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# A Problem on Ordinary Nested Radicals

### Problem

Prove or disprove that
$\sqrt {7+\sqrt{33}} + \sqrt {6+\sqrt{35} } = \sqrt {5+\sqrt{21}} + \sqrt {8+\sqrt{55}}$

### Solution

In order to simplify the radicals, the radicands should be forced to equal square numbers (e.g., $7+\sqrt{33}$ should be a square of some number). Numbers whose squares have a rational and radical part are usually in the form $x+y$.
So let $\sqrt{7 +\sqrt{33}} = x+y = \sqrt{(x+y)^2} = \sqrt {x^2+y^2+2xy}$
and set
$x^2+y^2=7$ and $2xy=\sqrt{33} \, , i.e., y=\sqrt{33}/2x$
Thus $x^2 + \left ( \frac {\sqrt{33}}{2x} \right )^2 =7$
which on simplification yields $x=\sqrt{22}/2$
And also $y=\sqrt{6}/2$

Thus,
$\mathbf {\sqrt{7+\sqrt{33}} = \frac {\sqrt{22}+\sqrt{6}}{2} }$
Using the same process for other radicals:
$\mathbf {\sqrt{6+\sqrt{35}} = \frac{\sqrt{10}+\sqrt{14}}{2} }$
$\mathbf {\sqrt{8+\sqrt{55}} = \frac{\sqrt{10}+\sqrt{22}}{2} }$
$\mathbf {\sqrt{5+\sqrt{21}} = \frac{\sqrt{6}+\sqrt{14}}{2} }$
Thus, now we can easily prove (by addition) that

$\sqrt {7+\sqrt{33}} + \sqrt {6+\sqrt{35}} =\sqrt {5+\sqrt{21}} + \sqrt {8+\sqrt{55}}$

## 1 Comment

1. quentin bennett says:

Thxs, this was good & sorely needed practice!