1 Introduction

A partition of a positive integer n is a nonincreasing sequence of positive integers, called parts, whose sum is n. Let p(n) denote the number of partitions of n. We follow the convention that \(p(0)= 1\). It is well known that the generating function for p(n) satisfies

$$\begin{aligned} \sum _{n=0}^\infty p(n)q^n=\frac{1}{(q;q)_\infty }. \end{aligned}$$

Throughout this note, we adopt the following notation:

$$\begin{aligned} (a;q)_\infty =\prod _{n=1}^\infty \left( 1-aq^{n-1} \right) ,\quad |q|<1. \end{aligned}$$

The most famous results for p(n) are the so called Ramanujan’s congruences: for \(n\ge 0\),

$$\begin{aligned} p(5n+4)\equiv & {} 0\ (\mathrm{mod\ }5),\end{aligned}$$
(1.1)
$$\begin{aligned} p(7n+5)\equiv & {} 0\ (\mathrm{mod\ }7),\end{aligned}$$
(1.2)
$$\begin{aligned} p(11n+6)\equiv & {} 0\ (\mathrm{mod\ }11). \end{aligned}$$
(1.3)

Ramanujan [25, Paper 30], Atkin and Swinnerton-Dyer [1], Winquist [26], Garvan [12], Garvan and Stanton [13], Hirschhorn [15–18], and Marivani [22] have given different proofs of the congruence (1.1 1.2 1.3). It is worth mentioning that Winquist [26] found an interesting identity which plays an important role in proving Ramanujan’s congruence modulo 11. In fact, Winquist used his identity to establish an identity for \((q;q)_\infty ^{10}\) from which the congruence (1.1 1.2 1.3) follows easily. Later several proofs of Winquist’s identity are found and new identities for \((q;q)_\infty ^{10}\) are established, see [3, 6, 8–10, 20, 21, 23], for example.

Recently, Hirschhorn [19] presented a most elementary, simple, beautiful proof of the congruence (1.1 1.2 1.3). Later, Gnang and Zeilberger [14] generalized and implemented Hirschhorn’s amazing algorithm for proving Ramanujan-type congruences. They considered \(p_{-a}(n)\), which is defined by

$$\begin{aligned} \sum _{n=0}^\infty p_{-a}(n)q^n=\frac{1}{(q;q)_\infty ^a}. \end{aligned}$$
(1.4)

There are many known Ramanujan-type congruences for \(p_{-a}(n)\). Boylan [4] has found all of them for a odd and \({\le }47\). For example,

$$\begin{aligned} p_{-3}(11n+7)\equiv 0\ (\mathrm{mod\ 11}). \end{aligned}$$
(1.5)

Every such congruence can be checked by using impressive algorithm of Radu [24]. Although Radu’s algorithm is powerful, it is not elementary. Based on this, Zeilberger said that “it is still interesting (at least to us!) to find a ‘Ramanujan-style,’ or ‘Hirschhorn-style’ proof.”

In this note, we aim to give one “Ramanujan-style” proof of the congruence (1.5). To this end, we will establish two identities for \((q;q)_\infty ^8\) by using Ramanujan’s two identities. Although the series for \((q;q)_\infty ^8\) have been considered by several authors, see [5, 7, 11] for example, our approach is more elementary.

2 Preliminaries

We first introduce two Ramanujan’s theta functions \(\varphi (q)\) and \(\psi (q)\), defined by

$$\begin{aligned} \varphi (q)= & {} \sum _{n=-\infty }^\infty q^{n^2},\\ \psi (q)= & {} \sum _{n=0}^\infty q^{n(n+1)/2}. \end{aligned}$$

Two lemmas related to \(\varphi (q)\) and \(\psi (q)\) are presented as follows.

Lemma 2.1

$$\begin{aligned} \varphi (q)= & {} \frac{(q^2;q^2)_\infty ^5}{(q;q)_\infty ^2(q^4;q^4)_\infty ^2},\end{aligned}$$
(2.1)
$$\begin{aligned} \psi (q)= & {} \frac{(q^2;q^2)_\infty ^2}{(q;q)_\infty },\end{aligned}$$
(2.2)
$$\begin{aligned} \varphi (-q)= & {} \frac{(q;q)_\infty ^2}{(q^2;q^2)_\infty }. \end{aligned}$$
(2.3)

Proof

See [2, Cor. 1.3.4] for a proof. \(\square \)

Lemma 2.2

$$\begin{aligned} \varphi (-q)=\varphi (q^4)-2q\psi (q^8). \end{aligned}$$
(2.4)

Proof

This identity can be derived by using series manipulations, and we omit the details here. \(\square \)

The rest of this section are Jacobi’s four-square theorem and two identities of Ramanujan, which are extremely useful to our later proof.

