Arithmetic properties arising from Ramanujan’s theta functions

Abstract

We prove some interesting arithmetic properties of theta function identities that are analogous to q-series identities obtained by Michael D. Hirschhorn. In addition, we find infinite family of congruences modulo powers of 2 for representations of a non-negative integer n as $\triangle _1+4\triangle _2$ and $\triangle +k\square$.

1 Introduction

Let $a_k(n)$ be the number of representations of n as $\triangle _1+2k\triangle _2$ and $b_k(n)$ be the number of representations of n as $\triangle +k\square$. Then

$$\begin{aligned} \sum \limits _{n=0}^{\infty }a_k\left( n\right) q^n&=\psi (q)\psi (q^{2k}),\\ \sum \limits _{n=0}^{\infty }b_k(n)q^n&=\psi (q)\varphi (q^{k}). \end{aligned}$$

Hirschhorn and Sellers [3] used some interesting q-series identities to obtain results of congruences modulo 5 for 4-colored Frobenius partitions, and Hirschhorn [2] proved equivalent results of q-series identities that are used in [3]: If

$$\begin{aligned} \sum \limits _{n\ge 0}g(n)q^n=\dfrac{(q^2;q^2)_{\infty }^2(q^4;q^4)_{\infty }^3}{(q;q)_{\infty }}\;\; \text {and}\;\; \sum \limits _{n\ge 0}h(n)q^n=\dfrac{(q;q)_{\infty }^3(q^2;q^2)_{\infty }^2}{(q^4;q^4)_{\infty }}, \end{aligned}$$

where $(a;q)_\infty =\prod \limits _{n\ge 0}(1-aq^n)$, then

$$\begin{aligned} \sum \limits _{n\ge 0}g(5n)q^n=\sum \limits _{n\ge 0}h(n)q^n-5\sum \limits _{n\ge 0}g(n)q^{5n+3} \end{aligned}$$

and

$$\begin{aligned} \sum \limits _{n\ge 0}h(5n+3)q^n=16\sum \limits _{n\ge 0}g(n)q^n-5\sum \limits _{n\ge 0}h(n)q^{5n}. \end{aligned}$$

Motivated by these works, in Sect. 3, we establish arithmetic properties of representations of a number $\triangle _1+2k\triangle _2$ and $\triangle +k\square$. In addition, we also find certain congruences modulo powers of 2 for number of representations of a non-negative integer n as $\triangle _1+4\triangle _2$ and $\triangle +k\square$. We list our main results in the following theorems and corollaries.

Theorem 1.1

Let $a_k(n)$ and $b_k(n)$ be the number of representations of n as $\triangle _1+2k\triangle _2$ and $\triangle +k\square$, respectively. Then

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

(1.1)

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_1\left( 3n+1\right) q^n&=4\psi (q)\psi (q^2)-\psi (q^3)\varphi (q^3),\end{aligned}$$

(1.2)

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_2\left( 5n+3\right) q^n&=4\psi (q)\psi (q^{4})-\psi (q^{5})\varphi (q^{10}),\end{aligned}$$

(1.3)

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_3\left( 7n+6\right) q^n&=4\psi (q)\psi (q^{6})-\psi (q^{7})\varphi (q^{21}), \end{aligned}$$

(1.4)

where k is a positive integer so that $2k+1$ is odd prime.

Theorem 1.2

Let $a_2(n)$ and $b_2(n)$ be the number of representations of n as $\triangle _1+4\triangle _2$ and $\triangle +2\square$, respectively. Then

$$\begin{aligned} a_2(625n+390)-2a_2(25n+15)+a_2(n)&=0,\end{aligned}$$

(1.5)

$$\begin{aligned} b_2(625n+78)-2b_2(25n+3)+b_2(n)&=0. \end{aligned}$$

(1.6)

The identities (1.5) and (1.6) are analogous to the Hirschhorn [2] identities

$$\begin{aligned} g(625n+390)-6g(25n+15)+25g(n)&=0,\\ h(625n+78)-6h(25n+3)+25h(n)&=0. \end{aligned}$$

Corollary 1.1

Let $b_{k}(n)$ be the number of representations of n as $\triangle +k\square$. Then for each n, $\alpha \ge 0$,

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_1\left( 3^{2\alpha }n+\dfrac{3^{2\alpha }-1}{8}\right) q^n&\equiv 3^\alpha \psi (q)\varphi (q) \pmod {2^2},\end{aligned}$$

(1.7)

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_2\left( 5^{2\alpha }n+\dfrac{5^{2\alpha }-1}{8}\right) q^n&\equiv 3^\alpha \psi (q) \varphi (q^2) \pmod {2^2},\end{aligned}$$

(1.8)

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_3\left( 7^{2\alpha }n+\dfrac{7^{2\alpha }-1}{8}\right) q^n&\equiv 3^\alpha \psi (q) \varphi (q^3) \pmod {2^2}. \end{aligned}$$

(1.9)

Corollary 1.2

Let $a_2(n)$ and $b_2(n)$ be the number of representations of n as $\triangle _1+4\triangle _2$ and $\triangle +2\square$, respectively. Then for all n, $\alpha \ge 0$,

$$\begin{aligned} a_2\left( 5^{4\alpha }n+\frac{5(5^{4\alpha }-1)}{8}\right)&\equiv a_2(n) \pmod {2},\end{aligned}$$

