Simplifying Ramanujan-type Nested Radicals

Landau's algorithm: http://ieeexplore.ieee.org/xpl/articleDetails.jsp?arnumber=63496 $\qquad\qquad$

https://en.wikipedia.org/wiki/Susan_Landau

Considering

$$(\sqrt[4]{a} \pm \sqrt[4]{b})^{4}= a+b+6\sqrt{ab} \pm 4\sqrt[4]{ab}(\sqrt{a}+\sqrt{b})$$

Making a factor in the form of $\sqrt{m}+\sqrt{n}$, $$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{a}}{\sqrt{b}} \implies \frac{a}{b}=5$$

Then $$\left( \frac{\sqrt[4]{5}+1}{\sqrt[4]{5}-1} \right)^{4}= \frac{(\sqrt{5}+1)(6+4\sqrt[4]{5})} {(\sqrt{5}+1)(6-4\sqrt[4]{5})}= \frac{3+2\sqrt[4]{5}} {3-2\sqrt[4]{5}}$$

Note on the symmetric roles of $a$ and $b$: $$\displaystyle \frac{6\sqrt{ab}}{a+b}=\frac{\sqrt{b}}{\sqrt{a}} \implies \frac{b}{a}=5$$ which gives the same result.

Hence, there's no other cases similar to $(2)$.

Once you have the formula, it's routine (if tedious) to check it by expanding out. Finding a nice formula is more challenging. In Maple, you could denest the left side of (1) as follows:

Q:= sqrt((1+4^(1/5))/5^(1/5));
convert(simplify(convert(Q,RootOf)),radical);

$$ \frac{1}{10}\,{5}^{2/5} \left( {4}^{4/5}+{4}^{3/5}+2\cdot {4}^{2/5}-2 \right) $$

EDIT:

Now let's try some reverse engineering. Say you wanted to find a nice formula for a square root where the numerator of the right side involved some linear combination of fifth roots of powers of $2$ with small integer coefficients. We might look at

$$R = a_0 + 2^{1/5} a_1 + 2^{2/5} a_2 + 2^{3/5} a_3 + 2^{4/5} a_4$$ We can square this and extract the coefficients of $2^{i/5}$, $i=0\ldots 4$: $R^2 = \sum_{i=0}^4 c_i 2^{i/5}$ where $$ \eqalign{c_0 &= {a_{{0}}}^{2}+4\,a_{{1}}a_{{4}}+4\,a_{{2}}a_{{3}}\cr c_1 &= 2\,a_{{0}}a_{{1}}+4\,a_{{2}}a_{{4}}+2\,{a_{{3}}}^{2}\cr c_2 &= 2\,a_{{0}}a_{{2}}+{a_{{1}}}^{2}+4\,a_{{3}}a_{{4}}\cr c_3 &= 2\,a_{{0}}a_{{3}}+2\,a_{{1}}a_{{2}}+2\,{a_{{4}}}^{2}\cr c_4 &= 2\,a_{{0}}a_{{4}}+2\,a_{{1}}a_{{3}}+{a_{{2}}}^{2} }$$ We want most of these (say at least $3$ of the $5$) to be $0$. It's not easy to solve such a system of Diophantine equations, but we can resort to brute force: try all cases where $a_i \in \{-2,-1,0,1,2\}$. In Maple it takes practically no time. Thus one such case is Ramanujan's $$ (a_0, \ldots, a_4) = (-1,1,0,1,1)$$ which makes $$ (c_0, \ldots, c_4) = (5,0,5,0,0)$$ Another is $$ (a_0, \ldots, a_4) = (2, 0, 2, 2, -1)$$ which makes $$ (c_0, \ldots, c_4) = (20,0,0,10,0) $$ i.e. $$ 20 + 10 \cdot 2^{3/5} = (2 + 2 \cdot 2^{2/5} + 2 \cdot 2^{3/5} - 2^{4/5})^2 $$ Divide by $100$ and take square roots: after checking the sign is right, this says $$ \sqrt{\dfrac{1 + 2^{-2/5}}{5}} = \dfrac{1 + 2^{2/5} + 2^{3/5} - 2^{-1/5}}{5}$$ I don't know if it's as nice as Ramanujan's formula, but I like it.

Or maybe you'd prefer

$$ \sqrt{\dfrac{4\cdot 2^{3/5} - 3 \cdot 2^{2/5}}{5}} = \dfrac{4 - 2^{1/5} - 2 \cdot 2^{2/5} + 2\cdot 2^{3/5}}{5} $$

or

$$ \sqrt{8+5 \cdot 3^{1/6}+3^{1/2}} = \frac{1+2 \cdot 3^{1/6}+3^{1/3} - 3^{1/2} + 3^{5/6}}{\sqrt{2}}$$