$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}.$
(6.31)

In Theorem 5.4, we further set

$\displaystyle\begin{array}[]{rclp{4.5cm}rcl}A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&(2m+1+2\tau)\mu\\\
w&\mapsto&\lx@intercol-(3\ell+m+2)k+\xi k+\tau k\hfil\lx@intercol\\\
\end{array}$

Then

$\displaystyle\check{w}=-(3\ell+m+2)k+(2\ell+1-\xi)k-\tau k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I},$
$\displaystyle\check{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2(A^{\prime}Bu+AB^{\prime}v)}{d^{*}AB}=\frac{(2m+1+2\xi+2\tau)k}{d^{*}}.$

Similarly, if $d^{*}\nmid(2m+1+2\xi+2\tau)k$, then $(\xi+\tau)k\not\equiv
0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv(\xi+\tau)k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\check{w}$ $\displaystyle\equiv-(\xi+\tau)k\not\equiv
0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid(2m+1+2\xi+2\tau)k$. Then the first assumption
in (5.24) is satisfied. We may also assume that $w$ is a multiple of
$d^{*}=d_{0}$ according to Theorem 5.4 (i). With the choice of $K=1$, we
compute that

$\displaystyle\frac{B(A-2A^{\prime})u^{2}+A(B-2B^{\prime})uv+2ABuw\cdot
K}{A(Av^{2}+Bu^{2})}$ $\displaystyle=-2,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.24) are satisfied. Finally, we compute that

$\displaystyle\varepsilon=(4\ell+2)\kappa+(2m+1)\lambda.$

Thus, by Theorem 5.4 (ii),

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\lambda}\mathbf{H}_{M}\big{(}\check{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}+(-1)^{\kappa}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa}\mathbf{H}_{M}\big{(}\check{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.31) is established.

####  6.2.2. Equation (1.15)

Recall that $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=2$. Also,
$M=2\ell+4m+3$ and $\sigma=-(2\ell+2m+2)k$. Further,
$(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ and $k$ is such that $\gcd(k,M)=1$.

In Corollary 5.2 and Theorems 5.3 and 5.4, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+1)\kappa&&\lambda&\mapsto&(2m+1)\lambda\\\
u&\mapsto&(2\ell+1)k&&v&\mapsto&2(2m+1)k\\\
A&\mapsto&(2\ell+1)\mu&&B&\mapsto&2(2m+1)\mu\\\ M&\mapsto&\lx@intercol
2\ell+4m+3\hfil\lx@intercol\\\ \end{array}$

Let

$\displaystyle
d_{0}=\gcd\big{(}2\ell+1,2(2m+1)\big{)}=\gcd\big{(}2\ell+1,2m+1\big{)}.$

Then

$\displaystyle d=\gcd(u,v)=d_{0}k.$

Noticing that $M=(2\ell+1)+2(2m+1)$, and that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=d_{0},$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=d_{0},$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=d_{0}.$

We compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$ $\displaystyle=1,$
$\displaystyle\frac{Bu\cdot dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=1.$

Thus, the second and third assumptions in (5.9), (5.16) and (5.24) are
satisfied.

  * $\blacktriangleright$

Examining (6.22). Let us keep in mind that we have made the substitutions in
(6.16) and (6.19): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=2$. Also,
$\sigma=-(2\ell+2m+2)k$. We want to show (6.22),

$\displaystyle H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$
(6.32)

In Corollary 5.2, we further set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&(2m+1)\mu\\\
w&\mapsto&\lx@intercol-(2\ell+2m+2)k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}.$

We may further assume that $w$ is a multiple of $d^{*}$. Otherwise, we know
from Theorem 5.1 (i) that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

which gives

$\displaystyle H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$

We further compute that

$\displaystyle\frac{-v\kappa+u\lambda}{d}=-\frac{(2\ell+1)(2m+1)}{d_{0}}(2\kappa-\lambda).$

The fact that $d_{0}=\gcd(2\ell+1,2m+1)$ then implies that
$(2\ell+1)(2m+1)/d_{0}$ is an odd integer. Thus, (5.9) is satisfied since
$(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$. Finally, we choose $J=0$ in (5.11).
Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\frac{d_{0}-(2\ell+1)}{2d_{0}},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\frac{d_{0}+(2m+1)}{d_{0}}.\end{aligned}$

Recall that $2\ell+1\equiv 2m+1\equiv d_{0}\pmod{2d_{0}}$ since
$d_{0}=\gcd(2\ell+1,2m+1)$ is odd. So (5.11) is also satisfied. We conclude by
Corollary 5.2 that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0.$

Thus, (6.32) is established.

  * $\blacktriangleright$

Examining (6.23). Let us keep in mind that we have made the substitutions in
(6.16), (6.20) and (6.21): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=2$.
Also, $\sigma=-(2\ell+2m+2)k$. We want to show (6.23),

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}.$
(6.33)

In Theorem 5.3, we further set

$\displaystyle\begin{array}[]{rclp{3.5cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&(2m+1+2\tau)\mu\\\
w&\mapsto&\lx@intercol-(2\ell+2m+2)k+2\tau k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(2\ell+2m+2)k-2\tau k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{2(2m+1+2\tau)k}{d^{*}}.$

If $d^{*}\nmid 2(2m+1+2\tau)k$, then $2\tau k\not\equiv 0\pmod{d_{0}}$, and
thus,

$\displaystyle w$ $\displaystyle=-(2\ell+1)k-(2m+1)k+2\tau k\equiv 2\tau
k\not\equiv 0\pmod{d_{0}},$ $\displaystyle\hat{w}$
$\displaystyle=-(2\ell+1)k-(2m+1)k-2\tau k\equiv-2\tau k\not\equiv
0\pmod{d_{0}}.$

So $w$ and $\hat{w}$ are nonmultiples of $d_{0}=d^{*}$. In this case we know
from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}=0,$

which gives

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2(2m+1+2\tau)k$. Then the first assumption in
(5.16) is satisfied. We may also assume that $w$ is a multiple of
$d^{*}=d_{0}$ according to Theorem 5.3 (i). It is easily seen that $d^{*}k$ is
a divisor of $\gcd(u,v,w)$. Thus, we solve the following stronger system

$\displaystyle\left\\{\begin{aligned} K&\equiv 1\pmod{M/d^{*}},\\\ K&\equiv
0\pmod{d/(d^{*}k)},\end{aligned}\right.$

and choose $K=1$. Finally, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=2,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(4\ell+2)\kappa+(2m+1)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\lambda}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}+H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.33) is established.