(1.10)

$$\begin{aligned} b_2\left( 5^{4\alpha }n+\dfrac{5^{4\alpha }-1}{8}\right)&\equiv b_2(n) \pmod {2}. \end{aligned}$$

(1.11)

2 Preliminaries

For $|ab|<1$, Ramanujan’s general theta function $f\left( a,b\right)$ is defined by

$$\begin{aligned} f\left( a,b\right)&=\sum \limits _{n=-\infty }^{\infty }a^{n\left( n+1\right) /2}b^{n\left( n-1\right) /2}\end{aligned}$$

(2.1)

$$\begin{aligned}&=\left( -a;ab\right) _\infty \left( -b;ab\right) _\infty \left( ab;ab\right) _\infty . \end{aligned}$$

(2.2)

Important special cases of $f\left( a,b\right)$ are

$$\begin{aligned} \varphi \left( q\right) :=f\left( q,q\right) =1+2\sum _{n=1}^{\infty }q^{n^2}=\sum \limits _{n=-\infty }^{\infty }q^{n^2}=(-q;q^2)^2_\infty (q^2;q^2)_\infty \end{aligned}$$

(2.3)

and

$$\begin{aligned} \psi \left( q\right) :=\dfrac{1}{2}f\left( 1,q\right) =f(q,q^3)=\sum \limits _{n=0}^{\infty }q^{(n^2+n)/2} =\sum \limits _{n=-\infty }^{\infty }q^{2n^2+n}=\dfrac{(q^2;q^2)_\infty }{(q;q^2)_\infty }. \end{aligned}$$

(2.4)

From (2.3) and (2.4), we can see that

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

(2.5)

3 Proofs of main results

Proof of Theorem 1.1

Let us write

$$\begin{aligned} E_k(q)=\psi (q)\psi (q^{2k}). \end{aligned}$$

Using (2.4) in $E_k\left( q\right)$, we obtain

$$\begin{aligned} E_k\left( q\right) =\sum \limits _{m=-\infty }^{\infty }q^{2m^2+m}\sum \limits _{n=-\infty }^{\infty }q^{2k\left( 2n^2+n\right) }. \end{aligned}$$

(3.1)

Replacing q by $q^8$ and then multiplying both sides of (3.1) by $q^{2k+1}$, we find that

$$\begin{aligned} q^{2k+1}E_k(q^8)&=\sum \limits _{m,n=-\infty }^{\infty }q^{\left( 4m+1\right) ^2 +2k\left( 4n+1\right) ^2}\nonumber \\&=\sum \limits _{x,y\equiv 1\ (\text {mod}\ 4)}^{}q^{x^2+2ky^2}. \end{aligned}$$

(3.2)

If we write

$$\begin{aligned} E_k\left( q\right) =E^0_k\left( q\right) +E^1_k\left( q\right) +\cdots +E^{2k}_k\left( q\right) , \end{aligned}$$

(3.3)

where $E^i_k\left( q\right)$ consists of all those terms in $E_k\left( q\right)$ in which powers of q are congruent to i modulo $2k+1$, then from (3.2) and (3.3), we see that

$$\begin{aligned} q^{2k+1}E^0_k(q^8)=\sum \limits _{x,y\equiv 1\ (\text {mod}\ 4)}q^{x^2+2ky^2}, \end{aligned}$$

(3.4)

where $x^2+2ky^2\equiv 0\ (\text {mod}\ 2k+1)$. The solution of $x^2+2ky^2\equiv 0\ (\text {mod}\ 2k+1)$ is $x\equiv \pm y \ (\text {mod}\ 2k+1)$, and from (3.4) it follows that

$$\begin{aligned} q^{2k+1}E^0_k(q^8)=&\sum \limits _{\begin{array}{c} x\equiv y\equiv 1 \ (\text {mod}\ 4)\\ x\equiv y \ (\text {mod}\ 2k+1) \end{array}}q^{x^2+2ky^2} +\sum \limits _{\begin{array}{c} x\equiv y\equiv 1 \ (\text {mod}\ 4)\\ x\equiv -y \ (\text {mod}\ 2k+1) \end{array}}q^{x^2+2ky^2}\nonumber \\&-\sum \limits _{\begin{array}{c} x\equiv y\equiv 1 \ (\text {mod}\ 4)\\ x\equiv y\equiv 0 \ (\text {mod}\ 2k+1) \end{array}}q^{x^2+2ky^2}. \end{aligned}$$

(3.5)

In the first sum of (3.5), we have $x\equiv y\ (\text {mod}\ 2k+1)$, $4m+1\equiv 4n+1 \ (\text {mod}\ 2k+1)$, $m-n\equiv 0 \ (\text {mod}\ 2k+1)$, $m-n=(2k+1)u$, $m+2kn=(2k+1)v$, $m=2ku+v$, $n=v-u$,

$$\begin{aligned} x=8ku+4v+1,\;\text {and}\;y=4v-4u+1. \end{aligned}$$

(3.6)

In the second sum of (3.5), $x\equiv -y \ (\text {mod}\ 2k+1)$, $4m+1\equiv -4n-1\ (\text {mod}\ 2k+1)$, $4(m+n)\equiv -2 \ (\text {mod}\ 2k+1)$, $4(m+n)\equiv -2+2(2k+1) \ (\text {mod}\ 2k+1)$, $m+n\equiv k \ (\text {mod}\ 2k+1)$, $m+n=(2k+1)v+k$, $m-2kn=(2k+1)u+k$, $m=u+2kv+k$, $n=v-u$,

$$\begin{aligned} x=4u+8kv+4k+1,\;\text {and}\; y=4v-4u+1. \end{aligned}$$

(3.7)

In the third sum of (3.5),

$$\begin{aligned} x=4(2k+1)u+2k+1\;\text {and}\;y=4(2k+1)v+2k+1. \end{aligned}$$

(3.8)

Employing (3.6), (3.7), and (3.8) in (3.5), we obtain

On simplification, the above equation reduces to

$$\begin{aligned} E^0_k(q)=&\sum \limits _{v=-\infty }^{\infty }q^{(2k+1)(2v^2+v)}\sum \limits _{u=-\infty }^{\infty }q^{4k(2k+1)u^2}\nonumber \\&+q^{k(2k+1)}\sum \limits _{u=-\infty }^{\infty }q^{(2k+1)(2u^2+u)} \sum \limits _{v=-\infty }^{\infty }q^{4k(2k+1)(v^2+v)}\nonumber \\&-q^{\frac{k(k+1)(2k+1)}{2}}\sum \limits _{u=-\infty }^{\infty }q^{(2k+1)^2(2u^2+u)}\sum \limits _{v=-\infty }^{\infty }q^{2k(2k+1)^2(2v^2+v)}. \end{aligned}$$

(3.9)

From (2.3), (2.4), and (3.9), we have

$$\begin{aligned} E^0_k(q)=&\,\psi (q^{2k+1})\left( \varphi (q^{4k(2k+1)}) +2q^{k(2k+1)}\psi (q^{8k(2k+1)})\right) \nonumber \\&-q^{\frac{k(k+1)(2k+1)}{2}}\psi (q^{(2k+1)^2})\psi (q^{2k(2k+1)^2}). \end{aligned}$$

(3.10)

From (2.5) and (3.10), we find that

$$\begin{aligned} E^0_k(q)=\psi (q^{2k+1})\varphi (q^{k(2k+1)})-q^{\frac{k(k+1)(2k+1)}{2}}\psi (q^{(2k+1)^2})\psi (q^{2k(2k+1)^2}). \end{aligned}$$

(3.11)

The identity (3.11) can be written as

$$\begin{aligned} \sum \limits _{n=0}^{\infty }a_k((2k+1)n)q^{(2k+1)n}&=\psi (q^{2k+1})\varphi (q^{k(2k+1)})\nonumber \\&\quad -q^{\frac{k(k+1)(2k+1)}{2}}\psi (q^{(2k+1)^2})\psi (q^{2k(2k+1)^2}). \end{aligned}$$

(3.12)

Now replacing $q^{2k+1}$ by q in (3.12), we arrive at (1.1).

Let us write $F_2(q)=\psi (q)\varphi (q)$, and using (2.5), we find that

$$\begin{aligned} F_2(q)&=\psi (q)\left( \varphi (q^{4})+2q\psi (q^{8})\right) \nonumber \\&=\sum \limits _{m,n=-\infty }^{\infty }q^{2m^2+m+4n^2}+q\sum \limits _{m,n=-\infty }^{\infty }q^{2m^2+m+4(n^2+n)}. \end{aligned}$$

(3.13)

Replacing q by $q^8$ and then multiplying both sides of (3.13) by q, we obtain

$$\begin{aligned} qF_2(q^8)&=\sum \limits _{m,n=-\infty }^{\infty }q^{(4m+1)^2+2(4n)^2}+\sum \limits _{m,n=-\infty }^{\infty }q^{(4m+1)^2+2(4n+2)^2}\nonumber \\&=\sum \limits _{x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)}q^{x^2+2y^2}+\sum \limits _{x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)}q^{x^2+2y^2}. \end{aligned}$$

(3.14)

If we write

$$\begin{aligned} F_2(q)=F^0_2(q)+F^1_2(q)+F^2_2(q), \end{aligned}$$

(3.15)

where $F^i_2(q)$ consists of all those terms in $F_2(q)$ in which powers of q are congruent to i modulo 3, then from (3.14) and (3.15), we see that

$$\begin{aligned} qF^1_2(q^8)=\sum \limits _{x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)}q^{x^2+2y^2}+\sum \limits _{x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)}q^{x^2+2y^2}, \end{aligned}$$

(3.16)

where, in both sums, $x^2+2y^2\equiv 0 \ (\text {mod}\ 3)$. The solution of $x^2+2y^2\equiv 0 \ (\text {mod}\ 3)$ is $x\equiv \pm y \ (\text {mod}\ 3)$, and from (3.16) it follows that

(3.17)

In the first sum of (3.17), we have $4m+1\equiv 4n \ (\text {mod}\ 3)$, $4(m-n)\equiv -1\ (\text {mod}\ 3)$, $4(m-n)\equiv -1+9\ (\text {mod}\ 3)$, $m-n\equiv 2 \ (\text {mod}\ 3)$, $m-n\equiv -1 \ (\text {mod}\ 3)$, $m+2n\equiv -1 \ (\text {mod}\ 3)$, $m-n=-3u-1$, $m+2n=-3v-1$, $m=-v-2u-1$, $n=u-v$,

$$\begin{aligned} x=-4v-8u-3,\;\text {and}\;y=4u-4v. \end{aligned}$$

(3.18)

In the second sum of (3.17), $4m+1\equiv -4n\ (\text {mod}\ 3)$, $4(m+n)\equiv -1\ (\text {mod}\ 3)$, $4(m+n)\equiv -1+9\ (\text {mod}\ 3)$, $m+n\equiv 2\ (\text {mod}\ 3)$, $m+n\equiv -1 \ (\text {mod}\ 3)$, $m-2n\equiv -1 \ (\text {mod}\ 3)$, $m+n=-3u-1$, $m-2n=-3v-1$, $m=-v-2u-1$, $n=v-u$,

$$\begin{aligned} x=-4v-8u-3,\;\text {and}\;y=4v-4u. \end{aligned}$$

(3.19)

In the third sum of (3.17),

$$\begin{aligned} x=12u+3\;\text {and}\;y=12v. \end{aligned}$$

(3.20)

In the fourth sum of (3.17), $4m+1\equiv 4n+2 \ (\text {mod}\ 3)$, $4(m-n)\equiv 1\ (\text {mod}\ 3)$, $m-n\equiv 1\ (\text {mod}\ 3)$, $m+2n\equiv -2 \ (\text {mod}\ 3)$, $m-n=3u+1$, $m+2n=-3v-2$, $m=2u-v$, $n=-u-v-1$,

$$\begin{aligned} x=8u-4v+1,\;\text {and}\;y=-4u-4v-2. \end{aligned}$$

(3.21)

In the fifth sum of (3.17), $4m+1\equiv -4n-2 \ (\text {mod}\ 3)$, $4(m+n)\equiv -3\ (\text {mod}\ 3)$, $m+n\equiv 0\ (\text {mod}\ 3)$, $m-2n\equiv 0 \ (\text {mod}\ 3)$, $m+n=3u$, $m-2n=-3v$, $m=2u-v$, $n=u+v$,

$$\begin{aligned} x=8u-4v+1,\;\text {and}\;y=4u+4v+2. \end{aligned}$$

(3.22)

In the sixth sum of (3.17),

$$\begin{aligned} x=12u+3\;\text {and}\;y=12v+6. \end{aligned}$$

(3.23)

Invoking (3.18), (3.19), (3.20), (3.21), (3.22), and (3.23) in (3.17), we arrive at

$$\begin{aligned} qF^1_2(q^8)=&\sum \limits _{u,v=-\infty }^{\infty }q^{(-4v-8u-3)^2+2(4u-4v)^2} +\sum \limits _{u,v=-\infty }^{\infty }q^{(-4v-8u-3)^2+2(4v-4u)^2}\\ {}&-\sum \limits _{u,v=-\infty }^{\infty }q^{(12u+3)^2+2(12v)^2} +\sum \limits _{u,v=-\infty }^{\infty }q^{(8u-4v+1)^2+2(-4u-4v-2)^2}\\ {}&+\sum \limits _{u,v=-\infty }^{\infty }q^{(8u-4v+1)^2+2(4u+4v+2)^2} -\sum \limits _{u,v=-\infty }^{\infty }q^{(12u+3)^2+2(12v+6)^2}\\ =&\sum \limits _{u,v=-\infty }^{\infty }q^{48(2u^2+u)+24(2v^2+v)+9} +\sum \limits _{u,v=-\infty }^{\infty }q^{48(2u^2+u)+24(2v^2+v)+9}\\ {}&-\sum \limits _{u,v=-\infty }^{\infty }q^{72(2u^2+u)+288v^2+9} +\sum \limits _{u,v=-\infty }^{\infty }q^{48(2u^2+u)+24(2v^2+v)+9}\\&+\sum \limits _{u,v=-\infty }^{\infty }q^{48(2u^2+u)+24(2v^2+v)+9} -\sum \limits _{u,v=-\infty }^{\infty }q^{72(2u^2+u)+288(v^2+v)+81}. \end{aligned}$$

Dividing throughout by q, replacing $q^8$ by q, and separating the terms of u and v in the above equation, we obtain

$$\begin{aligned} F^1_2(q)=&\,4q\sum \limits _{u=-\infty }^{\infty }q^{12 u^2+6u}\sum \limits _{v=-\infty }^{\infty }q^{6v^2+3v} -q\sum \limits _{u=-\infty }^{\infty }q^{18 u^2+9u}\sum \limits _{v=-\infty }^{\infty }q^{36v^2}\\ {}&-q^{10}\sum \limits _{u=-\infty }^{\infty }q^{18 u^2+9u}\sum \limits _{v=-\infty }^{\infty }q^{36v^2+36v}\\ =&\, 4q\psi (q^6)\psi (q^3)-q\psi (q^9)\varphi (q^{36}) -2q^{10}\psi (q^9)\psi (q^{72})\\ =&\, 4q\psi (q^6)\psi (q^3)-q\psi (q^9)\left( \varphi (q^{36}) +2q^{9}\psi (q^{72})\right) \\ =&\,4q\psi (q^6)\psi (q^3)-q\psi (q^9)\varphi (q^{9}). \end{aligned}$$

The above identity can be written as

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_1\left( 3n+1\right) q^{3n+1}=4q\psi (q^6)\psi (q^3)-q\psi (q^9)\varphi (q^{9}). \end{aligned}$$

(3.24)

Dividing throughout by q and then replacing $q^3$ by q in (3.24), we obtain (1.2).

Let $F_4(q)=\psi (q)\varphi (q^{2})$, and using (2.5), we find that

$$\begin{aligned} F_4(q)&=\psi (q)\left( \varphi (q^{8})+2q^{2}\psi (q^{16})\right) \nonumber \\&=\sum \limits _{m,n=-\infty }^{\infty }q^{2m^2+m+8n^2}+q^{2}\sum \limits _{m,n=-\infty }^{\infty }q^{2m^2+m+8(n^2+n)}. \end{aligned}$$

(3.25)

Replacing q by $q^8$ and then multiplying throughout by q in the above equation, we obtain

$$\begin{aligned} qF_4(q^8)&=\sum \limits _{m,n=-\infty }^{\infty }q^{(4m+1)^2+4(4n)^2}+\sum \limits _{m,n=-\infty }^{\infty }q^{(4m+1)^2+4(4n+2)^2}\nonumber \\&=\sum \limits _{x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)}q^{x^2+4y^2}+\sum \limits _{x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)}q^{x^2+4y^2}. \end{aligned}$$

(3.26)

Let us write

$$\begin{aligned} F_4(q)=F^0_4(q)+F^1_4(q)+F^2_4(q)+F^3_4(q)+F^4_4(q), \end{aligned}$$

(3.27)

where $F^i_4(q)$ consists of all those terms in $F_4(q)$ in which powers of q are congruent to i modulo 5. From (3.26) and (3.27), we see that

$$\begin{aligned} qF^3_4(q^8)=\sum \limits _{x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)}q^{x^2+4y^2}+\sum \limits _{x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)}q^{x^2+4y^2}, \end{aligned}$$

(3.28)

where, in both sums, $x^2+4y^2\equiv 0 \ (\text {mod}\ 5)$. The solution of $x^2+4y^2\equiv 0 \ (\text {mod}\ 5)$ is $x\equiv \pm y \ (\text {mod}\ 5),$ and it follows that

$$\begin{aligned}&qF^3_4(q^8)\nonumber \\&\quad =\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)\\ x\equiv y \ (\text {mod}\ 5) \end{array}}q^{x^2+4y^2} +\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)\\ x\equiv -y \ (\text {mod}\ 5) \end{array}}q^{x^2+4y^2}- \sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)\\ x\equiv y\equiv 0 \ (\text {mod}\ 5) \end{array}}q^{x^2+4y^2}\nonumber \\&\qquad +\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)\\ x\equiv y \ (\text {mod}\ 5) \end{array}}q^{x^2+4y^2} +\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)\\ x\equiv -y \ (\text {mod}\ 5) \end{array}}q^{x^2+4y^2} -\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)\\ x\equiv y\equiv 0 \ (\text {mod}\ 5) \end{array}}q^{x^2+4y^2}. \end{aligned}$$

(3.29)

In the first sum of (3.29), we have $x\equiv y \ (\text {mod}\ 5)$, $4m+1\equiv 4n \ (\text {mod}\ 5)$, $4(m-n)\equiv -1 \ (\text {mod}\ 5)$, $m-n\equiv 1 \ (\text {mod}\ 5)$, $m+4n\equiv 1 \ (\text {mod}\ 5)$, $m-n=5u+1$, $m+4n=5v+1$, $m=4u+v+1$, $n=v-u$,

$$\begin{aligned} x=16u+4v+5,\;\text {and}\;y=4v-4u. \end{aligned}$$

(3.30)

In the second sum of (3.29), $4m+1\equiv -4n \ (\text {mod}\ 5)$, $4(m+n)\equiv -1\ (\text {mod}\ 5)$, $4(m+n)\equiv -1+5\ (\text {mod}\ 5)$, $m+n\equiv 1 \ (\text {mod}\ 5)$, $m-4n\equiv 1 \ (\text {mod}\ 5)$, $m+n=5u+1$, $m-4n=5v+1$, $m=4u+v+1$, $n=u-v$,

$$\begin{aligned} x=16u+4v+5,\;\text {and}\;y=4u-4v. \end{aligned}$$

(3.31)

In the third sum of (3.29),

$$\begin{aligned} x=20u+5\;\text {and}\;y=20v. \end{aligned}$$

(3.32)

In the fourth sum of (3.29), $4m+1\equiv 4n+2 \ (\text {mod}\ 5)$, $4(m-n)\equiv 1\ (\text {mod}\ 5)$, $m-n\equiv -1\ (\text {mod}\ 5)$, $m+4n\equiv -1\ (\text {mod}\ 5)$, $m-n=-5u-1$, $m+4n=5v-1$, $m=v-4u-1$, $n=u+v$,

$$\begin{aligned} x=4v-16u-3,\;\text {and}\;y=4u+4v+2. \end{aligned}$$

(3.33)

In the fifth sum of (3.29), $4m+1\equiv -4n-2 \ (\text {mod}\ 5)$, $4(m+n)\equiv -3\ (\text {mod}\ 5)$, $m+n\equiv -2\ (\text {mod}\ 5)$, $m-4n\equiv 3\ (\text {mod}\ 5)$, $m+n=-5u-2$, $m-4n=5v+3$, $m=v-4u-1$, $n=-v-u-1$,

$$\begin{aligned} x=4v-16u-3,\;\text {and}\;y=-4u-4v-2. \end{aligned}$$

(3.34)

In the sixth sum of (3.29),

$$\begin{aligned} x=20u+5\;\text {and}\;y=20v+10. \end{aligned}$$

(3.35)

Employing (3.30), (3.31), (3.32), (3.33), (3.34), and (3.35) in (3.29), we obtain

$$\begin{aligned} qF^3_4(q^8)=&\sum \limits _{u,v=-\infty }^{\infty }q^{(16u+4v+5)^2+4(4v-4u)^2} +\sum \limits _{u,v=-\infty }^{\infty }q^{\left( 16u+4v+5\right) ^2+4(4u-4v)^2}\\&-\sum \limits _{u,v=-\infty }^{\infty }q^{\left( 20u+5\right) ^2+4\left( 20v\right) ^2} +\sum \limits _{u,v=-\infty }^{\infty }q^{\left( 4v-16u-3\right) ^2+4\left( 4v+4u+2\right) ^2}\\&+\sum \limits _{u,v=-\infty }^{\infty }q^{\left( 4v-16u-3\right) ^2+4\left( -4v-4u-2\right) ^2} -\sum \limits _{u,v=-\infty }^{\infty }q^{\left( 20u+5\right) ^2+4\left( 20v+10\right) ^2}\\ =&\sum \limits _{u,v=-\infty }^{\infty }q^{40(2v^2+v)+160(2u^2+u)+25} +\sum \limits _{u,v=-\infty }^{\infty }q^{40(2v^2+v)+160(2u^2+u)+25}\\&-\sum \limits _{u,v=-\infty }^{\infty }q^{200(2u^2+u)+1600v^2+25} +\sum \limits _{u,v=-\infty }^{\infty }q^{40(2v^2+v)+160(2u^2+u)+25}\\&+\sum \limits _{u,v=-\infty }^{\infty }q^{40(2v^2+v)+160(2u^2+u)+25} -\sum \limits _{u,v=-\infty }^{\infty }q^{200(2u^2+u)+1600(v^2+v)+425}. \end{aligned}$$

Dividing throughout by q, replacing $q^8$ by q, and then separating the terms containing u and v in the above equation, we find that

$$\begin{aligned} F^3_4(q)=&\, 4q^{3}\sum \limits _{v=-\infty }^{\infty }q^{10v^2+5v}\sum \limits _{u=-\infty }^{\infty }q^{40u^2+20u} -q^{3}\sum \limits _{u=-\infty }^{\infty }q^{50u^2+25u}\sum \limits _{v=-\infty }^{\infty }q^{200v^2}\nonumber \\&-q^{53}\sum \limits _{u=-\infty }^{\infty }q^{50u^2+25u}\sum \limits _{v=-\infty }^{\infty }q^{200v^2+200v}. \end{aligned}$$

(3.36)

Using (2.3) and (2.4) in (3.36), we obtain

$$\begin{aligned} {\begin{matrix} F^3_4(q)= 4q^{3}\psi (q^5)\psi (q^{20}) -q^{3}\psi (q^{25})\left( \varphi (q^{200}) +2q^{50}\psi (q^{400})\right) . \end{matrix}}\end{aligned}$$

(3.37)

From (2.5) and (3.37), we see that

$$\begin{aligned} F^3_4(q)=4q^{3}\psi (q^{5})\psi (q^{20})-q^{3}\psi (q^{25})\varphi (q^{50}). \end{aligned}$$

(3.38)

From (3.38), it follows that

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_2\left( 5n+3\right) q^{5n+3}=4q^{3}\psi (q^{5})\psi (q^{20})-q^{3}\psi (q^{25})\varphi (q^{50}). \end{aligned}$$

(3.39)

Dividing throughout by $q^3$ and then replacing $q^5$ by q in (3.39), we arrive at (1.3).

Let $F_6(q)=\psi \left( q\right) \varphi \left( q^3\right)$, and using (2.5), we find that

$$\begin{aligned} F_6(q)&=\psi (q)\left( \varphi (q^{12})+2q^3\psi (q^{24})\right) \nonumber \\&=\sum \limits _{m,n=-\infty }^{\infty }q^{2m^2+m+12n^2}+q\sum \limits _{m,n=-\infty }^{\infty }q^{2m^2+m+12(n^2+n)}. \end{aligned}$$

(3.40)

Replacing q by $q^8$ and then multiplying both sides of (3.40) by q, we see that

$$\begin{aligned} qF_6(q^8)&=\sum \limits _{m,n=-\infty }^{\infty }q^{(4m+1)^2+6(4n)^2}+\sum \limits _{m,n=-\infty }^{\infty }q^{(4m+1)^2+6(4n+2)^2}\nonumber \\&=\sum \limits _{x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)}q^{x^2+6y^2}+\sum \limits _{x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)}q^{x^2+6y^2}. \end{aligned}$$

