The solution to the principal quintic via the Brioschi and Rogers-Ramanujan cfrac $R(q)$

Question

In this interesting post, it quotes a paper that to write down the complete solution (not in radicals) of the general quintic, one would need "a piece of paper as big as a large asteroid". Probably an exaggeration, but perhaps we can reduce the size.

The quartic transformation that result in the Bring-Jerrard form can apply to any equation of deg $n>4$, and to undo everything later is a mess. But because of icosahedral symmetry, there is a transformation unique to the quintic which require only square roots, hence is much tidier.

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I. From general to principal quintic

One first transforms the general quintic $P(x) = 0$ to principal form $P(y) = 0$ using a quadratic Tschirnhausen transformation $x^2+mx+n = y$. Thus, to recover the roots $x$ from $y$ need only solving a quadratic.

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II. Solution of the principal quintic

Now that we have the principal form,

$$P(y)=y^5+5ay^2+5by+c = 0$$

then, like the Bring quintic, we can solve it using the Dedekind eta function $\eta(\tau)$ and hypergeometric function $_2F_1$ but without solving any cubic or quartic at all. Instead, the Rogers-Ramanujan continued fraction $R(q)$ appears. It can be done in just $9$ steps.

First, solve the quadratic for any root $\lambda$,

$$(a^4 + a b c - b^3)\lambda^2 - (11a^3b - a c^2 + 2b^2 c)\lambda + (64a^2b^2 - 27a^3c - b c^2) = 0\tag1$$

using the principal quintic's $(a,b,c)$. Then define $(B,C,j)$ as,

$$B = \frac{-8a \lambda^3 - 72b \lambda^2 - 72c \lambda + j a^2}{a \lambda^2 + b \lambda + c} \tag2$$

$$C = \frac{1}{1728 - j} \tag3$$

$$j = \frac{(a \lambda^2 - 3b \lambda - 3c)^3}{a^2(a c \lambda - b^2\lambda - b c)} \tag4$$

Let $q = e^{2\pi i\tau}$ and the quintic roots $y_n$ are,

$$y_n = \frac{Bz+\lambda}{C^{-1}z^2-3}\tag5$$

$$z = \small{ \pm\sqrt{\frac{-C(u^2 + 4)(u^2 - 2u - 4)^2}{u^5 + 5 u^3 + 5u - 11}} = \pm\left(\frac{1}{R(q)}+R(q)\right)\sqrt{\frac{-C(u^2 - 2u - 4)^2}{v - 11}} }\tag{6}$$

with Rogers-Ramanujan continued fraction $R(q)$ and where $(u,v)$ are,

$$u = \frac{1}{R(q)}-R(q) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1\tag{7a}$$

$$\quad v = \frac{1}{R^5(q)}-R^5(q) = \left(\frac{\eta(\tau)}{\eta(5\tau)}\right)^6+11\tag{7b}$$

$$\quad\quad \tau=\tau_n = n+ \frac{_2F_1\big(\tfrac16,\tfrac56,1,1-\alpha\big)}{_2F_1\big(\tfrac16,\tfrac56,1,\alpha\big)}\sqrt{-1}, \quad (n = 0,1,2,3,4) \tag8$$

$$\qquad\qquad \frac{1728C-1}{C}=\frac{1728}{4\alpha(1-\alpha)} = 0\quad(\text{for any root}\;\alpha)\\ \tag9$$

Steps $1-5$ transform the principal quintic into the Brioschi as described in this Dec 2015 post and can be found in Duke's and Tóth's "The Splitting of Primes in Division Fields of Elliptic Curves" in page 10. (Note: There is a stumbling block in Step 4 when $a=0$, but one can get around that by redefining the first four steps.) The steps afterwards are mine and a more efficient version of this Sept 2015 post.

Thus, by passing through the Brioschi quintic, it seems one may not need an asteroid-sized piece of paper after all.

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III. Example

Given,

$$y^5+5y^2+10y+2=0$$

so $(a,b,c) = (1,2,2).\,$ Solving $(1)$, we choose $\lambda = \tfrac{1}{3}\big({-17} + \sqrt{871}\big)$ and get,

$$B = \tfrac{16}{27}\big(5071 - 179\sqrt{871}\big)$$

$$C = \tfrac{1}{43312^2}\big(615193 + 20894 \sqrt{871}\big)$$

Substituting into Steps $5$-$9$, and choosing the correct sign of $z_n$, we find the five roots $y_n$.

