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, 2014 at 22:53
  • 6
    $\begingroup$ Since you are taking the $4$th root, it might be cleaner to just apply your process to square roots twice. $\endgroup$ May 31, 2014 at 22:54

6 Answers 6


Denesting $\sqrt w = \sqrt{a+b\sqrt{n}}\,$ can be done by a simple formula that I discovered as a teenager. $ {\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 $\ \ \ 9-4\sqrt 5$

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 $\,\ \ \ 2-\sqrt 5$

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

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

  • $\begingroup$ 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, 2015 at 17:35
  • 2
    $\begingroup$ @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. $\endgroup$ May 2, 2015 at 17:51
  • $\begingroup$ I can't understand how $\sqrt{1}=-1$ $\endgroup$
    – Lucas
    May 2, 2015 at 18:00
  • 1
    $\begingroup$ @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. $\endgroup$ Jun 21, 2016 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, 2021 at 14:45

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

  • 1
    $\begingroup$ Notice you can perform the same on $\sqrt{9-4\sqrt{5}}$. $\endgroup$
    – Sebastien
    May 31, 2014 at 22:56
  • $\begingroup$ I guess my brain isn't working this late,fixed $\endgroup$
    – kingW3
    May 31, 2014 at 23:03

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. $\endgroup$ Jul 31, 2016 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, 2016 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? $\endgroup$ Jul 31, 2016 at 1:49
  • 1
    $\begingroup$ @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, 2016 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, 2018 at 8:29
  • 1
    $\begingroup$ $2-\sqrt{5}<0$. Need the nonnegative root by definition of $\sqrt{}$. $\endgroup$ Feb 28, 2019 at 2:32
  • 2
    $\begingroup$ 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. $\endgroup$ Oct 28, 2019 at 18:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .