1 Introduction
For any complex number a, we define

and

We set
$E(q):=(q;q)_\infty =\prod _{n=1}^{\infty }(1-q^n)$
;
$E(q)$
is known as Euler’s product. We can view
$E(q)$
as a particular case of Ramanujan’s theta function (see [Reference Berndt1, page 34]): that is,

Jacobi’s triple product identity (see [Reference Hirschhorn5, page 5]) is

Replacing
$qz$
by a and
$q/z$
by b in Jacobi’s triple product identity, yields

In particular,

A t-dissection of a power series
$A(q)$
is given by
$A(q)=\sum _{k=0}^{t-1}q^kA_k(q^t)$
, where the
$A_k(q^t)$
are power series in
$q^t$
. The
$3$
-dissection of
$E(q)$
(see [Reference Berndt1, entry 31, page 48]) is

We also have

which are, respectively, 5-, 7- and 11- dissections of
$E(q)$
(see [Reference Berndt1, pages 82, 303 and 363]. A generalised form of these dissections is given in the following theorem.
Theorem 1.1 [Reference Berndt1, Theorem 12.1, page 274].
Suppose m is a positive integer with
$m\equiv 1 \ (\text {mod}~6)$
. If
$m=6t+1$
with t positive, then

If
$m=6t-1$
with t positive, then

Theorem 1.1 was found independently by Ramanathan [Reference Ramanathan7] and Evans [Reference Evans4]. Recently, it was reproved by McLaughlin [Reference McLaughlin6] while establishing some other general dissections involving infinite products.
Hirschhorn [Reference Hirschhorn5, page 332] gave the 2- and 4- dissections of
$E(q)$
as


where
$(a_1, a_2,\ldots ,a_n;q)_\infty :=(a_1;q)_\infty (a_2;q)_\infty \cdots (a_n;q)_\infty $
.
Identity (1.2) was obtained by 4-dissecting the products
$(q^6,q^{10},q^{16};q^{16})_\infty $
and
$(q^2,q^{14},q^{16};q^{16})_\infty $
in (1.1). Hirschhorn also remarked that we can continue in the same manner and find the 8-dissection, the 16-dissection and so on. He concluded by noting the following conjecture on the
$2^n$
-dissection of
$E(q)$
.
Conjecture 1.2.

where


and

Cao [Reference Cao3] discussed product identities for theta functions using integer matrix exact covering systems. In particular, he gave the following result involving the product of two theta functions.
Theorem 1.3 [Reference Cao3, Corollary 2.2].
If
$|ab|<1$
and
$cd=(ab)^{k_1k_2}$
, where both
$k_1$
and
$k_2$
are positive integers, then

Finally, we give the following version of the quintuple product identity.
Theorem 1.4 (Quintuple product identity [Reference Berndt2, page 19]).
For
$a\neq 0$
,

We use (1.6), (1.7) and other properties of Ramanujan’s theta function to prove Conjecture 1.2. In Section 2, we mention some preliminary results involving Ramanujan’s theta function and the pentagonal numbers
$P(n)$
. In Section 3, we provide the proof of Conjecture 1.2.
2 Preliminaries
Lemma 2.1. If
$ab=cd$
, then

Proof. Set
$k_1=k_2=1$
in (1.6).
Lemma 2.2 [Reference Berndt1, Entry 18, page 34].
For any integer n,

Lemma 2.3. We have

Proof. Set
$n=1$
in (2.2).
Lemma 2.4. Let
$P(n)$
be as in (1.5). For positive integers n and k,

Proof. We have

Replacing s and t by
$2^{n+1}+k-1$
and
$k-1$
, respectively, in (2.5), we find that

Lemma 2.5. If n is even and
$c_k=P(-{(2^{n+1}-2)}/{3}+(k-1))$
, then

Proof. We have
$c_k=P(((3k-2^{n+1})-1)/{3})$
, so

Similar calculations give the formulas in the following three lemmas.
Lemma 2.6. If n is even and
$c_k=P(-{(2^{n+1}-2)}/{3}+(k-1))$
, then

