Entry 200

There is a surprising relationship between solvable quintics and Pythagorean triples. Given

$$y^5+Ay^4+By^3+Cy^2+Dy+E=0$$

then there is an extremely broad solvable class with four free parameters \((A,B,C,D)\) and \(E\) is only a quadratic with respect to the others. For simplicity, transform the quintic into depressed form,

$$x^5 + 10c x^3 + 10d x^2 + 5 e x + f = 0$$

If \(c\neq0\) and the coefficients form the Pythagorean triple,

$$(c^3 - 25 c^4 - 16 c d^2 - c e + 10 c^2 e + d^2 - e^2)^2 + (2c^2 d - 2c f + 2d e)^2 = \\ (c^3 + 25 c^4 + 16 c d^2 - c e - 10 c^2 e + d^2 + e^2)^2$$ then the quintic is solvable in radicals. Expanding, the above becomes

$$(-25c^6 - 40c^3d^2 - 16d^4 + 35c^4e + 28c d^2e - 11c^2e^2 + e^3 - 2c^2d f - 2d e f + c f^2)\times c =W\times c = 0$$

where \(W\) is simply the constant term of Watson's sextic resolvent equated to zero. (We just multiplied it by \(c\) to get the Pythagorean form.)

Example. Let \((c,d,e)=(1,2,-3)\), then \(W\) factors as \((f-28)(f+36)=0\). So another nice feature of these quintics is they come in pairs. Hence

$$x^5 + 10 x^3 + 20 x^2 - 15x +28 = 0\\ x^5 + 10 x^3 + 20 x^2 - 15x - 36 = 0$$

are irreducible but solvable in radicals. And using the formulas above, their \((c,d,e,f)\) yield the triple,

$$120^2+64^2 = 136^2$$

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.

Entry 198

In Entry 190 and Entry 191, two relations between solvable quintics and \(G_m\) and \(g_m\) were proposed. Going higher,

Conjecture: The following sextics have a solvable Galois group

$$\begin{align}\frac{x^6+10x^3+5}x &= 4\left(\frac{4}{G_{m}^{16}}-G_{m}^{8}\right) \\ \frac{x^6+10x^3+5}x &= 4\left(\frac{4}{g_{m}^{16}}+g_{m}^{8}\right) \end{align}$$

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

$$\begin{align}x^6+10x^3+5 &= -2\left(-25+13\sqrt5\right) x \\ x^6+10x^3+5 &= -6\left(-65+27\sqrt5\right) x \end{align}$$

which are solvable in radicals, and so on.

Entry 197

From Entry 195

$$2^{-1/4}G_{71} = x,\quad \text{where}\, x^7 - 2x^6 - x^5 + x^4 + x^3 + x^2 - x - 1=0$$

We will now give the radical solution to this septic using the well-known cubic \(r^3+r^2-2r-1=0\) with roots $$r_1,\,r_2,\,r_3 = 2\cos\big(\tfrac{2\pi}7\big),\, 2\cos\big(\tfrac{4\pi}7\big),\, 2\cos\big(\tfrac{6\pi}7\big)$$Define the function

$$\begin{align}y_n &= P(r_n) =\frac{29323 r^2 - 20538 r - 15494 + (1193 r^2 - 1048 r - 730)\sqrt{7\times71}}2\\ y_{n+3} &= Q(r_n) =\frac{29323 r^2 - 20538 r - 15494 - (1193 r^2 - 1048 r - 730)\sqrt{7\times71}}2\end{align}$$

For example, \(y_1 = P(r_1) \approx 219.6454\). Then the real root of the septic is

$$\quad x = \frac{2 + y_1^{1/7}+y_2^{1/7}+y_3^{1/7}+y_4^{1/7}+y_5^{1/7}+y_6^{1/7}}7 = 2.1306068\dots$$

In fact, the \(y_n\) are the roots of a sextic with rather large coefficients,

$$y^6+ay^5+by^4+cy^3+dy^2+ey-461^7 = 0$$

and where the constant term is a \(7\)th power. This sextic is special since it can factor either over the square root extension \(\sqrt{7\times71}\) or the cubic extension \(2\cos\big(\tfrac{2\pi}7\big)\).

Entry 196

In Entry 194

$$2^{-1/4}G_{47} = x,\quad \text{where}\; x^5 - x^3 - 2x^2 - 2x - 1 = 0$$Ramanujan asked for the radical solution of this quintic. We give our version which partly uses the golden ratio \(\phi\)

$$\begin{align}y_1 &= \frac{1}{2\phi}\left(13+\frac{\sqrt{47(2+89\sqrt5)}}{5^{5/4}}\right)\\ y_2 &= \frac{1}{2\phi}\left(13-\frac{\sqrt{47(2+89\sqrt5)}}{5^{5/4}}\right)\\ y_3 &= \frac{\phi}{2}\left(13+\frac{\sqrt{47(-2+89\sqrt5)}}{5^{5/4}}\right)\\ y_4 &= \frac{\phi}{2}\left(13-\frac{\sqrt{47(-2+89\sqrt5)}}{5^{5/4}}\right) \end{align}$$

then the real root of quintic is

$$2^{-1/4}G_{47} = x = \frac{y_1^{1/5}+y_2^{1/5}+y_3^{1/5}+y_4^{1/5}}{\sqrt5} = 1.73469134\dots$$

In fact, the \(y_n\) are the four roots of the quartic

$$3125y^4 - 40625y^3 - 521250y^2 + 55250y - 1= 0$$

where the leading term is a \(5\)th power \(3125 = 5^5\).