Entry 199

Combining the work of Felix Klein and Ramanujan, given the Ramanujan \(G_m\) and \(g_m\)-functions. 

Conjecture: The following septics have a solvable Galois group

$$\begin{align}x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= 4\left(\frac{4}{G_{m}^{16}}-G_{m}^{8}\right) \\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3x &= 4\left(\frac{4}{g_{m}^{16}}+g_{m}^{8}\right) \end{align}$$

This is a version of Entry 148. For example, \(G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}\), \(G_{13} = \left(\tfrac{3+\sqrt{13}}2\right)^{1/4}\), and \(G_{37} = \left(6+\sqrt{37}\right)^{1/4}\)  yields

$$\begin{align}x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -2\left(-25+13\sqrt{5}\right)\\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -30\left(-31+9\sqrt{13}\right) \\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -60\left(-2837+468\sqrt{37}\right) \end{align}$$

while \(g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}\) and  \(g_{58} = \left(\tfrac{5+\sqrt{29}}2\right)^{1/2}\) yields

$$\begin{align}x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -6\left(-65+27\sqrt5\right)\\ x^7+7\Big(\tfrac{1-\sqrt{-7}}{2}\Big)x^4+7\Big(\tfrac{1+\sqrt{-7}}{2}\Big)^3 x &= -30\left(-140989 + 26163\sqrt{29}\right) \end{align}$$

which are all solvable in radicals, and so on.