  * $\blacktriangleright$

Examining (6.24). Let us keep in mind that we have made the substitutions in
(6.17), (6.18) and (6.19): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=2$.
Also, $\sigma=-(2\ell+2m+2)k$. Then (6.24) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}.$
(6.34)

To make use of Theorem 5.3, we need to swap the choice of $(\kappa,u,A)$ and
$(\lambda,v,B)$ in our initial setting. In other words, in Theorem 5.3, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2m+1)\lambda&&\lambda&\mapsto&(2\ell+1)\kappa\\\
u&\mapsto&2(2m+1)k&&v&\mapsto&(2\ell+1)k\\\
A&\mapsto&2(2m+1)\mu&&B&\mapsto&(2\ell+1)\mu\\\
A^{\prime}&\mapsto&(2m+1)\mu&&B^{\prime}&\mapsto&\xi\mu\\\
M&\mapsto&\lx@intercol 2\ell+4m+3\hfil\lx@intercol\\\
w&\mapsto&\lx@intercol-(2\ell+2m+2)k+\xi k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(2\ell+2m+2)k+(2\ell+1-\xi)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{2\xi k}{d^{*}}.$

If $d^{*}\nmid 2\xi k$, then $\xi k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv\xi k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\hat{w}$ $\displaystyle\equiv-\xi k\not\equiv 0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2\xi k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). With the choice of $K=1$, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=1,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(2\ell+1)\kappa+(2m+1)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\kappa+\lambda}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}+(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.34) is established.

  * $\blacktriangleright$

Examining (6.25). Let us keep in mind that we have made the substitutions in
(6.17), (6.18), (6.20) and (6.21): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and
$b=2$. Also, $\sigma=-(2\ell+2m+2)k$. Then (6.25) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}.$
(6.35)

In Theorem 5.4, we further set

$\displaystyle\begin{array}[]{rclp{4.5cm}rcl}A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&(2m+1+2\tau)\mu\\\
w&\mapsto&\lx@intercol-(2\ell+2m+2)k+\xi k+2\tau k\hfil\lx@intercol\\\
\end{array}$

Then

$\displaystyle\check{w}=-(2\ell+2m+2)k+(2\ell+1-\xi)k-2\tau k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I},$
$\displaystyle\check{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2(A^{\prime}Bu+AB^{\prime}v)}{d^{*}AB}=\frac{2(2m+1+\xi+2\tau)k}{d^{*}}.$

Similarly, if $d^{*}\nmid 2(2m+1+\xi+2\tau)k$, then $(\xi+2\tau)k\not\equiv
0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv(\xi+2\tau)k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\check{w}$ $\displaystyle\equiv-(\xi+2\tau)k\not\equiv
0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2(2m+1+\xi+2\tau)k$. Then the first
assumption in (5.24) is satisfied. We may also assume that $w$ is a multiple
of $d^{*}=d_{0}$ according to Theorem 5.4 (i). With the choice of $K=1$, we
compute that

$\displaystyle\frac{B(A-2A^{\prime})u^{2}+A(B-2B^{\prime})uv+2ABuw\cdot
K}{A(Av^{2}+Bu^{2})}$ $\displaystyle=-1,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.24) are satisfied. Finally, we compute that

$\displaystyle\varepsilon=(2\ell+1)\kappa+(2m+1)\lambda.$

Thus, by Theorem 5.4 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\kappa+\lambda}H_{M}\big{(}\check{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}+(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa}H_{M}\big{(}\check{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.35) is established.

###  6.3. Type I — Theorem 1.3

In this section, we treat the coefficient-vanishing results in Theorem 1.3.
The basic idea is similar to that for Theorem 1.2.

We still deduce from Corollary 3.6 with certain common factors extracted and
powers of $(-1)$ modified that, for any $\ell\geq 1$,

$\displaystyle
f\big{(}(-1)^{\kappa}q^{ak},(-1)^{\kappa}q^{-ak+M\mu}\big{)}^{\ell}=\sum_{s=0}^{\ell-1}\mathscr{A}_{s}\mathscr{F}_{s},$

where each $\mathscr{F}_{\star}$ is a series in $q^{M}$, and

$\displaystyle\mathscr{A}_{0}=f\big{(}(-1)^{\kappa\ell}q^{ak\ell},(-1)^{\kappa\ell}q^{-ak\ell+M\mu\ell}\big{)},$
(6.36)

and for $1\leq s\leq\ell-1$,

$\displaystyle\mathscr{A}_{s}$
$\displaystyle=q^{aks}f\big{(}(-1)^{\kappa\ell}q^{ak\ell+M\mu
s},(-1)^{\kappa\ell}q^{-ak\ell+M\mu(\ell-s)}\big{)}$
$\displaystyle\quad+(-1)^{\kappa\ell}q^{ak(\ell-s)}f\big{(}(-1)^{\kappa\ell}q^{ak\ell+M\mu(\ell-s)},(-1)^{\kappa\ell}q^{-ak\ell+M\mu
s}\big{)}$
$\displaystyle=:\mathscr{A}_{s,I}+(-1)^{\kappa\ell}\mathscr{A}_{s,I\\!I}.$

More generally, we consider

$\displaystyle\mathscr{A}_{I}$ $\displaystyle:=q^{a\xi
k}f\big{(}(-1)^{\kappa\ell}q^{ak\ell+M\mu\xi},(-1)^{\kappa\ell}q^{-ak\ell+M\mu(\ell-\xi)}\big{)},$
(6.37) $\displaystyle\mathscr{A}_{I\\!I}$
$\displaystyle:=q^{a(\ell-\xi)k}f\big{(}(-1)^{\kappa\ell}q^{ak\ell+M\mu(\ell-\xi)},(-1)^{\kappa\ell}q^{-ak\ell+M\mu\xi}\big{)},$
(6.38)

for generic $\xi\in\mathbb{Z}$.

Now in this occasion, by Corollary 3.6, we write for any $m\geq 1$,

$\displaystyle
f\big{(}(-1)^{\lambda}q^{bk},(-1)^{\lambda}q^{-bk+2M\mu}\big{)}^{m}=\sum_{t=0}^{m-1}\mathscr{B}_{t}\mathscr{G}_{t},$

where each $\mathscr{G}_{\star}$ is a series in $q^{M}$, and

$\displaystyle\mathscr{B}_{0}=f\big{(}(-1)^{\lambda m}q^{bkm},(-1)^{\lambda
m}q^{-bkm+2M\mu m}\big{)},$ (6.39)

and for $1\leq t\leq m-1$,

$\displaystyle\mathscr{B}_{t}$ $\displaystyle=q^{bkt}f\big{(}(-1)^{\lambda
m}q^{bkm+2M\mu t},(-1)^{\lambda m}q^{-bkm+2M\mu(m-t)}\big{)}$
$\displaystyle\quad+(-1)^{\lambda m}q^{bk(m-t)}f\big{(}(-1)^{\lambda
m}q^{bkm+2M\mu(m-t)},(-1)^{\lambda m}q^{-bkm+2M\mu t}\big{)}$
$\displaystyle=:\mathscr{B}_{t,I}+(-1)^{\lambda m}\mathscr{B}_{t,I\\!I}.$

