Entry 97

Level 5 In Entry 96, we asserted that $a=q^{11/60}H(q)$ and $b= q^{-1/60}\,G(q)$ are radicals for appropriate $\tau$. Proof: Given the j-function $j(q)$ and let $R(q)=r=a/b$

$$\begin{align}a &= q^{11/60}H(q) = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{11/20}}{j(q)^{11/60}\,(r^{10}+11r^5-1)}\\ b &= q^{-1/60}G(q) =  \frac{j(q)^{1/60}}{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{1/20}}\end{align}$$ Getting their ratio $a/b$, then it simplifies to

$$r = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3/5}}{j(q)^{1/5}\,(r^{10}+11r^5-1)}$$ Raising to the fifth power and moving $j(q)$ to the LHS 

$$j(q)  = - \frac{(r^{20}-228r^{15}+494r^{10}+228r^5+1)^{3}}{r^5(r^{10}+11r^5-1)^5}$$ and we get the icosahedral equation for the j-function in terms of the Rogers-Ramanujan continued fraction $R(q)=r$. Thus if $j(q)$ and $R(q)$ are radicals, then so are $(a, b)$ while the integer $60$ of $b= q^{-1/60}G(q)$ reflects the order $60$ of the icosahedral group.