Let $$K(k):=\int_{0}^{\frac{\pi}{2}}\frac{d\phi}{\sqrt{1-k^2\sin^2\phi}}=\frac{\pi}{2}{ _2F_1\bigg(\frac{1}{2},\frac{1}{2},1;k^2 \bigg)}$$

where $0<k<1$.

Let $K, K′, L$ and $L′$ denote the complete elliptic integrals of the first kind associated with the moduli $k, k′, l$ and $l′$, respectively. the complementary modulus is defined as $k':=\sqrt{1-k^2}$.

Then a modular equation of degree $n$ is a relation between the moduli $k$ and $l $

$$n\frac{K'}{K}=\frac{L'}{L}$$

If $\alpha=k^2$ and $\beta=l^2$, then we say $\beta$ is of degree $n$ over $\alpha$. The multiplier $m$ is defined by $m=\frac{K}{L}$.

Ramanujan’s class invariant $G_n$ is defined by

$$G_n:=\lbrace4\alpha(1-\alpha)\rbrace^{-1/24}$$

First I calculate $G_5=(\frac{1+\sqrt{5}}{2})^{1/4}$. To calculate $G_5$ we need to use a modular equation of degree 5

$$m=1+2^{4/3}\bigg(\frac{\beta^5(1-\beta)^5}{\alpha(1-\alpha)}\bigg)^{1/24}$$ $$\frac{5}{m}=1+2^{4/3}\bigg(\frac{\alpha^5(1-\alpha)^5}{\beta(1-\alpha)}\bigg)^{1/24} $$

apply the restriction $\beta=1-\alpha$ and $G_5=\lbrace4\alpha(1-\alpha)\rbrace^{-1/24}$. We find that $$G_5=\bigg(\frac{1+\sqrt{5}}{2}\bigg)^{1/4}$$

I tried to calculate $G_{125}$ by using Schläfli's modular equation of degree $5$.

$$\bigg(\frac{u}{v}\bigg)^3+\bigg(\frac{v}{u}\bigg)^3=2 \bigg(u^2v^2-\frac{1}{u^2v^2}\bigg)$$

where $u=G_{125}$ and $v=G_{5}$. this problem reduce to find the real solution of following sextic equation

$$G^6-2\varphi^5G^5+2\varphi G+\varphi^6=0$$ where $\varphi=\phi^{1/4}=\bigg(\frac{1+\sqrt{5}}{2}\bigg)^{1/4}$ and $G=G_{125}$

How to solve this equation in radicals? Or, does there exists easy way to calculate $G_{125}$ by using modular equations?

**P.S.** I use *Mathematica*, but there is no solution in terms of radicals.