Lemma 2.3

([Jacobi’s Four-Square Theorem])

$$\begin{aligned} \varphi (-q)^4=1-8\sum _{n=1}^\infty \left( \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}}-\frac{2nq^{2n}}{1+q^{2n}}\right) . \end{aligned}$$
(2.5)

Proof

See [2, Thm. 3.3.1] for a proof of (2.5). \(\square \)

Lemma 2.4

$$\begin{aligned} \sum _{n=-\infty }^\infty (6n+1)q^{3n^2+n}= & {} \frac{(q^2;q^2)_\infty ^5}{(q^4;q^4)_\infty ^2},\end{aligned}$$
(2.6)
$$\begin{aligned} \sum _{n=-\infty }^\infty (3n+1)q^{3n^2+2n}= & {} \frac{(q;q)_\infty ^2(q^4;q^4)_\infty ^2}{(q^2;q^2)_\infty }. \end{aligned}$$
(2.7)

Proof

For the proofs of (2.6) and (2.7), see [2, Cor. 1.3.21 and Cor. 1.3.22]. \(\square \)

3 Ramanujan-style proof

We first establish two identities of \((q;q)_\infty ^8\), either of which can be employed to produce the desired Ramanujan-style proof.

Theorem 3.1

$$\begin{aligned} 3(q;q)_\infty ^8= 4\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty q^{n^2}\right) \nonumber \\ -\left( \sum _{m=-\infty }^\infty (6m+1)^3q^{3m^2+m}\right) \times \left( \sum _{n=0}^\infty q^{n^2+n}\right) . \end{aligned}$$
(3.1)

Proof

Differentiating both sides of (2.6) with respect to q, we find that

$$\begin{aligned} \sum _{n=-\infty }^\infty (6n+1)\big (3n^2+n\big )q^{3n^2+n}=\frac{\big (q^2;q^2\big )_\infty ^5}{\big (q^4;q^4\big )_\infty ^2} \left( 5\sum _{n=0}^\infty \frac{-2nq^{2n}}{1-q^{2n}} -2\sum _{n=0}^\infty \frac{-4nq^{4n}}{1-q^{4n}}\right) , \end{aligned}$$
(3.2)

and thus,

$$\begin{aligned}&\sum _{n=-\infty }^\infty \left( (6n+1)^3-(6n+1)\right) q^{3n^2+n}\nonumber \\&\quad =\;12\frac{(q^2;q^2)_\infty ^5}{(q^4;q^4)_\infty ^2}\left( 5\sum _{n=0}^\infty \frac{-2nq^{2n}}{1-q^{2n}} -2\sum _{n=0}^\infty \frac{-4nq^{4n}}{1-q^{4n}}\right) . \end{aligned}$$
(3.3)

Applying (2.6), we have

$$\begin{aligned}&\sum _{n=-\infty }^\infty (6n+1)^3 q^{3n^2+n}\times \sum _{n=0}^\infty q^{n^2+n}\nonumber \\&\quad =\sum _{n=-\infty }^\infty (6n+1) q^{3n^2+n}\times \sum _{n=0}^\infty q^{n^2+n}\nonumber \\&\qquad +\,12\big (q^2;q^2\big )_\infty ^4 \left( 5\sum _{n=0}^\infty \frac{-2nq^{2n}}{1-q^{2n}} -2\sum _{n=0}^\infty \frac{-4nq^{4n}}{1-q^{4n}}\right) \nonumber \\&\quad =12\big (q^2;q^2\big )_\infty ^4\sum _{n=0}^\infty \left( \frac{-10nq^{2n}}{1-q^{2n}}+ \frac{8nq^{4n}}{1-q^{4n}}\right) \nonumber \\&\qquad +\,\big (q^2;q^2\big )_\infty ^4. \end{aligned}$$
(3.4)

Differentiating both sides of (2.7) with respect to q, we obtain

$$\begin{aligned}&\sum _{m=-\infty }^\infty (3m+1) \big (3m^2+2m\big )q^{3m^2+2m}\nonumber \\&\qquad =\frac{(q;q)_\infty ^2\big (q^4;q^4\big )_\infty ^2}{\big (q^2;q^2\big )_\infty } \sum _{n=1}^\infty \left( \frac{-2nq^n}{1-q^n} -\frac{8nq^{4n}}{1-q^{4n}}+\frac{2nq^{2n}}{1-q^{2n}}\right) , \end{aligned}$$

and thus

$$\begin{aligned} \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}= & {} \sum _{m=-\infty }^\infty (3m+1)q^{3m^2+2m}+ 3\frac{(q;q)_\infty ^2(q^4;q^4)_\infty ^2}{(q^2;q^2)_\infty }\\ \quad&\times \sum _{n=1}^\infty \left( \frac{-2nq^n}{1-q^n} -\frac{8nq^{4n}}{1-q^{4n}}+\frac{2nq^{2n}}{1-q^{2n}}\right) . \end{aligned}$$

