TITLE: Converting an infinite product to sum; Ramanujan $\tau$ function
QUESTION [13 upvotes]: I've gotten what seems most of the way, but I'm quite stuck at this point.

Define $\tau(n)$ by
  \begin{align*}
   q\prod_{n=1}^\infty (1-q^n)^{24} = \sum_{n=1}^\infty\tau(n)q^n.
  \end{align*}
  Prove that $\tau(n)$ is odd if and only if $n = (2m+1)^2$ for some $m$.

-

Edit: Changed due to errors pointed out in the comments. Old version below.

Recall Euler's pentagonal number theorem,
\begin{align*}
 \prod_{n=1}^\infty (1-q^n) = \sum_{k=-\infty}^\infty (-1)^kq^{k(3k-1)/2}.
\end{align*}
From this we can see
\begin{align*}
 \left(\prod_{n=1}^\infty (1-q^n)\right)^{24} = \left(\sum_{k=-\infty}^\infty (-1)^kq^{k(3k-1)/2}\right)^{24}.
\end{align*}
Now notice if $k = -j$ for some integer $j$, we have
\begin{align*}
 \frac{k(3k-1)}{2} = \frac{j(3j+1)}{2},
\end{align*}
so we may rewrite this as
\begin{align*}
\left(\prod_{n=1}^\infty (1-q^n)\right)^{24} = \left(1 + \left(\sum_{k=1}^\infty (-1)^kq^{k(3k-1)/2} + (-1)^{-k}q^{k(3k+1)/2}\right)\right)^{24}
\end{align*}
Reducing the right side modulo 2 gives
\begin{align*}
 \left(\prod_{n=1}^\infty (1-q^n)\right)^{24} &= \left(1 + \left(\sum_{k=1}^\infty (-1)^kq^{k(3k-1)/2} + (-1)^{-k}q^{k(3k+1)/2}\right)\right)^{8\cdot 3}\newline
 &\equiv \left(1 + \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)\right)^3\newline
 &\equiv 1 + \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k-1)} + q^{8k(3k+1)}\right)\newline
 &\ \ + \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)^3.
\end{align*}
Now we would like a nice expression for the last sum. Notice
\begin{align*}
 &\left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)^3 = \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)^2\left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)\newline
 &\equiv \left(\sum_{k=1}^\infty q^{8k(3k-1)} + q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)(\bmod\ 2)\newline
 &\equiv \left(\sum_{k=1}^\infty q^{8k(3k-1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k-1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k+1)}\right)\newline
 &\ \ + \left(\sum_{k=1}^\infty q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k+1)}\right) (\bmod\ 2).
\end{align*}
Plugging this back into our expression above, we have
\begin{align*}
 \left(\prod_{n=1}^\infty (1-q^n)\right)^{24} &\equiv 1 + \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k-1)} + q^{8k(3k+1)}\right)\newline
 &\ \  + \left(\sum_{k=1}^\infty q^{8k(3k-1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k-1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k+1)}\right)\newline
 &\ \  + \left(\sum_{k=1}^\infty q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k+1)}\right)\newline
 &\equiv 1 + \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k-1)} + q^{8k(3k+1)}\right)\newline
 &\ \  + \left(\sum_{k=1}^\infty q^{8k(3k-1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)} + \sum_{k=1}^\infty q^{4k(3k+1)}\right)\newline
 &\ \  + \left(\sum_{k=1}^\infty q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)} + \sum_{k=1}^\infty q^{4k(3k+1)}\right)\newline
 &\equiv 1 + \left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right) + \left(\sum_{k=1}^\infty q^{8k(3k-1)} + q^{8k(3k+1)}\right)\newline
 &\ \  + \left(\sum_{k=1}^\infty q^{8k(3k-1)} + q^{8k(3k+1)}\right)\left(\sum_{k=1}^\infty q^{4k(3k-1)} + q^{4k(3k+1)}\right)(\bmod\ 2).
\end{align*}
Now I'm not too sure if there's much I can do with this. The fact that it looks like $A + B + AB$ makes me curious though.

Edit: Applying the comment fixes to the method I tried before posting here actually solves the problem.

Recall Jacobi's formula, namely
\begin{align*}
 \prod_{n=1}^\infty (1-q^n)^3 = \sum_{k=0}^\infty (-1)^k(2k+1)q^{k(k+1)/2}.
\end{align*}
Looking modulo 2, we have
\begin{align*}
 \left[\prod_{n=1}^\infty (1-q^n)^3\right]^8 \equiv \prod_{n=1}^\infty (1-q^n)^{24} (\bmod\ 2),
\end{align*}
and so we can look at the coefficients of
\begin{align*}
 \left[\sum_{k=0}^\infty (-1)^k(2k+1)q^{k(k+1)/2}\right]^8
\end{align*}
to see where the odd coefficients lie. To this end, notice
\begin{align*}
 \left[\sum_{k=0}^\infty (-1)^k(2k+1)q^{k(k+1)/2}\right]^8 \equiv \sum_{k=0}^\infty q^{4k(k+1)} (\bmod\ 2)
\end{align*}
and so
\begin{align*}
 q\prod_{n=1}^\infty (1-q^n)^{24} &\equiv q\left[\sum_{k=0}^\infty (-1)^kq^{k(k+1)/2}\right]^8\newline
  &\equiv \sum_{k=0}^\infty q^{4k(k+1)+1}\newline
  &\equiv \sum_{k=0}^\infty q^{4k^2 + 4k + 1}\newline
  &\equiv \sum_{k=0}^\infty q^{(2k+1)^2} (\bmod\ 2).
\end{align*}
As $k\geq 0$ it is clear each term in the sum is distinct (i.e., we don't have problems modulo 2), hence the only non-zero coefficients modulo two (i.e., the odd ones) are the coefficients $2k+1$ for some non-negative integer $k$.
In case anyone makes it further than the massive chunk of equations above, I'm still interested in seeing if that method pans out, as I remember dealing with equations of the form $A + B + AB$ before, and really expect a solution that way, though my memory is faint, so I could just be crazy. Of course I'm quite happy with the simple version either way.

