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Let $G_n = \tfrac{1}{2}n(3n-1)$ be the pentagonal number for all $n\in \mathbb{Z}$ and $p(n)$ be the partition function. I was trying to prove one of the Ramanujan's congruences: $$p(5n-1) = 0 \pmod 5,$$ and my "brute force" proof reduces to show the following identity: $$\left(\sum_{G_n = 0 \pmod 5}(-1)^nq^{G_n}\right)\left(\sum_{G_n = 2 \pmod 5}(-1)^nq^{G_n}\right) + \left(\sum_{G_n = 1 \pmod 5}(-1)^nq^{G_n}\right)^2 = 0 \pmod 5,$$ where the sums are over all $n\in \mathbb{Z}$. After expanding a few terms (up to $q^{15000}$) of the left hand side of the identity, I strongly believe the following identity holds:

$$\left(\sum_{G_n = 0 \pmod 5}(-1)^nq^{G_n}\right)\left(\sum_{G_n = 2 \pmod 5}(-1)^nq^{G_n}\right) + \left(\sum_{G_n = 1 \pmod 5}(-1)^nq^{G_n}\right)^2 = 0.$$

I am in particular interested in a proof of the last identity.

Update: Here is what I tried. Denote by $$F_k(q) := \sum_{G_n=k\pmod 5}(-1)^nq^{G_n}.$$ Notice that $G_n = 1 \pmod 5$ if and only if $n = 1 \pmod 5$, and $G_{5k+1} = 1 + 25G_{-k}$. This would give $F_1(q) = q\prod_{m=1}^{\infty}(1-q^{25m})$ using the pentagonal number theorem. However, I do not know a way to factor $F_0(q)$ and $F_2(q)$.

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  • $\begingroup$ we have $G_{5k}\equiv G_{5k+2}\equiv0\mod5$ and $G_{5k+3}\equiv G_{5k+4}\equiv2\mod5$ $\endgroup$ – JonMark Perry Mar 16 '16 at 6:51
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It turns out that Ramanujan has asserted a result, from which the identity in the question follows. In order to state his result, recall the $q$-Pochhammer symbols $(a;q)_\infty = \prod_{k \ge 0}(1-aq^n)$. For convenience, we always suppress the subscript $(a;q) = (a;q)_\infty$.

Ramanujan asserted that,

$$(q;q) = \frac{(q^{10};q^{25})(q^{15};q^{25})(q^{25};q^{25})}{(q^{5};q^{25})(q^{20};q^{25})} - q(q^{25};q^{25})-q^2\frac{(q^{5};q^{25})(q^{20};q^{25})(q^{25};q^{25})}{(q^{10};q^{25})(q^{15};q^{25})}.$$

The pentagonal number theorem says that the left hand side $(q;q) = \sum_n (-1)^nq^{G_n}$. However the right hand side decomposes the sum based on $G_n\bmod 5$. Using the notation $F_k$ defined in the question, we obtain $$F_0 = \frac{(q^{10};q^{25})(q^{15};q^{25})(q^{25};q^{25})}{(q^{5};q^{25})(q^{20};q^{25})}, F_1 = - q(q^{25};q^{25}), F_2=-q^2\frac{(q^{5};q^{25})(q^{20};q^{25})(q^{25};q^{25})}{(q^{10};q^{25})(q^{15};q^{25})}.$$ Clearly, $F_0F_2 + F_1^2 = 0$.

Inspired by the proof of the Ramanujan's identity on $(q;q)$ in Ramanujan's "most beautiful identity" by Hirschhorn, here is a slightly direct proof of $F_0F_2 + F_1^2=0$.

Recall the Jacobi triple product: $(a^{-1}; q)(aq; q)(q;q)=\sum_n(-1)^na^nq^{n(n+1)/2}$. If $a \neq 1$, then we have $$(a^{-1}q;q)(aq;q)(q;q)=\sum_n(-1)^n\frac{a^n}{1-a^{-1}}q^{n(n+1)/2}.$$ Replacing $a$ by $a^{-1}$, we get $$(aq;q)(a^{-1}q;q)(q;q)=\sum_n(-1)^n\frac{a^{-n}}{1-a}q^{n(n+1)/2}.$$

Averaging the last two identities, we obtain $$(a^{-1}q;q)(aq;q)(q;q)=\sum_n(-1)^n\frac{1}{2}\left(\frac{a^n}{1-a^{-1}}+\frac{a^{-n}}{1-a}\right)q^{n(n+1)/2}.$$

