Given the [*Ramanujan theta function*](https://en.wikipedia.org/wiki/Ramanujan_theta_function),


$$f(a,b) = \sum\_{n=-\infty}^\infty a^{n(n+1)/2} \; b^{n(n-1)/2}$$


Let $q = e^{2\pi i \tau}$ and assume $\tau = \sqrt{-d}$.




---


**I. Degree 5**


\begin{align}
a &= q^{11/60}\;\frac{f(-q,-q^4)}{f(-q)} \,=\, q^{11/60}\, \prod\_{n=1}^\infty \frac1{(1-q^{5n-2})(1-q^{5n-3})}\\
b &= q^{-1/60}\,\frac{f(-q^2,-q^3)}{f(-q)} = q^{-1/60}\prod\_{n=1}^\infty \frac1{(1-q^{5n-1})(1-q^{5n-4})}\\
\end{align}


where $a,b$ surprisingly are ***radicals***. They obey the nice relation,


$$a\,b^{11}-a^{11}\,b = 11a^6\,b^6 +1$$


Equivalently,


$$\Big(\frac{b}{a}\Big)^5-\Big(\frac{a}{b}\Big)^5 = \frac1{(ab)^6}+11$$


The integer $60$ reflects the order of the icosahedron.




---


**II. Degree 7**


\begin{align}
a &= -q^{61/168}\;\frac{f(-q,-q^6)}{f(-q^2)} = -q^{61/168}\prod\_{n=1}^\infty \frac{(1-q^{7n-1})(1-q^{7n-6})(1-q^{7n})}{(1-q^{2n})}\\
b &= \,q^{13/168}\;\frac{f(-q^2,-q^5)}{f(-q^2)} \,=\; q^{13/168}\;\prod\_{n=1}^\infty\frac{(1-q^{7n-2})(1-q^{7n-5})(1-q^{7n})}{(1-q^{2n})}\\ 
c &= q^{-11/168}\;\frac{f(-q^3,-q^4)}{f(-q^2)} = q^{-11/168}\,\prod\_{n=1}^\infty \frac{(1-q^{7n-3})(1-q^{7n-4})(1-q^{7n})}{(1-q^{2n})}\\ 
\end{align}


where $a,b,c$ again are radicals. They obey,


$$a^3b+b^3c+c^3a = 0$$


and the integer $168$ reflects the order of the Klein quartic. (*See also [this post](https://mathoverflow.net/questions/437994/using-the-rogers-selberg-identities-to-solve-certain-septics).*)




---


**III. Degree 9**


\begin{align}
a &= \;q^{5/9}\;\frac{f(-q,-q^8)}{f(-q^3)} = q^{5/9}\prod\_{n=1}^\infty \frac{(1-q^{9n-1})(1-q^{9n-8})(1-q^{9n})}{(1-q^{3n})}\\
b &= \;q^{2/9}\;\frac{f(-q^2,-q^7)}{f(-q^3)} \,=\, q^{2/9}\prod\_{n=1}^\infty \frac{(1-q^{9n-2})(1-q^{9n-7})(1-q^{9n})}{(1-q^{3n})}\\
c &= -q^{-1/9}\;\frac{f(-q^4,-q^5)}{f(-q^3)} = -q^{-1/9}\prod\_{n=1}^\infty \frac{(1-q^{9n-4})(1-q^{9n-5})(1-q^{9n})}{(1-q^{3n})} 
\end{align}


where $a,b,c$ still are radicals. They now obey,


$$\color{blue}{a^2b+b^2c+c^2a = 0}$$
$$a^2c+b^2a+c^2b = 1$$


(*Added later.*) The first is equivalent to,


$$\frac{b}{a}+\frac{c}{b}+\frac{a}{c} = 0$$


so the ratio of the proper functions sum to zero which does not happen with the lower degrees. More on "[*Nonic Analogues of the Rogers-Ramanujan functions*](https://core.ac.uk/download/pdf/82473664.pdf)" (though they didn't affix the factor $q^{m/9}$ to make them into radicals).




---


**IV. Degree 11**


Studied by G. Andrews in 1975, the [first of five](https://mathworld.wolfram.com/Rogers-RamanujanIdentities.html) has the form,


$$a\_1=\frac{f(-q,-q^{10})}{f(-q^4)}$$


To make it a radical, I affixed a factor $q^{m/n}.\,$ I hoped it would involve $n = 660$ thus manifesting [*V. Arnold's trinities*](https://en.wikipedia.org/wiki/ADE_classification#Trinities) regarding PSL(2,5), PSL(2,7), and PSL(2,11) with orders $60, 168, 660$, but it was only $n = 264 = 11\times24$. (*Sigh*.)




---


**V. Questions**


1. Is there anything else special about $a^2b+b^2c+c^2a = 0$?
2. In what other contexts does it appear?



On the Klein quartic and the similar $a^2b+b^2c+c^2a$?
The questions are:



> 
> 1. Is there anything else special about $a^2b+b^2c+c^2a = 0$?
> 2. In what other contexts does it appear?
> 
> 
> 


For question 1, Given the $q$-series of degree $9$,
$$ \frac ac=a^3-b^3,\quad\frac ba=b^3-c^3,\quad\frac cb=c^3-a^3 $$
which leads to the telescoping sum
$$ \frac ac+\frac ba+\frac cb = 0 $$
as noticed already. Another feature is multi-section. Let
$$ A := -q^{-1/9}\frac{f(-q^{1/3})}{f(-q^3)} = a + b + c $$
where $a,b,c$ form a 3-section of the $q$-series $A$.


For question 2, refer to
L. E. Dickson, *History of the Theory of Numbers*, Volume II, Chapter XXI, section "Impossibility of $x^3+y^3=z^3$",
pp. $545$-$550$. On page $546$ it states



> 
> $\quad$ J. A. Euler$^9$ noted that, if $\,p^3+q^3+r^3=0\,$ is
> possible, $\,x=p^2q,$ $y=q^2r,$ $z=r^2p\,$ satisfy
> $\,x/y+y/z+z/x=0\,$ or $\,x^2z+y^2x+z^2y=0.$
> 
> 
> 


Something similar happens with the Klein quartic
curve and the Fermat septic curve.


