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Define the Ramanujan theta function $f(a,b)$ as,

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

and the Dedekind eta function,

$$\eta(\tau) = q^{1/24}\prod_{n=1}^{\infty} \left(1-q^n\right)$$

A. In Ramanujan's Notebooks III, p. 372, Entry 8 he gives a series of identities which can be summarized as follows: let $q = e^{2\pi i \tau}$ and

$$h_k = (-1)^{k-1}q^{k(-13+3k)/26}\,\frac{f(-q^{2k},-q^{13-2k})}{f(-q^{k},-q^{13-k})} \;\;\text{for}\;k = 1,\dots,6.$$

Then the 3 roots of the cubic

$$\left(\frac{\eta(\tau)}{\eta(13\tau)}\right)^2=\frac{y^3+y^2-4y+1}{y(1-y)}\tag{0}$$

are

$$y_1 = h_1h_5,\;\;y_2 = h_2h_3,\;\;y_3 = h_4h_6.$$

B. I suspected this might have an analogue for $n=25$, the last integer such that $n-1$ divides $24$. Define

$$h_k = (-1)^{k-1}q^{k(-25+3k)/50}\,\frac{f(-q^{2k},-q^{25-2k})}{f(-q^{k},-q^{25-k})}\;\;\text{for}\;k = 1,\dots,12.\tag{1}$$

One pair is a constant, $h_5 h_{10} = -1$. After some experimentation with Mathematica’s integer relations, the other five pairs were the roots of a quintic which, to my surprise, turned out to be the Emma Lehmer quintic for real cyclic fields,

$$1+(10+10v+4v^2+v^3)x+(5+15v+11v^2+5v^3+v^4)x^2\\ -2(5+5v+3v^2+v^3)x^3+v^2x^4+x^5=0.\tag{2}$$

This is solvable in radicals for any $v$. Moreover, if we let

$$v=\frac{1}{R(q^5)}-R(q^5)-1= \frac{\eta(\tau)}{\eta(25\tau)},\tag{3}$$

where $R(q)$ is the Rogers-Ramanujan continued fraction, then it can be empirically observed that the five roots are

$$ x_1,\, \dots,\, x_5 = h_1 h_7,\; h_2 h_{11},\; h_3 h_4,\; h_6 h_8,\; h_9 h_{12},\tag{4} $$

hence Emma Lehmer's quintic is analogous to Ramanujan's cubic.

Question: Anybody knows how to prove that if $v$ is defined by $(3)$, then these $x_i$ are indeed the roots?

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  • $\begingroup$ I think it would be helpful for you to check Henri Darmon's paper "Note on a polynomial of Emma Lehmer" $\endgroup$
    – user41102
    Nov 28, 2013 at 7:21
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    $\begingroup$ In B., where you say "the other five pairs", do you mean you came up with a grouping of the 10 remaining $h_k$ into five pairs that yield a "beautiful" quintic? (I see that with indices mod 25, all roots have the form e.g. $h_kh_{7k}$, noting $h_k=h_{-k}=h_{25-k}$.) $\endgroup$
    – Wolfgang
    Feb 6, 2015 at 17:41
  • $\begingroup$ @Wolfgang: Yes. I meant, "...the five other pairs". :) (P.S. if you're interested, this also has a $n=49$ version. It has $h_k$ from $k = 1,\dots24$ where one now groups the $h_k$ into triplets. One is $h_7 h_{14} h_{21} =-1$, but the seven other triplets form a septic where the coefficients involve a square root.) $\endgroup$ Feb 7, 2015 at 1:39
  • $\begingroup$ So this means the pattern is not limited to divisors of 24. Rather to multiplicative groups like $\mathbb Z_{13}^*/\pm,\mathbb Z_{25}^*/\pm,\mathbb Z_{49}^*/\pm$ etc. Maybe all $\mathbb Z_{6k+1}^*$? In which way would it depend on the group structure of $\mathbb Z_{6k+1}^*$ if the $h_k$ "must" (?) be grouped into pairs, triplets or other tuples? $\endgroup$
    – Wolfgang
    Feb 7, 2015 at 10:26
  • $\begingroup$ @Wolfgang: I used $r=\frac{24}{n-1}$ as a heuristic. For $n=49$ hence $r = 24/48 = 1/2$ it was my clue that the coefficients might involve a square root of an expression. And it did. It probably will generalize to other $n=6k+1$, but will be quite complicated. $\endgroup$ Feb 7, 2015 at 13:47

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