By (2.1 2.2 2.3) and (2.7), we deduce that

$$\begin{aligned}&\sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\times \sum _{n=-\infty }^\infty q^{n^2}\nonumber \\&\quad =\sum _{m=-\infty }^\infty (3m+1)q^{3m^2+2m}\times \sum _{n=-\infty }^\infty q^{n^2}\nonumber \\&\qquad +\,3(q^2;q^2)_\infty ^4\sum _{n=1}^\infty \left( \frac{-2nq^n}{1-q^n} -\frac{8nq^{4n}}{1-q^{4n}}+\frac{2nq^{2n}}{1-q^{2n}}\right) \nonumber \\&\quad =(q^2;q^2)_\infty ^4\sum _{n=1}^\infty \left( \frac{-6nq^n}{1-q^n} -\frac{24nq^{4n}}{1-q^{4n}}+\frac{6nq^{2n}}{1-q^{2n}}\right) .\nonumber \\&\qquad +\,(q^2;q^2)_\infty ^4. \end{aligned}$$
(3.5)

From (3.4) and (3.5), we find that

$$\begin{aligned}&4\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty q^{n^2}\right) \nonumber \\&\quad -\left( \sum _{m=-\infty }^\infty (6m+1)^3q^{3m^2+m}\right) \times \left( \sum _{n=0}^\infty q^{n^2+n}\right) \nonumber \\&\qquad =3(q^2;q^2)_\infty ^4\left( 1-8\sum _{n=1}^\infty \left( \frac{nq^n}{1-q^n}+\frac{8nq^{4n}}{1-q^{4n}}-\frac{6nq^{2n}}{1-q^{2n}} \right) \right) \nonumber \\&\qquad =3(q^2;q^2)_\infty ^4\left( 1-8\sum _{n=1}^\infty \left( \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}}-\frac{2nq^{2n}}{1+q^{2n}} \right) \right) , \end{aligned}$$
(3.6)

where the last equation follows from the following fact:

$$\begin{aligned}&\frac{nq^n}{1-q^n}+\frac{8nq^{4n}}{1-q^{4n}}-\frac{6nq^{2n}}{1-q^{2n}}\nonumber \\&\quad = \frac{(2n-1)q^{2n-1}}{1-q^{2n-1}}+\frac{8nq^{4n}}{1-q^{4n}}-\frac{4nq^{2n}}{1-q^{2n}} \\&\quad = \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}}+\frac{(4n-2)q^{4n-2}}{1-q^{4n-2}}+ \frac{8nq^{4n}}{1-q^{4n}}-\frac{4nq^{2n}}{1-q^{2n}}\\&\quad = \frac{(2n-1)q^{2n-1}}{1+q^{2n-1}}+ \frac{4nq^{4n}}{1-q^{4n}}-\frac{2nq^{2n}}{1-q^{2n}}\\&\quad =\frac{(2n-1)q^{2n-1}}{1+q^{2n-1}}-\frac{2nq^{2n}}{1+q^{2n}}. \end{aligned}$$

Applying (2.5) in (3.6), we conclude that

$$\begin{aligned}&4\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty q^{n^2}\right) \nonumber \\&\quad -\,\left( \sum _{m=-\infty }^\infty (6m+1)^3q^{3m^2+m}\right) \times \left( \sum _{n=0}^\infty q^{n^2+n}\right) \\&\qquad =3(q^2;q^2)_\infty ^4 \varphi ^4(-q)\\&\qquad =3(q;q)_\infty ^8. \end{aligned}$$

This completes the proof. \(\square \)

It is interesting to present another identity for \((q;q)_\infty ^8\) which can be derived from (3.1).

