<s>[INST] About equality of nested radicals. Allow me, please, reformulate this problem.
The equal numbers $$a=\sqrt{13}+\sqrt{10+2\sqrt{13}}$$
$$b=\sqrt{5+2\sqrt3}+\sqrt{18-2\sqrt3+2\sqrt{65-26\sqrt3}}$$ have the same minimal polynomial (over $\mathbb Z$) $$f(x)=x^4-46x^2-104x-43$$ in which it appears  just one sign change so the equation $f(x)=0$ has at most a positive root (Descartes' Sign Rule). Thus, since $a>0$ and $b>0$ we verify that $a=b$.
This elegant way is not possible to be apply to the equal numbers $$\alpha=\sqrt {29}+\sqrt{14+2\sqrt {29}}$$ $$\beta=\sqrt{7+2\sqrt 5}+\sqrt{36-2\sqrt 5+2\sqrt{203-58\sqrt 5}}$$ whose common minimal polynomial is $$f(x)=x^4-86x^2-232x+109$$
How to verify using $f(x)$ that $\alpha=\beta$ ?
It is not hard but, is there a method something elegant (like the first given example) to do it?
 [/INST] The two examples are of the following form:
Take $a,b,c\in \mathbb{R}$ such that $a=b^2-4c$, and suppose that all the square roots below are well defined. Then if
$$\alpha=\sqrt{a}+\sqrt{2b+2\sqrt{a}}$$
and 
$$\beta=\sqrt{b+2\sqrt{c}}+\sqrt{a+b-2\sqrt{c}+2\sqrt{ab-2a\sqrt{c}}}$$
we have $\alpha=\beta$.
We have only to see that $$a+b-2\sqrt{c}+2\sqrt{ab-2a\sqrt{c}}=\big (\sqrt{a}+\sqrt{b-2\sqrt{c}} \big)^2$$
to simplify by $\sqrt{a}$, and then to take $\sqrt{b-2\sqrt{c}}+ \sqrt{b+2\sqrt{c}}$ and $\sqrt{2b+2\sqrt{a}}$  to the power two.
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