Entry 155

It turns out we can use the previous entries to parameterize the equation $$x^n+y^n = z^n$$ for \(n=2,3,4\). Given the square of the nome \(q = e^{2\pi i\tau}\) and assume the functions \(A(q), B(q), C(q)\). Define

$$\gamma = \left(\frac{8}{\left(\tfrac{\eta(\tau/2)}{\eta(2\tau)}\right)^8+8}\right)^2 = \left(\frac{C(q)}{A(q)}\right)^2$$

Then we conjecture,

$$\begin{align}\left(\frac{C(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=\gamma\\ \left(\frac{B(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1-\gamma\\ \left(\frac{A(q)}{_2F_1\big(\tfrac14,\tfrac34,1,\gamma\big)^2}\right)^2 &\overset{\color{red}?}=1\end{align}$$

where \(C(q), B(q), A(q)\) are defined by the relation and eta quotients below,

$$\big(C(q)\big)^2+\big(B(q)\big)^2=\big(A(q)\big)^2$$

$$\left(\frac{8\eta^8(2\tau)}{\eta^4(\tau)}\right)^2+\left(\frac{\eta^8(\tau)}{\eta^4(2\tau)}\right)^2=\left(\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\right)^2$$ Note also that $$\begin{align}A(q) &= 1+24\sum_{n=1}^\infty\frac{n q^n}{1+q^n}\\ &=\big(\vartheta_2(0,q)\big)^4+\big(\vartheta_2(0,q)\big)^4\\ &=\frac{\eta^8(\tau)+32\eta^8(4\tau)}{\eta^4(2\tau)}\end{align}$$ where, in this particular instance, we let the Jacobi theta functions \(\vartheta_n(0,q)\) use \(q = e^{2\pi i\tau}\).