Old error-y version:

Recall Euler's pentagonal number theorem,
\begin{align*}
 \prod_{n=1}^\infty (1-q^n) = \sum_{k=-\infty}^\infty (-1)^kq^{k(3k-1)/2}.
\end{align*}
From this we can see
\begin{align*}
 \left(\prod_{n=1}^\infty (1-q^n)\right)^{24} = \left(\sum_{k=-\infty}^\infty (-1)^kq^{k(3k-1)/2}\right)^{24}.
\end{align*}
Now notice if $k = -j$ for some integer $j$, we have
\begin{align*}
 \frac{k(3k-1)}{2} = \frac{j(3j+1)}{2},
\end{align*}
so we may rewrite this as
\begin{align*}
 \left(\prod_{n=1}^\infty (1-q^n)\right)^{24} = 1 + \left(\sum_{k=1}^\infty (-1)^kq^{k(3k-1)/2} + (-1)^{-k}q^{k(3k+1)/2}\right)^{24}.
\end{align*}
Reducing the right side modulo 2 gives
\begin{align*}
 \left(\prod_{n=1}^\infty (1-q^n)\right)^{24} &\equiv 1 + \left(\sum_{k=1}^\infty (-1)^kq^{k(3k-1)/2} + (-1)^kq^{k(3k+1)/2}\right)^{24}(\bmod\ 2)\newline
 &\equiv 1 + \left(\sum_{k=1}^\infty q^{k(3k-1)/2} + q^{k(3k+1)/2}\right)^{8\cdot 3}(\bmod\ 2)\newline
 &\equiv 1 + \left(\sum_{k=1}^\infty (q^{k(3k-1)/2} + q^{k(3k+1)/2})^{8}\right)^3(\bmod\ 2)\newline
 &\equiv 1 + \left(\sum_{k=1}^\infty (q^{4k(3k-1)} + q^{4k(3k+1)})\right)^3 (\bmod\ 2).
\end{align*}
Hence
\begin{align*}
 q\left(\prod_{n=1}^\infty (1-q^n)\right)^{24} &\equiv q + q\left(\sum_{k=1}^\infty (q^{4k(3k-1)} + q^{4k(3k+1)})\right)^3 (\bmod\ 2)\newline
 &\equiv q+q\left(\sum_{k=1}^\infty q^{12k(3k-1)} + q^{(8k(3k-1)) + (4k(3k+1))}\right.\newline
 &\ \ \ \ + \left.q^{(4k(3k-1)) + (8k(3k+1))}  + q^{12k(3k+1)}\right) (\bmod\ 2)\newline
 &\equiv q+q\left(\sum_{k=1}^\infty q^{36k^2-12k} + q^{36k^2-4k} + q^{36k^2+4k}  + q^{36k^2+12k}\right) (\bmod\ 2)\newline
 &\equiv q+\left(\sum_{k=1}^\infty q^{36k^2-12k+1} + q^{36k^2-4k+1} + q^{36k^2+4k+1}  + q^{36k^2+12k+1}\right) (\bmod\ 2)\newline
 &\equiv q+\left(\sum_{k=1}^\infty q^{(6k-1)^2} + q^{36k^2-4k+1} + q^{36k^2+4k+1} + q^{(6k+1)^2}\right) (\bmod\ 2)
\end{align*}
so...?
We've got all the odd numbers congruent to 1 and 5 mod 6, but I don't see how we can get the numbers congruent to 3 mod 6... which makes me think there's an error I cannot find somewhere.
I should add, I do (vaguely) know of another way to solve this problem, I just enjoyed this approach, so wanted to make it work if possible.
Thanks!

REPLY [9 votes]: You fixed the main error I pointed out in the comments and used Jacobi's formula.
To summarize:
$$\sum_{n=1}^\infty\tau(n)q^n=q\prod_{n=1}^\infty(1-q^n)^{24}$$
$$=q \left(\sum_{k=0}^\infty(-1)^k(2k+1)q^{k(k+1)/2}\right)^8$$
$$\equiv q\sum_{k=0}^\infty \left((-1)^k(2k+1)q^{k(k+1)/2}\right)^8\mod2 $$
$$\equiv\sum_{m=0}^\infty q^{(2m+1)^2}\mod2,$$
QED.