More generally, we consider

$\displaystyle\mathscr{B}_{I}$ $\displaystyle:=q^{b\tau
k}f\big{(}(-1)^{\lambda m}q^{bkm+2M\mu\tau},(-1)^{\lambda
m}q^{-bkm+2M\mu(m-\tau)}\big{)},$ (6.40) $\displaystyle\mathscr{B}_{I\\!I}$
$\displaystyle:=q^{b(m-\tau)k}f\big{(}(-1)^{\lambda
m}q^{bkm+2M\mu(m-\tau)},(-1)^{\lambda m}q^{-bkm+2M\mu\tau}\big{)},$ (6.41)

for generic $\tau\in\mathbb{Z}$.

With all parameters chosen in each case below, our generic target is to show
the following:

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0,$
(6.42)
$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda
m}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)},$
(6.43)
$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa\ell}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)},$
(6.44)

and

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa\ell+\lambda
m}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}.$
(6.45)

Upon setting $\tau\mapsto m-\tau$ in (6.45), we automatically have

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I\\!I}\big{)}=-(-1)^{\kappa\ell+\lambda
m}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I}\big{)}.$
(6.46)

Once the above relations are established, it is safe to conclude by our
pairing that

$\displaystyle\mathbf{H}_{M}\Big{(}q^{\sigma}f\big{(}(-1)^{\kappa}q^{ak},(-1)^{\kappa}q^{-ak+M\mu}\big{)}^{\ell}f\big{(}(-1)^{\lambda}q^{bk},(-1)^{\lambda}q^{-bk+2M\mu}\big{)}^{m}\Big{)}=0.$
(6.47)

####  6.3.1. Equation (1.16)

Recall that $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=1$. Also,
$M=4\ell+2m+3$ and $\sigma=-(\ell+m+1)k$. Further,
$(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ and $k$ is such that $\gcd(k,M)=1$.

In Corollary 5.2 and Theorems 5.3 and 5.4, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+1)\kappa&&\lambda&\mapsto&(2m+1)\lambda\\\
u&\mapsto&(2\ell+1)k&&v&\mapsto&(2m+1)k\\\
A&\mapsto&(2\ell+1)\mu&&B&\mapsto&2(2m+1)\mu\\\ M&\mapsto&\lx@intercol
4\ell+2m+3\hfil\lx@intercol\\\ \end{array}$

Let

$\displaystyle d_{0}=\gcd\big{(}2\ell+1,2m+1\big{)}.$

Then

$\displaystyle d=\gcd(u,v)=d_{0}k.$

Noticing that $M=2(2\ell+1)+(2m+1)$, and that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=d_{0},$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=d_{0},$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=d_{0}.$

We compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$ $\displaystyle=1,$
$\displaystyle\frac{Bu\cdot dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=2.$

Thus, the second and third assumptions in (5.9), (5.16) and (5.24) are
satisfied.

  * $\blacktriangleright$

Examining (6.42). Let us keep in mind that we have made the substitutions in
(6.36) and (6.39): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=1$. Also,
$\sigma=-(\ell+m+1)k$. We want to show (6.42),

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$
(6.48)

In Corollary 5.2, we further set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&0\\\
w&\mapsto&\lx@intercol-(\ell+m+1)k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}.$

We may further assume that $w$ is a multiple of $d^{*}$. Otherwise, we know
from Theorem 5.1 (i) that

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

which gives

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$

We further compute that

$\displaystyle\frac{-v\kappa+u\lambda}{d}=-\frac{(2\ell+1)(2m+1)}{d_{0}}(\kappa-\lambda).$

The fact that $d_{0}=\gcd(2\ell+1,2m+1)$ then implies that
$(2\ell+1)(2m+1)/d_{0}$ is an odd integer. Thus, (5.9) is satisfied since
$(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$. Finally, we choose $J=0$ in (5.11).
Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\frac{d_{0}-(2\ell+1)}{2d_{0}},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\frac{d_{0}+(2m+1)}{2d_{0}}.\end{aligned}$

Recall that $2\ell+1\equiv 2m+1\equiv d_{0}\pmod{2d_{0}}$ since
$d_{0}=\gcd(2\ell+1,2m+1)$ is odd. So (5.11) is also satisfied. We conclude by
Corollary 5.2 that

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=0.$

Thus, (6.48) is established.

  * $\blacktriangleright$

Examining (6.43). Let us keep in mind that we have made the substitutions in
(6.36), (6.40) and (6.41): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=1$.
Also, $\sigma=-(\ell+m+1)k$. Then (6.43) becomes

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}.$
(6.49)

In Theorem 5.3, we further set

$\displaystyle\begin{array}[]{rclp{3.5cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&2\tau\mu\\\
w&\mapsto&\lx@intercol-(\ell+m+1)k+\tau k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(\ell+m+1)k+(2m+1-\tau)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{2\tau k}{d^{*}}.$

If $d^{*}\nmid 2\tau k$, then $\tau k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$
$\displaystyle=-\tfrac{1}{2}(2\ell+1)k-\tfrac{1}{2}(2m+1)k+\tau k\equiv\tau
k\not\equiv 0\pmod{d_{0}},$ $\displaystyle\hat{w}$
$\displaystyle=-\tfrac{1}{2}(2\ell+1)k+\tfrac{1}{2}(2m+1)k-\tau k\equiv-\tau
k\not\equiv 0\pmod{d_{0}}.$

So $w$ and $\hat{w}$ are nonmultiples of $d_{0}=d^{*}$. In this case we know
from Theorem 5.1 (i) that

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=\mathbf{H}_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}=0,$

which gives

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2\tau k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). It is easily seen that $d^{*}k$ is a divisor of
$\gcd(u,v,w)$. Thus, we solve the following stronger system

$\displaystyle\left\\{\begin{aligned} K&\equiv 1\pmod{M/d^{*}},\\\ K&\equiv
0\pmod{d/(d^{*}k)},\end{aligned}\right.$

and choose $K=1$. Finally, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=1,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=0.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(2\ell+1)\kappa.$

Thus, by Theorem 5.3 (ii),

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\kappa}\mathbf{H}_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}+(-1)^{\lambda}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\lambda}\mathbf{H}_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.49) is established.