Lemma 2.7. If n is odd and
$c_k=P({(2^{n+1}-1)}/{3}-(k-1))$
, then

Lemma 2.8. If n odd is and
$c_k=P({(2^{n+1}-1)}/{3}-(k-1))$
, then

3 Proof of the main result
Theorem 3.1. Conjecture 1.2 is true.
Proof. We set
$a=-q$
,
$b=-q^2$
,
$c=-q^{2^{2n+2}} d=-q^{2^{2n+3}}$
,
$k_1=k_2=2^{n+1}$
and
$r=k-1$
in (1.6) to obtain

Since
$f(-q,-q^2)=E(q)$
,
$f(-q^{2^{2n+2}},-q^{2^{2n+3}})=E(q^{2^{2n+2}})$
,
$P(k-1)={(k-1)(3k-4)}/{2}$
, we can rewrite (3.1) as

where

Now

Employing (2.4) and simplifying the exponents of the arguments of the theta functions, we can rewrite (3.4) as


Setting
$a=q^{3\times 2^{2n+1}+2^n(6k-7)}$
,
$b=q^{3\times 2^{2n+1}-2^n(6k-7)}$
,
$c=-q^{2^{2n+2}}$
and
$d=-q^{2^{2n+3}}$
in (3.6) gives


From (3.2),


where
$H_k=(-1)^{k+1}q^{P(k-1)}f(q^{3\times 2^{2n+1}+2^n(6k-7)},q^{3\times 2^{2n+1}-2^n(6k-7)})$
. We now arrange the
$H_k$
in pairs so that each of these pairs can be reduced to quintuple products. For this purpose, we consider two separate cases according to whether n is even or odd.
Case I: n is even. We consider


Note that
$S_2$
is the sum of
$H_1,H_2,\ldots ,H_{{(2^{n+1}+4)}/{3}}$
and
$S_1$
is the sum of
$H_{{(2^{n+1}+7)}/{3}}$
,
$H_{{(2^{n+1}+10)}/{3}},\ldots ,H_{2^{n+1}}$
and so

We also note that
$c_k$
is given by (1.4).
Now

Employing Lemma 2.5 in (3.13) and simplifying the exponents of the arguments of the theta functions, we find that

Also

Employing Lemma 2.6 in (3.15) and simplifying the exponents,


Substituting the expressions for
$H_{{(2^{n+2}+2)}/{3}-k+1}$
and
$H_{{(2^{n+2}+5)}/{3}+k-1}$
from (3.14) and (3.17), respectively, in (3.10), we find that

Further,

Also,

Employing Lemma 2.5 in (3.20) and simplifying the exponents,

Using (3.19) and (3.21) in (3.11),

From (3.12), (3.18) and (3.22),

Case II: n is odd. In this case, we consider the sums


Note that
$S_3$
is the sum of
$H_1,H_2,\ldots ,H_{{(2^{n+2}+4)}/{3}}$
and
$S_4$
is the sum of
$H_{{(2^{n+2}+7)}/{3}}$
,
$H_{{(2^{n+2}+10)}/{3}},\ldots ,H_{2^{n+1}}$
and so

We also note that
$c_k$
is given by (1.3).
We have

Similarly,

Using Lemma 2.7 in (3.28) and simplifying the exponents,

Now we employ (2.3) in (3.29) to obtain

Using (3.27) and (3.30) in (3.24),

Also,

Using Lemma 2.7 in (3.32) and simplifying the exponents,

Employing (2.2) in (3.33), we find that

Further,

Using Lemma 2.8 in (3.35), we find that


From (3.34), (3.37) and (3.25),

Substituting the expressions for
$S_3$
and
$S_4$
from (3.31) and (3.38), respectively, in (3.26),

From (3.23) and (3.39) we find that, for any positive integer n,

where the appropriate expression for
$c_k$
is chosen according to whether n is even or odd.
Setting
$A=-q^{-2^{2n+1}+2^n(2k-1)}$
and
$Q=q^{2^{2n+2}}$
in (3.40), we obtain

Employing the quintuple product identity in (3.41),

Thus we have completed the proof of Conjecture 1.2.