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.

• This looks like a nice question. It would look even better with your own workings and thoughts. Feb 11 at 1:17
• What have you tried? I see some conjugates here. Feb 11 at 1:17
• I would continue to square both sides because the right side keeps on simplifying itself. Feb 11 at 1:52
• Now start from the right, make the 2 enters in the square root, put the whole RHS to power 4 at least.
– EDX
Feb 11 at 3:10

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

We can write that:

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

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 $$\ (8a^5-4a^3)\sqrt{a^2-1} = 1-8a^6+8a^4$$

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 $$x=y$$.