  * $\blacktriangleright$

Examining (6.44). Let us keep in mind that we have made the substitutions in
(6.37), (6.38) and (6.39): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=1$.
Also, $\sigma=-(\ell+m+1)k$. Then (6.44) becomes

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}.$
(6.50)

To make use of Theorem 5.3, we need to swap the choice of $(\kappa,u,A)$ and
$(\lambda,v,B)$ in our initial setting. In other words, in Theorem 5.3, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2m+1)\lambda&&\lambda&\mapsto&(2\ell+1)\kappa\\\
u&\mapsto&(2m+1)k&&v&\mapsto&(2\ell+1)k\\\
A&\mapsto&2(2m+1)\mu&&B&\mapsto&(2\ell+1)\mu\\\
A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&\xi\mu\\\ M&\mapsto&\lx@intercol
4\ell+2m+3\hfil\lx@intercol\\\ w&\mapsto&\lx@intercol-(\ell+m+1)k+\xi
k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(\ell+m+1)k+(2\ell+1-\xi)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{2\xi k}{d^{*}}.$

If $d^{*}\nmid 2\xi k$, then $\xi k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv\xi k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\hat{w}$ $\displaystyle\equiv-\xi k\not\equiv 0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2\xi k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). With the choice of $K=1$, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=1,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=0.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(2m+1)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\lambda}\mathbf{H}_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}+(-1)^{\kappa}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}$
$\displaystyle=\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa}\mathbf{H}_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.50) is established.

  * $\blacktriangleright$

Examining (6.45). Let us keep in mind that we have made the substitutions in
(6.37), (6.38), (6.40) and (6.41): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and
$b=1$. Also, $\sigma=-(\ell+m+1)k$. Then (6.45) becomes

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa+\lambda}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}.$
(6.51)

In Theorem 5.4, we further set

$\displaystyle\begin{array}[]{rclp{4cm}rcl}A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&2\tau\mu\\\
w&\mapsto&\lx@intercol-(\ell+m+1)k+\xi k+\tau k\hfil\lx@intercol\\\
\end{array}$

Then

$\displaystyle\check{w}=-(\ell+m+1)k+(2\ell+1-\xi)k+(2m+1-\tau)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I},$
$\displaystyle\check{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2(A^{\prime}Bu+AB^{\prime}v)}{d^{*}AB}=\frac{2(\xi+\tau)k}{d^{*}}.$

Similarly, if $d^{*}\nmid 2(\xi+\tau)k$, then $(\xi+\tau)k\not\equiv
0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv(\xi+\tau)k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\check{w}$ $\displaystyle\equiv-(\xi+\tau)k\not\equiv
0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa+\lambda}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2(\xi+\tau)k$. Then the first assumption in
(5.24) is satisfied. We may also assume that $w$ is a multiple of
$d^{*}=d_{0}$ according to Theorem 5.4 (i). With the choice of $K=1$, we
compute that

$\displaystyle\frac{B(A-2A^{\prime})u^{2}+A(B-2B^{\prime})uv+2ABuw\cdot
K}{A(Av^{2}+Bu^{2})}$ $\displaystyle=0,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=0.$

So the remaining assumptions in (5.24) are satisfied. Finally, we compute that

$\displaystyle\varepsilon=0.$

Thus, by Theorem 5.4 (ii),

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=\mathbf{H}_{M}\big{(}\check{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ that

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}+(-1)^{\kappa+\lambda}\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa+\lambda}\mathbf{H}_{M}\big{(}\check{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.51) is established.

####  6.3.2. Equation (1.17)

Recall that $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=4$. Also,
$M=2\ell+16m+9$ and $\sigma=-(2\ell+12m+7)k$. Further,
$(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ and $k$ is such that $\gcd(k,M)=1$.

In Corollary 5.2 and Theorems 5.3 and 5.4, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+1)\kappa&&\lambda&\mapsto&(2m+1)\lambda\\\
u&\mapsto&(2\ell+1)k&&v&\mapsto&4(2m+1)k\\\
A&\mapsto&(2\ell+1)\mu&&B&\mapsto&2(2m+1)\mu\\\ M&\mapsto&\lx@intercol
2\ell+16m+9\hfil\lx@intercol\\\ \end{array}$

Let

$\displaystyle
d_{0}=\gcd\big{(}2\ell+1,4(2m+1)\big{)}=\gcd\big{(}2\ell+1,2m+1\big{)}.$

Then

$\displaystyle d=\gcd(u,v)=d_{0}k.$

Noticing that $M=(2\ell+1)+8(2m+1)$, and that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=d_{0},$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=d_{0},$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=d_{0}.$

We compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$ $\displaystyle=2,$
$\displaystyle\frac{Bu\cdot dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=1.$

Thus, the second and third assumptions in (5.9), (5.16) and (5.24) are
satisfied.

  * $\blacktriangleright$

Examining (6.42). Let us keep in mind that we have made the substitutions in
(6.36) and (6.39): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=4$. Also,
$\sigma=-(2\ell+12m+7)k$. We want to show (6.42),

$\displaystyle H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$
(6.52)

In Corollary 5.2, we further set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&0\\\
w&\mapsto&\lx@intercol-(2\ell+12m+7)k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}.$

We may further assume that $w$ is a multiple of $d^{*}$. Otherwise, we know
from Theorem 5.1 (i) that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

which gives

$\displaystyle H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$

We further compute that

$\displaystyle\frac{-v\kappa+u\lambda}{d}=-\frac{(2\ell+1)(2m+1)}{d_{0}}(4\kappa-\lambda).$

The fact that $d_{0}=\gcd(2\ell+1,2m+1)$ then implies that
$(2\ell+1)(2m+1)/d_{0}$ is an odd integer. Thus, (5.9) is satisfied since
$(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$. Finally, we choose $J=0$ in (5.11).
Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\frac{3d_{0}-(2\ell+1)}{2d_{0}},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\frac{d_{0}+2(2m+1)}{d_{0}}.\end{aligned}$

Recall that $2\ell+1\equiv 2m+1\equiv d_{0}\pmod{2d_{0}}$ since
$d_{0}=\gcd(2\ell+1,2m+1)$ is odd. So (5.11) is also satisfied. We conclude by
Corollary 5.2 that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0.$

Thus, (6.52) is established.