(3.41)

If we write

$$\begin{aligned} F_6(q)=F^0_6(q)+F^1_6(q)+\cdots +F^6_6(q), \end{aligned}$$

(3.42)

where $F^i_6(q)$ consists of all those terms in $F_6(q)$ in which powers of q are congruent to i modulo 7, then from (3.41) and (3.42), we find that

$$\begin{aligned} qF^6_6(q^8)=\sum \limits _{x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)}q^{x^2+6y^2}+\sum \limits _{x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)}q^{x^2+6y^2}, \end{aligned}$$

(3.43)

where, in both sums, $x^2+6y^2\equiv 0 \ (\text {mod}\ 7)$. The solution of $x^2+6y^2\equiv 0 \ (\text {mod}\ 7)$ is $x\equiv \pm y \ (\text {mod}\ 7),$ and from (3.43), it follows that

$$\begin{aligned}&qF^6_6(q^8)\nonumber \\&=\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)\\ x\equiv y \ (\text {mod}\ 7) \end{array}}q^{x^2+6y^2} +\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)\\ x\equiv -y \ (\text {mod}\ 7) \end{array}}q^{x^2+6y^2}- \sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 0 \ (\text {mod}\ 4)\\ x\equiv y\equiv 0 \ (\text {mod}\ 7) \end{array}}q^{x^2+6y^2}\nonumber \\&+\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)\\ x\equiv y \ (\text {mod}\ 7) \end{array}}q^{x^2+6y^2} +\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)\\ x\equiv -y \ (\text {mod}\ 7) \end{array}}q^{x^2+6y^2} -\sum \limits _{\begin{array}{c} x\equiv 1,\;y\equiv 2 \ (\text {mod}\ 4)\\ x\equiv y\equiv 0 \ (\text {mod}\ 7) \end{array}}q^{x^2+6y^2}. \end{aligned}$$

(3.44)

In the first sum of (3.44), we have $4m+1\equiv 4n \ (\text {mod}\ 7)$, $4(m-n)\equiv -1\ (\text {mod}\ 7)$, $4(m-n)\equiv -8\ (\text {mod}\ 7)$, $m-n\equiv -2\ (\text {mod}\ 7)$, $m+6n\equiv -2\ (\text {mod}\ 7)$, $m-n=-7v-2$, $m+6n=-7u-2$, $m=-u-6v-2$, $n=v-u$,

$$\begin{aligned} x=-4u-24v-7,\;\text {and}\;y=4v-4u. \end{aligned}$$

(3.45)

In the second sum of (3.44), $4m+1\equiv -4n \ (\text {mod}\ 7)$, $4(m+n)\equiv -1\ (\text {mod}\ 7)$, $4(m+n)\equiv -8\ (\text {mod}\ 7)$, $m+n\equiv -2\ (\text {mod}\ 7)$, $m-6n\equiv -2\ (\text {mod}\ 7)$, $m+n=-7v-2$, $m-6n=-7u-2$, $m=-6v-u-2$, $n=u-v$,

$$\begin{aligned} x=-24v-4u-7,\;\text {and}\;y=4u-4v. \end{aligned}$$

(3.46)

In the third sum of (3.44),

$$\begin{aligned} x=28u+7\;\text {and}\;y=28v. \end{aligned}$$

(3.47)

In the fourth sum of (3.44), $4m+1\equiv 4n+2 \ (\text {mod}\ 7)$, $4(m-n)\equiv 1\ (\text {mod}\ 7)$, $m-n\equiv 2\ (\text {mod}\ 7)$, $m+6n\equiv -5\ (\text {mod}\ 7)$, $m-n=7v+2$, $m+6n=-7u-5$, $m=6v-u+1$, $n=-v-u-1$,

$$\begin{aligned} x=24v-4u+5,\;\text {and}\;y=-4u-4v-2. \end{aligned}$$

(3.48)

In the fifth sum of (3.44), $4m+1\equiv -4n-2 \ (\text {mod}\ 7)$, $4(m+n)\equiv -3\ (\text {mod}\ 7)$, $m+n\equiv 1\ (\text {mod}\ 7)$, $m-6n\equiv 1\ (\text {mod}\ 7)$, $m+n=7v+1$, $m-6n=-7u+1$, $m=6v-u+1$, $n=u+v$,

$$\begin{aligned} x=24v-4u+5,\;\text {and}\;y=4u+4v+2. \end{aligned}$$

(3.49)

In the sixth sum of (3.44),

$$\begin{aligned} x=28u+7\;\text {and}\;y=28v+14. \end{aligned}$$

(3.50)

Invoking (3.45), (3.46), (3.47), (3.48), (3.49), and (3.50) in (3.44), we find that

$$\begin{aligned} qF^6_6(q^8)=&\sum \limits _{u,v=-\infty }^{\infty }q^{(-4u-24v-7)^2+6(4v-4u)^2} +\sum \limits _{u,v=-\infty }^{\infty }q^{(-24v-4u-7)^2+6(4u-4v)^2}\nonumber \\ {}&\nonumber -\sum \limits _{u,v=-\infty }^{\infty }q^{(28u+7)^2+6(28v)^2} +\sum \limits _{u,v=-\infty }^{\infty }q^{(24v-4u+5)^2+6(-4u-4v-2)^2}\nonumber \\ {}&+\sum \limits _{u,v=-\infty }^{\infty }q^{(24v-4u+5)^2+6(4u+4v+2)^2} -\sum \limits _{u,v=-\infty }^{\infty }q^{(28u+7)^2+6(28v+14)^2}. \end{aligned}$$

