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Given the Dedekind eta function $\eta(\tau)$ and complex numbers a,b with imaginary part > 0, anybody knows how to prove the proposed identity,

$$\sum_{k=0}^{p-1} e^{2\pi i k/4}\eta^3\big(\tfrac{a+k}{p}\big)\eta^3\big(\tfrac{b+k}{p}\big) = p^3\eta^3(p a)\eta^3(p b)$$

where p is ANY prime of form $p = 4n-1$.

(This was inspired by Berndt and Hart's paper "An Identity for the Dedekind eta function involving two independent complex variables" (2007) wherein they discussed the case p = 3 but not p = 7, 11, etc.)

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    $\begingroup$ The expansion of $\eta^3$ involves only terms $\pm n q^{n^2/8}$ with $n$ odd. The sum over $k$ seems to isolate terms with $n^2 + {n'}^2$ divisible by $p$, and if $p\equiv -1 \bmod 4$ then $p|n^2 + {n'}^2$ if and only if $p|n$ and $p|n'$, from which the identity should follow. $\endgroup$
    – Noam D. Elkies
    Sep 11, 2012 at 17:16
  • $\begingroup$ Thank you, Dr. Elkies. Hart was kind enough to reply to my query, and said that a more recent paper, homepages.warwick.ac.uk/~masfaw/newidentity.pdf (Example 1) showed it was a special case true for all odd prime p. $\endgroup$
    – Tito Piezas III
    Sep 12, 2012 at 16:14

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