  * $\blacktriangleright$

Examining (6.43). Let us keep in mind that we have made the substitutions in
(6.36), (6.40) and (6.41): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=4$.
Also, $\sigma=-(2\ell+12m+7)k$. Then (6.43) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}.$
(6.53)

In Theorem 5.3, we further set

$\displaystyle\begin{array}[]{rclp{4cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&2\tau\mu\\\
w&\mapsto&\lx@intercol-(2\ell+12m+7)k+4\tau k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(2\ell+12m+7)k+4(2m+1-\tau)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{8\tau k}{d^{*}}.$

If $d^{*}\nmid 8\tau k$, then $4\tau k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle=-(2\ell+1)k-6(2m+1)k+4\tau k\equiv 4\tau
k\not\equiv 0\pmod{d_{0}},$ $\displaystyle\hat{w}$
$\displaystyle=-(2\ell+1)k-2(2m+1)k-4\tau k\equiv-4\tau k\not\equiv
0\pmod{d_{0}}.$

So $w$ and $\hat{w}$ are nonmultiples of $d_{0}=d^{*}$. In this case we know
from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}=0,$

which gives

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 8\tau k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). It is easily seen that $d^{*}k$ is a divisor of
$\gcd(u,v,w)$. Thus, we solve the following stronger system

$\displaystyle\left\\{\begin{aligned} K&\equiv 1\pmod{M/d^{*}},\\\ K&\equiv
0\pmod{d/(d^{*}k)},\end{aligned}\right.$

and choose $K=1$. Finally, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=2,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-2.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(4\ell+2)\kappa+(4m+2)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}+(-1)^{\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\lambda}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.53) is established.

  * $\blacktriangleright$

Examining (6.44). Let us keep in mind that we have made the substitutions in
(6.37), (6.38) and (6.39): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and $b=4$.
Also, $\sigma=-(2\ell+12m+7)k$. Then (6.44) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}.$
(6.54)

To make use of Theorem 5.3, we need to swap the choice of $(\kappa,u,A)$ and
$(\lambda,v,B)$ in our initial setting. In other words, in Theorem 5.3, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2m+1)\lambda&&\lambda&\mapsto&(2\ell+1)\kappa\\\
u&\mapsto&4(2m+1)k&&v&\mapsto&(2\ell+1)k\\\
A&\mapsto&2(2m+1)\mu&&B&\mapsto&(2\ell+1)\mu\\\
A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&\xi\mu\\\ M&\mapsto&\lx@intercol
2\ell+16m+9\hfil\lx@intercol\\\ w&\mapsto&\lx@intercol-(2\ell+12m+7)k+\xi
k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(2\ell+12m+7)k+(2\ell+1-\xi)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{2\xi k}{d^{*}}.$

If $d^{*}\nmid 2\xi k$, then $\xi k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv\xi k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\hat{w}$ $\displaystyle\equiv-\xi k\not\equiv 0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2\xi k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). With the choice of $K=1$, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=3,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(2\ell+1)\kappa+(6m+3)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\kappa+\lambda}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}+(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.54) is established.

  * $\blacktriangleright$

Examining (6.45). Let us keep in mind that we have made the substitutions in
(6.37), (6.38), (6.40) and (6.41): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=1$ and
$b=4$. Also, $\sigma=-(2\ell+12m+7)k$. Then (6.45) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa+\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}.$
(6.55)

In Theorem 5.4, we further set

$\displaystyle\begin{array}[]{rclp{4.5cm}rcl}A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&2\tau\mu\\\
w&\mapsto&\lx@intercol-(2\ell+12m+7)k+\xi k+4\tau k\hfil\lx@intercol\\\
\end{array}$

Then

$\displaystyle\check{w}=-(2\ell+12m+7)k+(2\ell+1-\xi)k+4(2m+1-\tau)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I},$
$\displaystyle\check{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2(A^{\prime}Bu+AB^{\prime}v)}{d^{*}AB}=\frac{2(\xi+4\tau)k}{d^{*}}.$

Similarly, if $d^{*}\nmid 2(\xi+4\tau)k$, then $(\xi+4\tau)k\not\equiv
0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv(\xi+4\tau)k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\check{w}$ $\displaystyle\equiv-(\xi+4\tau)k\not\equiv
0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa+\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2(\xi+4\tau)k$. Then the first assumption in
(5.24) is satisfied. We may also assume that $w$ is a multiple of
$d^{*}=d_{0}$ according to Theorem 5.4 (i). With the choice of $K=1$, we
compute that

$\displaystyle\frac{B(A-2A^{\prime})u^{2}+A(B-2B^{\prime})uv+2ABuw\cdot
K}{A(Av^{2}+Bu^{2})}$ $\displaystyle=-1,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-2.$

So the remaining assumptions in (5.24) are satisfied. Finally, we compute that

$\displaystyle\varepsilon=(2\ell+1)\kappa+(4m+2)\lambda.$

Thus, by Theorem 5.4 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\kappa}H_{M}\big{(}\check{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}+(-1)^{\kappa+\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa+\lambda}H_{M}\big{(}\check{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.55) is established.

####  6.3.3. Equation (1.18)

Recall that $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=2$ and $b=1$. Also,
$M=16\ell+2m+9$ and $\sigma=-(10\ell+2m+6)k$. Further,
$(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$ and $k$ is such that $\gcd(k,M)=1$.

In Corollary 5.2 and Theorems 5.3 and 5.4, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+1)\kappa&&\lambda&\mapsto&(2m+1)\lambda\\\
u&\mapsto&2(2\ell+1)k&&v&\mapsto&(2m+1)k\\\
A&\mapsto&(2\ell+1)\mu&&B&\mapsto&2(2m+1)\mu\\\ M&\mapsto&\lx@intercol
16\ell+2m+9\hfil\lx@intercol\\\ \end{array}$

Let

$\displaystyle
d_{0}=\gcd\big{(}2(2\ell+1),2m+1\big{)}=\gcd\big{(}2\ell+1,2m+1\big{)}.$

Then

$\displaystyle d=\gcd(u,v)=d_{0}k.$

Noticing that $M=8(2\ell+1)+(2m+1)$, and that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=d_{0},$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=d_{0},$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=d_{0}.$

We compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$ $\displaystyle=1,$
$\displaystyle\frac{Bu\cdot dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=4.$

Thus, the second and third assumptions in (5.9), (5.16) and (5.24) are
satisfied.

  * $\blacktriangleright$

Examining (6.42). Let us keep in mind that we have made the substitutions in
(6.36) and (6.39): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=2$ and $b=1$. Also,
$\sigma=-(10\ell+2m+6)k$. We want to show (6.42),

$\displaystyle H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$
(6.56)

In Corollary 5.2, we further set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&0\\\
w&\mapsto&\lx@intercol-(10\ell+2m+6)k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}.$

We may further assume that $w$ is a multiple of $d^{*}$. Otherwise, we know
from Theorem 5.1 (i) that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

which gives

$\displaystyle H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{0}\big{)}=0.$

We further compute that

$\displaystyle\frac{-v\kappa+u\lambda}{d}=-\frac{(2\ell+1)(2m+1)}{d_{0}}(\kappa-2\lambda).$

The fact that $d_{0}=\gcd(2\ell+1,2m+1)$ then implies that
$(2\ell+1)(2m+1)/d_{0}$ is an odd integer. Thus, (5.9) is satisfied since
$(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$. Finally, we choose $J=0$ in (5.11).
Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\frac{d_{0}-(2\ell+1)}{d_{0}},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\frac{5d_{0}+(2m+1)}{2d_{0}}.\end{aligned}$

Recall that $2\ell+1\equiv 2m+1\equiv d_{0}\pmod{2d_{0}}$ since
$d_{0}=\gcd(2\ell+1,2m+1)$ is odd. So (5.11) is also satisfied. We conclude by
Corollary 5.2 that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0.$

Thus, (6.56) is established.

