I have been working problems out of Ram Murty's "Problems in the Theory of Modular Forms," which has been marvelous. His solutions are complete, but are way too slick for satisfaction (probably by his design). So, I'm taking the exercise of filling in the gaps of understanding.
Exercise 1.5.1 has had me exploring for some time. The intent appears to be merely an application of Jacobi's Triple Product Formula. It just seems like overkill to me and I'm wondering if there's a more forward way using triangular numbers. Here's the problem.
Define $\tau(n)$ by $$q\Pi_{n=1}^\infty(1-q^n)^{24}=\sum_{n=1}^\infty\tau(n)q^n$$
Show that $\tau(n)$ is odd iff $n=(2m+1)^2$ for some $m$.
The first thing I did was change the $-$ to a $+$, since we're working $(\text{mod }2)$. Then, it's natural to use Kummer's Theorem to weed out the even binomial coefficients. This can be done in two ways,
$$\Pi_{n=1}^\infty(1+q^n)^{24}=\Pi_{n=1}^\infty(1+q^{8n})^3 (\text{mod }2)$$
or
$$\Pi_{n=1}^\infty(1+q^n)^{24}=\Pi_{n=1}^\infty(1+q^{8n}+q^{16n}+q^{24n})(\text{mod }2)$$
The first way is the one Murty uses before he wraps up summarily with Jacobi's Triple Product Formula
$$\Pi_{n=1}^\infty(1-q^n)^3=\sum_{k=0}^\infty (-1)^k(2k+1)q^{k(k+1)/2}$$
Both ways should end up using the nice touch that odd squares are one more than $8$ times triangular numbers (substitute $q^{8}$ for $q$).
More Basic Approach
If we instead want to use the top product, all that's left is to expand in another way. I'm hoping the intricacy of Jacobi's formula will unravel some, since we are working in $(\text{mod }2)$.
Particularly, my hope is to set $y=(1,2,3,4,\dots)$ and use $$\tau(8n+1) = \text{card}(X_n)$$ where $$X_n=\{x\in\{0,1,2,3\}^\infty: x\cdot y=n\}$$ Next, I think it is natural to look at sequences in $$S=\{x\in\mathbb{Z}^\infty: x\cdot y=0\}$$ which are generated by addition/subtraction of elements of the form $$(\dots, 0, 0, 1,-1, 0, 0, \dots, 0, 0, -1, 1, 0, 0, \dots)$$ Then, $\tau(8n+1)$ is something like the size of "orbits" of $S$ acting on $X_n$ (of course $S$ is not technically an action because it's not closed).
Interest
The reason I'm interested is that $X_n$ has an odd number of elements precisely when $n$ is a triangular number, i.e. when $(1,1,\dots,1,1,0,0,0,\dots)\in X_n$. So, I'm hoping there is a nice involution of $X_n$ that only fixes the point beginning with $n$ 1's, then all zeros, $(1,1,\dots,1,1,0,0,0,\dots)$.
The parity problem would be solved immediately. Please let me know if you find such an involution, or if you have other comments or approaches regarding this problem. Thank you very much for reading!