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IV. Question

Note that Step $6$ solves the Brioschi quintic,

$$z^5-10Cz^3+45C^2z-C^2 = 0$$

$$z = \pm\sqrt{\frac{-C(u^2 + 4)(u^2 - 2u - 4)^2}{u^5 + 5 u^3 + 5u - 11}}\tag{6a}$$

Q. But is there a way to simplify the expression in $u$ under the square root sign? Or solve for $z$ without square roots at all? (The polynomials are so familiar I feel the solution must already have been tackled by Brioschi, Klein, or their contemporaries.)

Update: As pointed out by @dxiv and @ParamanandSingh in the comments, we can indeed simplify the root $z$ as,

$$\quad z = \pm\sqrt{-C\;}\,(u^2 - 2u - 4)\left(\frac{1}{R(q)}+R(q)\right) \left(\frac{\eta(5\tau)}{\eta(\tau)}\right)^{3}\tag{6b}$$

where $u = \frac{1}{R(q)}-R(q) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1,$ so only $C$ remains under the square root, confirming my belief that $z$ must have a simpler form. (But this can't be the solution found by Brioschi or his contemporaries since they lived before Ramanujan.)

Answer 1

The question asks if the expression under the square root can be simplified. Turns out one way to answer this is to get rid of the square root in the first place.

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I. Solution of the Brioschi quintic using $R(q)$

Using lower-case $c$ for convenience,

$$z^5-10cz^3+45c^2z-c^2=0$$

Almost by accident, I found back in 2013 that if $c$ is a special rational form, then the Brioschi factors. Define,

\begin{align} c &= \frac{1}{1728-j}\\ j &= \frac{-(r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1)^3}{r^5(r^{10} + 11r^5 - 1)^5}\\ u &= \frac1r-r \end{align}

then a root of the Brioschi is,

\begin{align} z &= \frac{r^2(r^{10} + 11r^5 - 1)^2(r^6 + 2r^5 - 5r^4 - 5r^2 - 2r + 1)}{(r^{30} + 522r^{25} - 10005r^{20} - 10005r^{10} - 522r^5 + 1)}\\[7pt] &=\pm\sqrt{\frac{-c (u^2 + 4)\,(u^2 - 2u - 4)^2}{u^5 + 5u^3 + 5u - 11}} \end{align}

While the second form seems simpler, the first is preferable since there is no sign ambiguity. And all three polynomial invariants of the icosahedron appear, namely,

\begin{align} f &= r(r^{10} + 11r^5 - 1)\\[4pt] H &= r^{20} - 228r^{15} + 494r^{10} + 228r^5 + 1\\[4pt] T &= r^{30} + 522r^{25} - 10005r^{20} - 10005r^{10} - 522r^5 + 1 \end{align}

which obey the beautiful relationship,

$$1728 f^5 + H^3 = T^2$$

Let $q = e^{2\pi i \tau}$. Note that if $r = R(q)$ is the Rogers-Ramanujan continued fraction, then $j = j(\tau)$ is the j-function.

Given the Brioschi's single parameter $c=\frac{1}{1728-j(\tau)}$, the problem then is to determine $\tau$. Fortunately, we can use Ramanujan's theory of alternative elliptic functions, namely the one for signature $6$,

$$j(\tau)=\frac{1728c-1}{c}=\frac{1728}{4\alpha(1-\alpha)}$$

Easily solving for the quadratic in $\alpha$, one can find $\tau$ as,

$$\tau = \frac{{}_2F_1 \left (\tfrac{1}{6},\tfrac{5}{6},1;1-\alpha \right)}{{}_2F_1 \left(\tfrac{1}{6},\tfrac{5}{6},1;\alpha \right )}\sqrt{-1}$$

then find $r=R(q)$ using its well-known relationship to the Dedekind eta function $\eta(\tau)$,

$$\frac1r-r = \frac{\eta(\tau/5)}{\eta(5\tau)}+1$$

and all five roots $z_n$ of the Brioschi are,

$$z_n(\tau+n) = \frac{r^2(r^{10} + 11r^5 - 1)^2(r^6 + 2r^5 - 5r^4 - 5r^2 - 2r + 1)}{(r^{30} + 522r^{25} - 10005r^{20} - 10005r^{10} - 522r^5 + 1)}$$

for $n=0,1,2,3,4.$

Hence the Brioschi (and by extension the general quintic) can be solved by the Rogers-Ramanujan continued fraction $R(q)$, as was stated in a Sept 7, 2015 post. Nikos Bagis will arrive at a similar result in a Sept 30, 2015 paper.