  * $\blacktriangleright$

Examining (6.43). Let us keep in mind that we have made the substitutions in
(6.36), (6.40) and (6.41): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=2$ and $b=1$.
Also, $\sigma=-(10\ell+2m+6)k$. Then (6.43) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}.$
(6.57)

In Theorem 5.3, we further set

$\displaystyle\begin{array}[]{rclp{3.5cm}rcl}A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&2\tau\mu\\\
w&\mapsto&\lx@intercol-(10\ell+2m+6)k+\tau k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(10\ell+2m+6)k+(2m+1-\tau)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{2\tau k}{d^{*}}.$

If $d^{*}\nmid 2\tau k$, then $\tau k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle=-5(2\ell+1)k-(2m+1)k+\tau k\equiv\tau
k\not\equiv 0\pmod{d_{0}},$ $\displaystyle\hat{w}$
$\displaystyle=-5(2\ell+1)k-\tau k\equiv-\tau k\not\equiv 0\pmod{d_{0}}.$

So $w$ and $\hat{w}$ are nonmultiples of $d_{0}=d^{*}$. In this case we know
from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}=0,$

which gives

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}=-(-1)^{\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2\tau k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). It is easily seen that $d^{*}k$ is a divisor of
$\gcd(u,v,w)$. Thus, we solve the following stronger system

$\displaystyle\left\\{\begin{aligned} K&\equiv 1\pmod{M/d^{*}},\\\ K&\equiv
0\pmod{d/(d^{*}k)},\end{aligned}\right.$

and choose $K=1$. Finally, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=5,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(10\ell+5)\kappa+(2m+1)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\kappa+\lambda}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I}\big{)}+(-1)^{\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{0}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\lambda}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.57) is established.

  * $\blacktriangleright$

Examining (6.44). Let us keep in mind that we have made the substitutions in
(6.37), (6.38) and (6.39): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=2$ and $b=1$.
Also, $\sigma=-(10\ell+2m+6)k$. Then (6.44) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}.$
(6.58)

To make use of Theorem 5.3, we need to swap the choice of $(\kappa,u,A)$ and
$(\lambda,v,B)$ in our initial setting. In other words, in Theorem 5.3, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2m+1)\lambda&&\lambda&\mapsto&(2\ell+1)\kappa\\\
u&\mapsto&(2m+1)k&&v&\mapsto&2(2\ell+1)k\\\
A&\mapsto&2(2m+1)\mu&&B&\mapsto&(2\ell+1)\mu\\\
A^{\prime}&\mapsto&0&&B^{\prime}&\mapsto&\xi\mu\\\ M&\mapsto&\lx@intercol
16\ell+2m+9\hfil\lx@intercol\\\ w&\mapsto&\lx@intercol-(10\ell+2m+6)k+2\xi
k\hfil\lx@intercol\\\ \end{array}$

Then

$\displaystyle\hat{w}=-(10\ell+2m+6)k+2(2\ell+1-\xi)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I},$
$\displaystyle\hat{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2B^{\prime}v}{d^{*}B}=\frac{4\xi k}{d^{*}}.$

If $d^{*}\nmid 4\xi k$, then $2\xi k\not\equiv 0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv 2\xi k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\hat{w}$ $\displaystyle\equiv-2\xi k\not\equiv 0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}=-(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 4\xi k$. Then the first assumption in (5.16)
is satisfied. We may also assume that $w$ is a multiple of $d^{*}=d_{0}$
according to Theorem 5.3 (i). With the choice of $K=1$, we compute that

$\displaystyle\frac{(A-2A^{\prime})v^{2}-(B-2B^{\prime})uv-2Buw\cdot
K}{Av^{2}+Bu^{2}}$ $\displaystyle=2,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-4.$

So the remaining assumptions in (5.16) are satisfied. Finally, we compute that

$\displaystyle\epsilon=(8\ell+4)\kappa+(4m+2)\lambda.$

Thus, by Theorem 5.3 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I}\big{)}+(-1)^{\kappa}H_{M}\big{(}q^{\sigma}\mathscr{B}_{0}\mathscr{A}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa}H_{M}\big{(}\hat{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.58) is established.

  * $\blacktriangleright$

Examining (6.45). Let us keep in mind that we have made the substitutions in
(6.37), (6.38), (6.40) and (6.41): $(\ell,m)\mapsto(2\ell+1,2m+1)$, $a=2$ and
$b=1$. Also, $\sigma=-(10\ell+2m+6)k$. Then (6.45) becomes

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa+\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}.$
(6.59)

In Theorem 5.4, we further set

$\displaystyle\begin{array}[]{rclp{4.5cm}rcl}A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&2\tau\mu\\\
w&\mapsto&\lx@intercol-(10\ell+2m+6)k+2\xi k+\tau k\hfil\lx@intercol\\\
\end{array}$

Then

$\displaystyle\check{w}=-(10\ell+2m+6)k+2(2\ell+1-\xi)k+(2m+1-\tau)k.$

So,

$\displaystyle\mathscr{H}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I},$
$\displaystyle\check{\mathscr{H}}(q)$
$\displaystyle=q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}.$

Now we compute that

$\displaystyle\frac{2(A^{\prime}Bu+AB^{\prime}v)}{d^{*}AB}=\frac{2(2\xi+\tau)k}{d^{*}}.$

Similarly, if $d^{*}\nmid 2(2\xi+\tau)k$, then $(2\xi+\tau)k\not\equiv
0\pmod{d_{0}}$, and thus,

$\displaystyle w$ $\displaystyle\equiv(2\xi+\tau)k\not\equiv 0\pmod{d_{0}},$
$\displaystyle\check{w}$ $\displaystyle\equiv-(2\xi+\tau)k\not\equiv
0\pmod{d_{0}},$

implying from Theorem 5.1 (i) that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}=-(-1)^{\kappa+\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}=0.$

Below, we assume that $d^{*}\mid 2(2\xi+\tau)k$. Then the first assumption in
(5.24) is satisfied. We may also assume that $w$ is a multiple of
$d^{*}=d_{0}$ according to Theorem 5.4 (i). With the choice of $K=1$, we
compute that