Theorem 3.2

$$\begin{aligned} 3(q;q)_\infty ^8= & {} 4\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty (-q)^{\frac{n^2}{4}}\right) \nonumber \\ \quad&-\,\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{\frac{3m^2+2m}{4}}\right) \times \left( \sum _{n=0}^\infty q^{n^2+n}\right) . \end{aligned}$$
(3.7)

Proof

Applying (2.4) with q replaced by \(q^{1/4}\), we arrive at

$$\begin{aligned}&4\varphi \big (-q^{1/4}\big )\sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m} -\psi (q^2)\nonumber \\&\,\,\times \sum _{m=-\infty }^\infty (3m+1)^3q^{\frac{3m^2+2m}{4}}\\&\quad = 4\Big (\varphi (q)-2q^{1/4}\psi \big (q^2\big )\Big )\sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\\&\qquad -\,\psi (q^2)\left( \sum _{m=-\infty }^\infty (6m+1)^3q^{3m^2+m}-8q^{1/4}\sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \\&\quad =4\varphi (q)\sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}-\psi (q^2)\sum _{m=-\infty }^\infty (6m+1)^3q^{3m^2+m}\\&\quad =3(q;q)_\infty ^8, \end{aligned}$$

where the last equation follows from Theorem 3.1. This finishes the proof. \(\square \)

Now we are ready to prove the following theorem by using Theorem 3.1.

Theorem 3.3

For \(n\ge 0\),

$$\begin{aligned} p_{-3}(11n+7)\equiv 0\ (\mathrm{mod\ 11}). \end{aligned}$$
(3.8)

Proof

Let us define a(n) to be

$$\begin{aligned} 3(q;q)_\infty ^8=\sum _{n=0}^\infty a(n)q^n. \end{aligned}$$

Then,

$$\begin{aligned} \sum _{n=0}^\infty 3p_{-3}(n)q^n\equiv & {} \frac{3(q;q)_\infty ^8}{\big (q^{11};q^{11}\big )_\infty }\\= & {} \frac{1}{\big (q^{11};q^{11}\big )_\infty }\sum _{n=0}^\infty a(n)q^n\ (\mathrm{mod\ }11). \end{aligned}$$

Extracting those terms with powers of the form \(11n + 7\), we conclude that

$$\begin{aligned} \sum _{n=0}^\infty 3p_{-3}(11n+7)q^{11n+7}\equiv \frac{1}{\big (q^{11};q^{11}\big )_\infty }\sum _{n=0}^\infty a(11n+7)q^{11n+7}\ (\mathrm{mod\ }11). \end{aligned}$$

To prove \(p_{-3}(11n+7)\equiv 0\ (\mathrm{mod\ }11)\), we only need to show that

$$\begin{aligned} a(11n+7)\equiv 0\ (\mathrm{mod\ }11). \end{aligned}$$

Consider the congruence equation

$$\begin{aligned} 3m^2+2m+n^2\equiv 7\ (\mathrm{mod\ }11), \end{aligned}$$

which can be rewritten as

$$\begin{aligned} 4(3m+1)^2+n^2\equiv 0\ (\mathrm{mod\ }11). \end{aligned}$$

Since \(-4\) is a quadratic nonresidue modulo 11, we see that \(3m+1\) is divisible by 11.

Similarly, if we consider the congruence equation \(3m^2+m+n^2+n\equiv 7\ (\mathrm{mod\ }11)\), we can deduce that \(6m+1\) is divisible by 11.

By Theorem 3.1, we see that

$$\begin{aligned} a(11n+7)\equiv 0\ (\mathrm{mod\ }11^3), \end{aligned}$$

which completes the proof. \(\square \)

Remark (3.7) can also be used to prove (3.8), and we leave the details to readers. Multiplying on both sides of (3.1) or (3.7) by \((q;q)_\infty ^2\), and using Ramanujan’s identities (2.6) and (2.7), we obtain a new proof of the following two identities of \((q;q)_\infty ^{10}\) due to Chu [8, Cor. 4.2] and Chan [6, Thm. 3.4]:

$$\begin{aligned} 3(q;q)_\infty ^{10}= & {} 4\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty (6n+1) q^{3n^2+n}\right) \nonumber \\ \quad&-\,\left( \sum _{m=-\infty }^\infty (3m+1)q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty (6n+1)^3q^{3n^2+n}\right) ,\end{aligned}$$
(3.9)
$$\begin{aligned} 3(q;q)_\infty ^{10}= & {} 4\left( \sum _{m=-\infty }^\infty (3m+1)^3q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty (3n+1) q^{(3n^2+2n)/4}\right) \nonumber \\ \quad&-\,\left( \sum _{m=-\infty }^\infty (3m+1)q^{3m^2+2m}\right) \times \left( \sum _{n=-\infty }^\infty (3n+1)^3q^{(3n^2+2n)/4}\right) .\nonumber \\ \end{aligned}$$
(3.10)