Entry 190

From Entry 184, the quintic \(y(y-5)^4 =  j_2(\tau)\) has a solvable Galois group. For example

$$\begin{align}y(y-5)^4 &= -(4\sqrt2)^4\\ y(y-5)^4 &=  -(12\sqrt2)^4\\ y(y-5)^4 &= 12^4\\ y(y-5)^4 &= 396^4\qquad\end{align}$$

But these can also be expressed as

$$\begin{align}z^4(z+5) &= (4\sqrt2)i\\ z^4(z+5) &=  (12\sqrt2)i \\ z^4(z-5) &= 12\\ z^4(z+5) &= 396\qquad\end{align}$$

In general, since there is a relationship between \(j_2(\alpha)\) and \(j_4(\beta)\) where the latter is expressible by Ramanujan's \(G_m\)-function, then we conjecture that the special Bring-Jerrard quintics below are solvable

$$\begin{align}z^5+5z &= \sqrt{2^3\left(\frac1{G_m^{12}}-G_m^{12}\right)}\\ z^5-5z &= \sqrt{2^3\left(\frac1{g_m^{12}}+g_m^{12}\right)}\end{align}$$

For example, \(G_5 = \left(\tfrac{1+\sqrt5}2\right)^{1/4}\,\) and \(g_{10} = \left(\tfrac{1+\sqrt{5}}2\right)^{1/2}\) yields

$$\begin{align}z^5+5z &= (4\sqrt2)i\\ z^5-5z &= 12\qquad\end{align}$$

respectively, where the latter is quite well-known as an example of a solvable quintic with small integer coefficients. And so on for all Ramanujan \(G\) and \(g\)-functions, with explicit examples in the previous entries.