$\displaystyle\frac{B(A-2A^{\prime})u^{2}+A(B-2B^{\prime})uv+2ABuw\cdot
K}{A(Av^{2}+Bu^{2})}$ $\displaystyle=-4,$
$\displaystyle\frac{B(A-2A^{\prime})uv+A(B-2B^{\prime})v^{2}+2ABvw\cdot
K}{B(Av^{2}+Bu^{2})}$ $\displaystyle=-1.$

So the remaining assumptions in (5.24) are satisfied. Finally, we compute that

$\displaystyle\varepsilon=(8\ell+4)\kappa+(2m+1)\lambda.$

Thus, by Theorem 5.4 (ii),

$\displaystyle
H_{M}\big{(}\mathscr{H}(q)\big{)}=(-1)^{\lambda}H_{M}\big{(}\check{\mathscr{H}}(q)\big{)}.$

It follows from $(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$ that

$\displaystyle
H_{M}\big{(}q^{\sigma}\mathscr{A}_{I}\mathscr{B}_{I}\big{)}+(-1)^{\kappa+\lambda}H_{M}\big{(}q^{\sigma}\mathscr{A}_{I\\!I}\mathscr{B}_{I\\!I}\big{)}$
$\displaystyle=H_{M}\big{(}\mathscr{H}(q)\big{)}+(-1)^{\kappa+\lambda}H_{M}\big{(}\check{\mathscr{H}}(q)\big{)}$
$\displaystyle=0.$

Thus, (6.59) is established.

###  6.4. Type II — Theorem 1.4

This section is devoted to the three coefficient-vanishing results in Theorem
1.4. Here we will use a different strategy in comparison to how we treat Type
I. We summarize the basic idea as follows.

We deduce from Theorem 3.5 with certain common factors extracted and powers of
$(-1)$ modified that, for any $\ell\geq 1$,

$\displaystyle
f\big{(}(-1)^{\kappa}q^{ak},(-1)^{\kappa}q^{-ak+M\mu}\big{)}^{\ell}=\sum_{s=0}^{\ell-1}\mathscr{A}_{s}\mathscr{F}_{s},$

where each $\mathscr{F}_{\star}$ is a series in $q^{M}$, and for $0\leq
s\leq\ell-1$,

$\displaystyle\mathscr{A}_{s}=q^{aks}f\big{(}(-1)^{\kappa\ell}q^{ak\ell+M\mu
s},(-1)^{\kappa\ell}q^{-ak\ell+M\mu(\ell-s)}\big{)}.$

More generally, we consider

$\displaystyle\mathscr{A}:=q^{a\xi
k}f\big{(}(-1)^{\kappa\ell}q^{ak\ell+M\mu\xi},(-1)^{\kappa\ell}q^{-ak\ell+M\mu(\ell-\xi)}\big{)},$
(6.60)

for generic $\xi\in\mathbb{Z}$.

Similarly, by Theorem 3.5, we write for any $m\geq 1$,

$\displaystyle
f\big{(}(-1)^{\lambda}q^{bk},(-1)^{\lambda}q^{-bk+2M\mu}\big{)}^{m}=\sum_{t=0}^{m-1}\mathscr{B}_{t}\mathscr{G}_{t},$

where each $\mathscr{G}_{\star}$ is a series in $q^{M}$, and for $0\leq t\leq
m-1$,

$\displaystyle\mathscr{B}_{t}=q^{bkt}f\big{(}(-1)^{\lambda m}q^{bkm+2M\mu
t},(-1)^{\lambda m}q^{-bkm+2M\mu(m-t)}\big{)}.$

More generally, we consider

$\displaystyle\mathscr{B}:=q^{b\tau k}f\big{(}(-1)^{\lambda
m}q^{bkm+2M\mu\tau},(-1)^{\lambda m}q^{-bkm+2M\mu(m-\tau)}\big{)},$ (6.61)

for generic $\tau\in\mathbb{Z}$.

With all parameters chosen in each case below, our generic target is to show
the following by Corollary 5.2:

$\displaystyle\mathbf{H}_{M}\big{(}q^{\sigma}\mathscr{A}\mathscr{B}\big{)}=0.$
(6.62)

Once the above relation is established, it is safe to conclude that

$\displaystyle\mathbf{H}_{M}\Big{(}q^{\sigma}f\big{(}(-1)^{\kappa}q^{ak},(-1)^{\kappa}q^{-ak+M\mu}\big{)}^{\ell}f\big{(}(-1)^{\lambda}q^{bk},(-1)^{\lambda}q^{-bk+2M\mu}\big{)}^{m}\Big{)}=0.$
(6.63)

####  6.4.1. Equation ((i))

Recall that $(\ell,m)\mapsto(2\ell+1,2m+1)$ with $\gcd(2\ell+1,2m+1)=1$,
$a=2m+1$ and $b=2\ell+1$. Also, $M=2\ell+4m+3$ and $\sigma=2(2m+1)^{2}k$.
Further, $(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$ and $k$ is such that
$\gcd(k,M)=1$.

In Corollary 5.2, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+1)\kappa&&\lambda&\mapsto&(2m+1)\lambda\\\
u&\mapsto&(2m+1)(2\ell+1)k&&v&\mapsto&(2\ell+1)(2m+1)k\\\
A&\mapsto&(2\ell+1)\mu&&B&\mapsto&2(2m+1)\mu\\\
A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&2\tau\mu\\\
M&\mapsto&\lx@intercol 2\ell+4m+3\hfil\lx@intercol\\\ w&\mapsto&\lx@intercol
2(2m+1)^{2}k+(2m+1)\xi k+(2\ell+1)\tau k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}\mathscr{B}.$

We find that

$\displaystyle d=\gcd(u,v)=(2\ell+1)(2m+1)k.$

Also, noticing that $M=(2\ell+1)+2(2m+1)$ with $\gcd(2\ell+1,2m+1)=1$, and
that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=1,$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=1,$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=1.$

Then $d^{*}\mid w$. We further compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$
$\displaystyle=2\ell+1,$ $\displaystyle\frac{Bu\cdot
dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=4m+2,$
$\displaystyle\frac{-v\kappa+u\lambda}{d}$
$\displaystyle=-(2\ell+1)\kappa+(2m+1)\lambda.$

Thus, (5.9) and (5.10) are satisfied by recalling that
$(\kappa,\lambda)\in\\{(0,1),(1,0)\\}$. Finally, we choose $J=J_{0}k$ in
(5.11). Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\dfrac{J_{0}-2m-1-\tau}{2m+1},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\dfrac{2J_{0}+2\ell-4m-1-\xi}{2\ell+1}.\end{aligned}$

We may further choose $J_{0}$ so that

$\displaystyle\left\\{\begin{aligned} J_{0}-2m-1-\tau&\equiv 0\pmod{2m+1},\\\
2J_{0}+2\ell-4m-1-\xi&\equiv 0\pmod{2\ell+1}.\end{aligned}\right.$

Since $\gcd(2\ell+1,2m+1)=1$, we know that the above system is solvable. So
(5.11) is also satisfied. We conclude by Corollary 5.2 that

$\displaystyle\mathbf{H}_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

and therefore confirm (6.62).