(3.51)

After simplification of (3.51), we obtain

$$\begin{aligned} qF^6_6(q^8)=&\sum \limits _{u,v=-\infty }^{\infty }q^{56(2u^2+u)+336(2v^2+v)+49} +\sum \limits _{u,v=-\infty }^{\infty }q^{56(2u^2+u)+336(2v^2+v)+49}\nonumber \\ {}&\nonumber -\sum \limits _{u,v=-\infty }^{\infty }q^{392(2u^2+u)+4704v^2+49} +\sum \limits _{u,v=-\infty }^{\infty }q^{56(2u^2+u)+336(2v^2+v)+49}\nonumber \\&+\sum \limits _{u,v=-\infty }^{\infty }q^{56(2u^2+u)+336(2v^2+v)+49} -\sum \limits _{u,v=-\infty }^{\infty }q^{392(2u^2+u)+4704(v^2+v)+1225}. \end{aligned}$$

(3.52)

Dividing throughout by q, replacing $q^8$ by q, and separating the terms of u and v in (3.52), we obtain

$$\begin{aligned} F^6_6(q)=&\,4q^6\sum \limits _{u=-\infty }^{\infty }q^{14u^2+7u}\sum \limits _{v=-\infty }^{\infty }q^{84v^2+42v} -q^6\sum \limits _{u=-\infty }^{\infty }q^{98u^2+49u}\sum \limits _{v=-\infty }^{\infty }q^{588v^2}\nonumber \\ {}&-q^{153}\sum \limits _{u=-\infty }^{\infty }q^{98u^2+49u}\sum \limits _{v=-\infty }^{\infty }q^{588v^2+588v}. \end{aligned}$$

(3.53)

From (2.3), (2.4), and (3.53), we see that

$$\begin{aligned} F^6_6(q)=4q^6\psi (q^7)\psi (q^{42})-q^6\psi (q^{49})\left( \varphi (q^{588}) +2q^{147}\psi (q^{1176})\right) . \end{aligned}$$

(3.54)

From (2.5) and (3.54), we obtain

$$\begin{aligned} F^6_6(q)=4q^6\psi (q^7)\psi (q^{42})-q^6\psi (q^{49})\varphi (q^{147}). \end{aligned}$$

(3.55)

The above equation can be written as

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_3\left( 7n+6\right) q^{7n+6}=4q^6\psi (q^7)\psi (q^{42})-q^6\psi (q^{49})\varphi (q^{147}). \end{aligned}$$

(3.56)

Dividing throughout by $q^6$ and then replacing $q^7$ by q in (3.56), we arrive at (1.4). This completes the proof. $\square$

Proof of Theorem 1.2

Put $k=2$ in (1.1), we find that

$$\begin{aligned} \sum \limits _{n=0}^{\infty }a_2(5n)q^{n}=\sum \limits _{n=0}^{\infty }b_2(n)q^{n}-\sum \limits _{n=0}^{\infty }a_2(n)q^{5n+3}, \end{aligned}$$

(3.57)

which yields

$$\begin{aligned} a_2(n)+a_2(25n+15)=b_2(5n+3). \end{aligned}$$

(3.58)

From (1.3), it follows that

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_2(5n+3)q^{n}=4\sum \limits _{n=0}^{\infty }a_2(n)q^{n}-\sum \limits _{n=0}^{\infty }b_2(n)q^{5n}. \end{aligned}$$

(3.59)

Equating coefficients of $q^{5n}$ in (3.59), we obtain

$$\begin{aligned} b_2\left( n\right) +b_2\left( 25n+3\right) =4a_2\left( 5n\right) . \end{aligned}$$

(3.60)

Equations (1.5) and (1.6) follow from (3.58) and (3.60). This completes the proof. $\square$

Proof of Corollary 1.1

From Eq. (1.2), we see that

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_1(3n+1)q^n\equiv 3\psi (q^3)\varphi (q^3) \pmod {4}. \end{aligned}$$

(3.61)

Equating coefficients of $q^{3n}$ on both sides of (3.61), we find that

$$\begin{aligned} \sum \limits _{n=0}^{\infty }b_1(9n+1)q^n\equiv 3\psi (q)\varphi (q) \pmod {4}. \end{aligned}$$

(3.62)

From (3.62) and by mathematical induction, we arrive at (1.7). Similarly, we obtain (1.8) and (1.9) from (1.3) and (1.4), respectively. This completes the proof. $\square$

Proof of Corollary 1.2

In view of Eq. (1.5), we see that

$$\begin{aligned} a_2\left( 625n+390\right) \equiv a_2(n) \ (\text {mod}\ 2). \end{aligned}$$

(3.63)

By mathematical induction on n in (3.63), we arrive at (1.10). Similarly, we obtain (1.11) from (1.6). This completes the proof. $\square$