Combinatorial interpretation of this identity of Gauss?

Solution 1:

There is a classical proof by Andrews which you can find in my survey here (section 5.5). There is also a bijective proof of a more general identity I gave in this paper (section 2.2). Enjoy!

Solution 2:

Here's a combinatorial interpretation, but I have no idea how to turn it into a combinatorial proof.

$\prod(1+q^k)$ is the generating function that counts the number of partitions into distinct parts. $\prod(1-q^k)$ is the generating function that counts the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts. Thus

$$\sum_{n\in\mathbb{Z}}(-1)^nq^{n^2}\prod_{k\geq 1}\left(1+q^k\right)=\prod_{k\geq 1}\left(1-q^k\right)$$

considers partitions into a square and distinct parts and states that the excess of such partitions with an odd square over such partitions with an even square (where each non-zero square occurs in two colours, positive and negative) is equal to the excess of partitions into an even number of distinct parts over the partitions into an odd number of distinct parts.

Solution 3:

The typical analytic proof is not difficult and is an easy consequence of Jacobi's triple product $$\sum_{n=-\infty} ^{\infty} z^{n} q^{n^{2}}=\prod_{n=1}^{\infty}(1-q^{2n})(1+zq^{2n-1})(1+z^{-1}q^{2n-1})$$ for all $z, q$ with $z\neq 0,|q|<1$. Let's put $z=-1$ to get the sum in question. The corresponding product is equal to $$\prod(1-q^{2n})(1-q^{2n-1})^{2}=\prod(1-q^{n})(1-q^{2n-1})=\prod \frac{(1-q^{2n})(1-q^{2n-1})} {1+q^{n}}=\prod\frac{1-q^{n}} {1+q^{n}}$$ which completes the proof.

The proof for Jacobi's triple product is non-trivial / non-obvious and you may have a look at the original proof by Jacobi in this blog post.

On the other hand Franklin obtained a nice and easy combinatorial proof of the Euler's Pentagonal theorem which is equivalent to Jacobi Triple Product.