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

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