Entry 25

Let \(\eta(\tau)\) be the Dedekind eta function. In his Lost Notebook, Ramanujan came up with the nice evaluations$$\begin{aligned}\eta(i) &= \frac{\Gamma\big(\tfrac{1}{4}\big)}{2\,\pi^{3/4}}\\\eta(2i) &= \frac{\eta(i)}{2^{3/8}}\\ \eta(3i) &= \frac{\eta(i)}{3^{3/8}\,(2+\sqrt{3})^{1/12}} \\ \eta(4i) &= \frac{\eta(i)}{2^{13/16}\,(1+\sqrt{2})^{1/4}} \end{aligned}$$Extending this list, we find $$\begin{aligned}\eta(5i) &= \frac{\eta(i)}{\sqrt{5}}\left(\tfrac{1+\sqrt{5}}{2}\right)^{-1/2} \\\eta(6i) &=\frac{\eta(i)}{6^{3/8}} \Big(\tfrac{5-\sqrt{3}}{2}-\tfrac{3^{3/4}}{\sqrt{2}}\Big)^{1/6} \\ \eta(7i) &= \frac{\eta(i)}{\sqrt{7}}\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)^{{1/4}}\\ \eta(8i) &= \frac{\eta(i)}{2^{41/32}} \frac{(-1+\sqrt[4]{2})^{1/2}}{(1+\sqrt{2})^{1/8}}\\ \eta(16i) &= \frac{\eta(i)}{2^{113/64}} \frac{(-1+\sqrt[4]{2})^{1/4}}{(1+\sqrt{2})^{1/16}} \left(-2^{5/8}+\sqrt{1+\sqrt{2}}\right)^{1/2}\end{aligned}$$It seems that for prime \(p=4m+1\), then \(x=\Big(\sqrt{p}\,\tfrac{\eta(p\,i)}{\eta(i)}\Big)^\color{blue}2\) is an algebraic number of degree \(\frac{p-1}{2}\), while if \(p=4m+3\) for \(p>3\), then \(y=\Big(\sqrt{p}\,\tfrac{\eta(p\,i)}{\eta(i)}\Big)^\color{blue}4\)  has degree \(\frac{p+1}{2}\). For the special case \(p=7\), note that $$y =\left(-\tfrac{7}{2}+\sqrt{7}+\tfrac{1}{2}\sqrt{-7+4\sqrt{7}} \right)\approx0.092192$$is one of the four quartic roots such that the polynomial on the \(\text{RHS}\) vanishes $$\frac{(y^2 + 5y + 1)^3(y^2 + 13y + 49)}{y} = 12^3+\frac{(y^4 + 14y^3 + 63y^2 + 70y - 7)^2}{y}$$ and the \(\text{LHS}\) assumes the value \(12^3=1728\).