# Converting an infinite product to sum; Ramanujan $\tau$ function

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 $m$.

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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!

• Question: When you say "we may rewrite this as," why is the "$1+$" (i.e. the $k=0$ indexed summand) outside of the 24th power? Also, after "Hence," how did you go from $(\Sigma)^3$ to $\Sigma()^3$? (I'm not really familiar with modulo reducing power series, to be honest.)
– anon
Oct 9, 2011 at 22:31
• @anon, Oh! Well that is an error I missed. You can't pull the 1 out in the first step. I'll work on fixing that... also when combining modulo reductions with series, the binomial theorem is very useful. As a simple example, notice $(x+y)^2 = x^2 + 2xy + y^2$, so modulo 2, the cross terms disappear. Similarly $(x+y)^3 = x^3 + 3x^2y + 3xy^2 + 3y^3$, and the cross terms disappear again!
– Alex
Oct 9, 2011 at 23:39
• Wait, $(a+b)^3\equiv a^3+a^2b+ab^2+b^3$, I don't see how the cross terms disappear modulo 2...
– anon
Oct 10, 2011 at 0:20
• Yes, I just realized that... the sum is far more messy... It looks like letting (sum)^3 = (sum^2)*sum may be helpful, but I haven't quite worked it out yet
– Alex
Oct 10, 2011 at 0:27
• That was my idea too. It would seem we need to prove that for any $n$, the number of integer pairs $k,j$, such that $n$ can be written as $8k(3k-1)+4j(3j-1)$, is odd if and only if $n$ is the square of some odd integer.
– anon
Oct 10, 2011 at 0:45

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.