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II. Solution of the Bring quintic using $R(q)$

As a nice consequence, an alternative solution to the Bring quintic using $R(q)$ can be found. Given,

$$x^5-x+A = 0\tag1$$

this can be transformed to the Brioschi,

$$z^5-10cz^3+45c^2z-c^2=0\tag2$$

using the rational Tschirnhausen transformation,

$$x = \frac{5(a+bz)A}{c^{-1}z^2-3}\tag3$$

Eliminating $z$ between $(2),(3)$ using resultants yields $(1)$ if $(a,b,c)$ satisfy,

\begin{align} a &= 1\\[4pt] b &= \frac{d+112+25A^2\sqrt{5d}}2\\[4pt] c &= \frac{-125A^2(d+49)+(d+175)\sqrt{5d}}{250A^2(d-392)^2}\\[4pt] d &= 3125A^4-256 \end{align}

Once the five roots $z_n$ of the Brioschi are found using the procedure in the first section, the relation yields the five roots $x_n$ of the Bring.

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III. Example

Let $x^5-x+1=0,\,$ so $A=1$. The relation,

$$x = \frac{5(a+bz)A}{c^{-1}z^2-3}$$

needs the variables,

\begin{align} a &= 1\\[4pt] b &= \frac{2981+25\sqrt{14345}}2\\[4pt] c &= \frac{-364750+3044\sqrt{14345}}{10\times12385^2} \end{align}

and relates the Bring to the Brioschi. The Brioschi's solution using $r=R(q)$ needs the variables,

\begin{align} & \alpha = \frac12\left(1+\sqrt{\frac1{1-1728c}}\right) \approx 0.9999526\\[4pt] & \tau = \frac{{}_2F_1 \left (\tfrac{1}{6},\tfrac{5}{6},1;1-\alpha \right)}{{}_2F_1 \left(\tfrac{1}{6},\tfrac{5}{6},1;\alpha \right )}\sqrt{-1} \approx 0.3920688\sqrt{-1}\\[4pt] &q = e^{2\pi i\tau}\\[4pt] & \frac1{R(q)}-R(q) = \frac{\eta(\tau/5)}{\eta(5\tau)}+1 \end{align}

and, with the formula for $z_n$, then all the needed variables are complete.

Answer 2

Too long for a comment :

We can play with mixing nested radicals and continued fraction :

Example

Let :

$$x=\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\cdot\cdot\cdot}}}}+\frac{1}{1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\cdot\cdot\cdot}}}}}+\frac{1}{1+\frac{1}{1+\frac{1}{\sqrt{1+\frac{1}{\sqrt{1+\cdot\cdot\cdot}}}}}}}}$$

Then $x$ is near from the solution of :

$$x^{2}\left(1+x+\frac{1}{x+1}+\frac{1}{1+\frac{1}{x+1}}\right)-1\simeq 0$$

Or :

$$x^{5}+5x^{4}+8x^{3}+4x^{2}-3x-2\simeq 0$$

Then use Newton's method and the general solution of the quartic .

Hope it helps .

Edit : I misread the question here there is no square roots but square :

we have :

$$\frac{1}{\left(1+\frac{1}{\left(1+\frac{1}{\left(1+\frac{1}{\left(1+1\right)^{2}}\right)^{2}}\right)^{2}}\right)^{2}+\frac{1}{\left(1+\frac{1}{\left(1+\frac{1}{\left(1+\frac{1}{\left(1+1\right)}\right)^{2}}\right)^{2}}\right)^{2}}}=x$$

Then we have :

$$x\left(\left(1+x\right)^{2}+\frac{1}{\left(1+x\right)^{2}}\right)-1\simeq 0$$

Or :

$$x^{5}+4x^{4}+7x^{3}+5x^{2}+x-2=0$$

Then use Newton's method and quartic general formula .