Nested Radicals from Brazil

Question

Show that $$ \frac{ \sqrt{2} }{ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} - \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1}} = \sqrt[8]{ 1 + 2 \sqrt{ \sqrt{5} -2 } }. $$

What I've tried so far. $$ \begin{align*} E & = \frac{ \sqrt{2} }{ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} - \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1}} \\ & = \frac{1}{\sqrt{2}} \cdot \left[ \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} + 1} + \sqrt{\sqrt[4]{ \frac{\sqrt{5}+2}{4}} - 1} \right] \\ & = \frac{1}{\sqrt{2}} \cdot \left[ \sqrt{ a + 1 } + \sqrt{a -1} \right] \end{align*} $$ Squaring $E$, $$ \begin{align*} E^2 & = \frac{1}{2} \left[ a + 1 + a - 1 + 2\sqrt{a+1} \sqrt{a-1} \right] \\ & = \frac{1}{2} \left[ 2a + 2 \sqrt{a^2-1} \right] \\ & = a + \sqrt{a^2-1} \end{align*} $$ Then, isolating the radical $$ \begin{align*} E^2 - a & = \sqrt{a^2-1} \\ E^4 + a^2 - 2aE^2 & = a^2 - 1 \end{align*} $$ We obtain $a$, $$ a = \frac{1}{2} \left( E^2 + \frac{1}{E^2} \right). $$ And then, nothing useful comes up.

Answer 1

With your notations:

$E^2 = a +\sqrt{a^2-1}\\ E^8 =a^4+4a^3 \sqrt{a^2-1} +6a^2(a^2-1)+4a(a^2-1)\sqrt{a^2-1}+(a^2-1)^2\\ E^8 = 8a^4-8a^2+1+(8a^3-4a)\sqrt{a^2-1}$

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We can write that:

$2\sqrt{\sqrt{5}-2} = \dfrac{2 \sqrt{\sqrt{5}-2} \sqrt{\dfrac{\sqrt{5}+2}{4}}} {\sqrt{\dfrac{\sqrt{5}+2}{4}}} =\dfrac{1}{a^2} $

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So, we want to show that $\ E^8 = 1+\dfrac{1}{a^2}$

We want to show that $\ 8a^4-8a^2 + (8a^3-4a)\sqrt{a^2-1}=\dfrac{1}{a^2}$

We want to show that $\ (8a^5-4a^3)\sqrt{a^2-1} = 1-8a^6+8a^4$

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Let $\ x=(8a^5-4a^3)\sqrt{a^2-1}\ $ and $\ y=1-8a^6+8a^4\ $. It's easy to check that $x$ and $y$ are positive.

$x^2-y^2 = (8a^5-4a^3)^2 (a^2-1) - (1-8a^6+8a^4)^2\\ x^2-y^2 = (64a^{10}-64a^8+16a^6)(a^2-1)-(64a^{12}-128a^{10}+64a^8-16a^6+16a^4+1)\\ x^2-y^2 = 16a^8-16a^4-1$

And, now, it's easy to check that $x=y$.