TITLE: Direct proof of the level four eta identity
QUESTION [4 upvotes]: Of Somos' over 6200 eta identities, the one of smallest level (maybe the simplest?) is the eta identity of level $4$ $$\eta(z)^{16} \eta(4z)^8 + 16 \eta(4z)^{16} \eta(z)^8 = \eta(2z)^{24}.$$
As with all identities among modular forms, we can prove it by checking the first few coefficients. I wonder if it is possible to give a "real" proof of this straight from the product $$\eta(z) = q^{1/24} \prod_{n=1}^{\infty} (1 - q^n)$$ in the style of this answer. The higher exponents in this problem seem to make it harder to use Jacobi's triple product.

REPLY [2 votes]: I am surprised at the idea that this needs "real" proof. The identity is the very first one in my list and the very first formulation of it is the "aequatio identica satis abstrusa" straight from Jacobi's Fundamenta Nova Theoriae Functionum Ellipticarum, page 90 equation (14) published in 1829. There is a reference to Bruce Berndt's editing of Ramanujan's Notebooks, Part III, page 40, Entry 25 (vii) which is exactly the same that is used in the parametrization of the Arithmetico-Geometric Mean by squares of the theta null functions. That is, $\;\theta_3^4(q) = \theta_4^4(q) + \theta_2^4(q)\;$ is the $a^2=b^2+c^2$ of the AGM and this goes back to Gauss even before Jacobi.
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