For the Dedekind eta function, defined as usual by $\eta(q) = q^{\frac1{24}} \prod\limits_{n=1}^{\infty} (1-q^{n}) $, let for brevity $e_k:=\eta(q^k)$. With this notation, a blog entry of Michael Somos gives three beautiful identities for sums of $\eta$-products where all exponents are only $0$ or $1$:

$I_{60}:\qquad \ \ e_{1}e_{12}e_{15}e_{20} + e_{3}e_{4}e_{5}e_{60}=e_{2}e_{6}e_{10}e_{30} $

$I_{210}:\qquad e_{1}e_{30}e_{35}e_{42 }+ e_{3}e_{10}e_{14}e_{105} = e_{2}e_{15}e_{21}e_{70} +e_{5}e_{6}e_{7}e_{210}$

$I_{30}:\qquad\ \ e_{1}e_{3}e_{5}e_{15}+ 2e_{2}e_{6}e_{10}e_{30}=e_{1}e_{2}e_{15}e_{30}+ e_{3}e_{5}e_{6}e_{10} $

For $I_{60}$ and $I_{30}$ the structure is clear at one glance if we write the divisors of $60$ as vertices of a Cayley-like graph (here: 'union' of two cube graphs $Q_3$ with the common face $(2,6,30,10)$ ):

```
4 20
2 10
1 5
3 15
6 30
12 60
```

Alternatively, if we define

$a_0:=e_{1} e_{15} \qquad b_0:= e_{3} e_{5} $

$a_1:=e_{2} e_{30} \qquad b_1:= e_{6} e_{10} $

$a_2:=e_{4} e_{60} \qquad b_2:= e_{12} e_{20} $

then

$I_{60}\iff a_0b_2+b_0a_2=a_1b_1$

$I_{30}\iff a_0a_1+b_0b_1=a_0b_0+2a_1b_1$.

For $I_{210}$ the symmetry is a bit less obvious to see. We can identify the divisors of $210$ with the vertices of a tesseract graph $Q_4$ or write the factors of the four products as lines of a matrix and note $a_{i,j}a_{4-i,j}=210$ as well as the factor $3$ between the two pairs of lines:

```
1 30 35 42
3 10 105 14
70 21 2 15
210 7 6 5
```

I'd suggest to call identities of this type

linear eta product identities. Their linearity seems to enforce a high degree of symmetry in the way these three identities $I_n$ feature the divisors of $n$, which makes them very special among the thousands of known eta product identities.It looks like there is something deeper behind. And:Why do all products have exactly $4$ factors?

So, more precisely, for naturals $a\ge b$ let's define a **linear eta product identity of type** $\mathbf{(a,b)}$ as an identity $L_1+\cdots+L_a=R_1+\cdots+R_b$, where each $L_i$ and each $R_i$ is a finite product of pairwise different terms of form $\eta(q^{\lambda})$ with $\lambda\in\mathbb N$. (The products $L_i$ and $R_i$ don't need to be all different, e.g. the above $I_{30}$ is of type $(3,2)$ with $L_2=L_3$. But of course we want $\{L_i\}\cap\{R_i\}=\emptyset$, and also that the gcd of all the $\lambda$'s is $1$.)

Somos conjectures that $I_{60}$ is the only linear identity of type $(2,1)$.

Is it possible that the three above identities are only the first ones of a whole (infinite?) set of linear eta product identities, and/or that for naturals $a\ge b$, there is at most one such identity of type $(a,b)$?

Mathematicacouldn't seem to find anything. Sigh. – Tito Piezas III Feb 1 '14 at 1:15that seems to cry for generalization, but in between I have gained the impression that in spite of the thousands of existing eta-identities, everything is finite there in terms of re-occuring patterns. – Wolfgang Feb 1 '14 at 14:32