Entry 94

Level 4. Given \(q = e^{2\pi i \tau}\), Ramanujan's theta function \(f(a,b)\) and its one-parameter version as \(f(-q) = f(-q,-q^2)\). We have the sum-product identities,

$$\begin{align}A(q) &= \frac{f(-q,-q^3)}{f(-q)} \,=\, \sum_{n=0}^\infty \frac {q^{n^2+n}} {(q^2;q^2)_n} = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-2})(1-q^{4n-2})} = q^{-1/12}\frac{\eta(4\tau)}{\eta(2\tau)}\\ B(q) &= \frac{f(-q^2,-q^2)}{f(-q)} = \sum_{n=0}^\infty \frac {q^{n^2}} {(q^2;q^2)_n}  = \prod_{n=1}^\infty\frac{\alpha_4 }{(1-q^{4n-1})(1-q^{4n-3})} = q^{1/24}\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\end{align}$$

where \(\alpha_4 = \dfrac{(1-q^{2n})}{(1-q^{4n})}\). 

Let \(a = \sqrt2\,q^{1/12}A(q)\) and \(b= q^{-1/24}B(q)\), then \((a, b)\) are just eta quotients and for appropriate quadratic \(\tau\) are actually radicals. Furthermore, the difference of their \(8\)th powers is also an \(8\)th power $$-a^8+b^8 =-\left(\frac{\sqrt2\,\eta(4\tau)\,}{\;\eta(2\tau)}\right)^8+\left(\frac{\eta^2(2\tau)}{\eta(\tau)\eta(4\tau)}\right)^8=\left(\frac{\eta(\tau)}{\eta(2\tau)}\right)^8$$