# Simplifying $\sqrt{161-72 \sqrt{5}}$

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

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?

#### Solutions Collecting From Web of "Simplifying $\sqrt{161-72 \sqrt{5}}$"

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

Denestings of $\sqrt{a+b\sqrt{n}}\,$ can be found by a simple formula that I discovered in my youth.

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

\!\!\!\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}

Here $\:161-72\sqrt 5\:$ has norm $= 1.\:$ $\rm\ \color{blue}{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{blue}{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 to get the positive square-root yields the result. We chose the sign of the square-roots so that the arithmetic is simplest. Any choice will work, e.g. here. For further worked examples see many prior posts on denesting. Below is a sketch of a proof of the rule.

Lemma $\ \ \ \sqrt w\, =\, \dfrac{s}t,\quad \begin{eqnarray}s &=& w \pm \sqrt{ww’}\\ t &=& \pm\sqrt{s+s’}\end{eqnarray}\ \$ when $\ \ \sqrt{ww’}\in\Bbb Q$

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

Another approach. We can apply twice the following general algebraic identity involving nested
\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*}
\end{equation*}

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

i.e.

\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 $\sqrt{5}-2$.