Entry 95

Level 4. Using the eta quotients $(a,b)$ from Entry 94 we found that $-a^8+b^8=c^8$ while their ratio is

$$\begin{align}\text{K}_4(q) &= \frac{a}{b} = \sqrt2\,q^{1/8} \prod_{n=1}^\infty\frac{(1-q^{4n-1})(1-q^{4n-3})}{(1-q^{4n-2})(1-q^{4n-2})} = \frac{\sqrt2\,\eta(\tau)\,\eta^2(4\tau)}{\eta^3(2\tau)}\\ &= \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+\cfrac{q+q^2} {1+\cfrac{q^3} {1+\cfrac{q^2+q^4} {1+\ddots}}}}} = \cfrac{\sqrt2\,q^{1/8}} {1+\cfrac{q} {1+q+\cfrac{q^2} {1+q^2+\cfrac{q^3} {1+q^3+\cfrac{q^4} {1+q^4+\ddots}}}}}\end{align}$$

For appropriate $\tau$, then $(a, b, \text{K}_4)$ are radicals. The formula for the j-function using $\text{K}_4(q)$ employs polynomial invariants of the octahedron and the integer $24$ of $b= q^{-1/24}B(q)$ in Entry 94 reflects the order $24$ of the octahedral group. Note that the $8$th power of the reciprocal of $\text{K}_4(q)$ without the $\sqrt2$ is the McKay-Thompson series of class 4C of the Monster (A007248)

$$\left(\frac{\eta^3(2\tau)}{\eta(\tau)\,\eta^2(4\tau)}\right)^8-16=\left(\frac{\eta(\tau)}{\eta(4\tau)}\right)^8$$