Let $\omega := e^{2\pi i/5}$ be a (primitive) 5th root of unity. Take $a = \omega$, we have $$(\omega^{-1}q;q)(\omega q;q)(q;q)=\sum_n(-1)^nc_nq^{n(n+1)/2},$$ where $$c_n = \frac{1}{2}\left(\frac{\omega^n}{1-\omega^{-1}}+\frac{\omega^{-n}}{1-\omega}\right) = \begin{cases}\frac{1}{2} & \text{if }n=0\pmod 5 \\ \frac{1+\sqrt{5}}{4} & \text{if }n=1\pmod 5 \\ 0 & \text{if }n=2\pmod 5 \\ -\frac{1+\sqrt{5}}{4} & \text{if }n=3\pmod 5 \\ -\frac{1}{2} & \text{if }n=4\pmod 5\end{cases}.$$

So $(\omega^{-1}q;q)(\omega q;q)(q;q) = \frac{1}{2}(C_0 + \phi_1 C_1 - \phi_1 C_3 - C_4)$, where $C_i = \sum_{n=i\pmod 5}(-1)^nq^{n(n+1)/2}$ and $\phi_1 = \frac{1+\sqrt{5}}{2}$. Notice that by replacing $n$ by $-(n+1)$ in the sums, one can show that $C_0 = -C_4$ and $C_1 = - C_3$. So we get $$(\omega^{-1}q;q)(\omega q;q)(q;q) = C_0+\phi_1 C_1.$$

Similarly, take $a = \omega^2$ and obtain $$(\omega^{-2}q;q)(\omega^2 q;q)(q;q) = \frac{1}{2}(C_0+\phi_2C_1-\phi_2C_3-C_4)=C_0+\phi_2C_1,$$ where $\phi_2 = \frac{1-\sqrt{5}}{2}$. Multiply the last two identities, we have $$(\omega^{-2}q;q)(\omega^{-1}q;q)(\omega q;q)(\omega^{2}q;q)(q;q)(q;q) = C_0^2-C_0C_1-C_1^2.$$ Since $(\omega^{-2}q;q)(\omega^{-1}q;q)(\omega q;q)(\omega^{2}q;q)(q;q) = (q^5;q^5),$ we have $$(q;q)=\frac{C_0^2-C_0C_1-C_1^2}{(q^5;q^5)}.$$ By the same reasoning we used at the very beginning, we have $$F_0 = \frac{C_0^2}{(q^5;q^5)}, F_1 = \frac{-C_0C_1}{(q^5;q^5)}, F_2 = \frac{-C_1^2}{(q^5;q^5)}.$$

Remark: With a bit of extra work (using Jacobi triple product), one can show $C_0 = (q^{10};q^{25})(q^{15};q^{25})(q^{25};q^{25}), C_1 = (q^5;q^{25})(q^{20};q^{25})(q^{25};q^{25})$ and hence derive Ramanujan's identity.

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I can't seem to prove the identity, but I can show you what I've got.

$G_{5k}\equiv G_{5k+2}\equiv0\mod5$

which can easily be seen in the $5k$ case, and for $5k+2$ we have: $$\dfrac{(5k+2)(3(5k+2)-1)}{2}$$ $$=\dfrac{(5k+2)(15k+5)}{2}$$ $$=\dfrac{5(5k+2)(3k+1)}{2}$$

$G_{5k+3}\equiv G_{5k+4}\equiv2\mod5$

The $2$ cases have very similar proofs, here is for $5k+3$:

$$\dfrac{(5k+3)(3(5k+3)-1)}{2}$$ $$=\dfrac{(5k+3)(15k+8)}{2}$$ $$=\dfrac{75k^2+85k+24}{2}$$

As $5$ is not divisible by $2$, the remainder is $\dfrac{24}{2}\equiv2\mod5$.

Next we can expand the expressions for $G_n$ to get relations between them.

$$G_{5k}=\dfrac{75k^2-5k}{2}$$ $$G_{5k+2}=\dfrac{75k^2+55k+10}{2}$$

Adding the two gives:

$$G_{5k}+G_{5k+2}=\dfrac{150k^2+50k+10}{2}=25G_{-k}+5$$

$$G_{5k+3}=\dfrac{75k^2+85k+24}{2}$$ $$G_{5k+4}=\dfrac{75k^2+115k+44}{2}$$

Subtracting the two gives:

$$G_{5k+4}-G_{5k+3}=\dfrac{30k+20}{2}=15k+10$$

as I can't seem to get the square and the linear component to form a valid ratio either!

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