####  6.4.2. Equation ((ii))

Recall that $(\ell,m)\mapsto(2\ell+1,2m+2)$ with $\gcd(2\ell+1,2m+2)=1$,
$a=2m+2$ and $b=2\ell+1$. Also, $M=2\ell+4m+5$ and $\sigma=2(2m+2)^{2}k$.
Further, $(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$ and $k$ is such that
$\gcd(k,M)=1$.

In Corollary 5.2, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+1)\kappa&&\lambda&\mapsto&(2m+2)\lambda\\\
u&\mapsto&(2m+2)(2\ell+1)k&&v&\mapsto&(2\ell+1)(2m+2)k\\\
A&\mapsto&(2\ell+1)\mu&&B&\mapsto&2(2m+2)\mu\\\
A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&2\tau\mu\\\
M&\mapsto&\lx@intercol 2\ell+4m+5\hfil\lx@intercol\\\ w&\mapsto&\lx@intercol
2(2m+2)^{2}k+(2m+2)\xi k+(2\ell+1)\tau k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}\mathscr{B}.$

We find that

$\displaystyle d=\gcd(u,v)=(2\ell+1)(2m+2)k.$

Also, noticing that $M=(2\ell+1)+2(2m+2)$ with $\gcd(2\ell+1,2m+2)=1$, and
that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=1,$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=1,$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=1.$

Then $d^{*}\mid w$. We further compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$
$\displaystyle=2\ell+1,$ $\displaystyle\frac{Bu\cdot
dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=4m+4,$
$\displaystyle\frac{-v\kappa+u\lambda}{d}$
$\displaystyle=-(2\ell+1)\kappa+(2m+2)\lambda.$

Thus, (5.9) and (5.10) are satisfied by recalling that
$(\kappa,\lambda)\in\\{(1,0),(1,1)\\}$. Finally, we choose $J=J_{0}k$ in
(5.11). Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\dfrac{J_{0}-2m-2-\tau}{2m+2},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\dfrac{2J_{0}+2\ell-4m-3-\xi}{2\ell+1}.\end{aligned}$

We may further choose $J_{0}$ so that

$\displaystyle\left\\{\begin{aligned} J_{0}-2m-2-\tau&\equiv 0\pmod{2m+2},\\\
2J_{0}+2\ell-4m-3-\xi&\equiv 0\pmod{2\ell+1}.\end{aligned}\right.$

Since $\gcd(2\ell+1,2m+2)=1$, we know that the above system is solvable. So
(5.11) is also satisfied. We conclude by Corollary 5.2 that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

and therefore confirm (6.62).

####  6.4.3. Equation ((iii))

Recall that $(\ell,m)\mapsto(2\ell+2,2m+1)$ with $\gcd(2\ell+2,2m+1)=1$,
$a=2m+1$ and $b=4\ell+4$. Also, $M=4\ell+2m+5$ and $\sigma=3(2\ell+2)^{2}k$.
Further, $(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$ and $k$ is such that
$\gcd(k,M)=1$.

In Corollary 5.2, we set

$\displaystyle\begin{array}[]{rclp{3cm}rcl}\kappa&\mapsto&(2\ell+2)\kappa&&\lambda&\mapsto&(2m+1)\lambda\\\
u&\mapsto&(2m+1)(2\ell+2)k&&v&\mapsto&(4\ell+4)(2m+1)k\\\
A&\mapsto&(2\ell+2)\mu&&B&\mapsto&2(2m+1)\mu\\\
A^{\prime}&\mapsto&\xi\mu&&B^{\prime}&\mapsto&2\tau\mu\\\
M&\mapsto&\lx@intercol 4\ell+2m+5\hfil\lx@intercol\\\ w&\mapsto&\lx@intercol
3(2\ell+2)^{2}k+(2m+1)\xi k+(4\ell+4)\tau k\hfil\lx@intercol\\\ \end{array}$

So,

$\displaystyle\mathscr{H}(q)=q^{\sigma}\mathscr{A}\mathscr{B}.$

We find that

$\displaystyle d=\gcd(u,v)=(2\ell+2)(2m+1)k.$

Also, noticing that $M=2(2\ell+2)+(2m+1)$ with $\gcd(2\ell+2,2m+1)=1$, and
that $k$ is coprime to $M$, we have

$\displaystyle d^{*}$ $\displaystyle=\gcd(u,v,M)=1,$ $\displaystyle d_{u}$
$\displaystyle=\gcd(u,M)=1,$ $\displaystyle d_{v}$
$\displaystyle=\gcd(v,M)=1.$

Then $d^{*}\mid w$. We further compute that

$\displaystyle\frac{Av\cdot dM}{d_{u}(Av^{2}+Bu^{2})}$
$\displaystyle=2\ell+2,$ $\displaystyle\frac{Bu\cdot
dM}{d_{v}(Av^{2}+Bu^{2})}$ $\displaystyle=2m+1,$
$\displaystyle\frac{-v\kappa+u\lambda}{d}$
$\displaystyle=-(4\ell+4)\kappa+(2m+1)\lambda.$

Thus, (5.9) and (5.10) are satisfied by recalling that
$(\kappa,\lambda)\in\\{(0,1),(1,1)\\}$. Finally, we choose $J=J_{0}k$ in
(5.11). Then

$\begin{aligned} \dfrac{2dMAv\cdot J-2dAvw-
dAuv+dBu^{2}+2dA^{\prime}uv-2dB^{\prime}u^{2}-Auv^{2}-Bu^{3}}{2d(Av^{2}+Bu^{2})}&=\dfrac{J_{0}-3\ell-3-\tau}{2m+1},\\\
\dfrac{2dMBu\cdot
J-2dBuw+dAv^{2}-dBuv-2dA^{\prime}v^{2}+2dB^{\prime}uv+Av^{3}+Bu^{2}v}{2d(Av^{2}+Bu^{2})}&=\dfrac{J_{0}-\xi}{2\ell+2}.\end{aligned}$

We may further choose $J_{0}$ so that

$\displaystyle\left\\{\begin{aligned} J_{0}-3\ell-3-\tau&\equiv
0\pmod{2m+1},\\\ J_{0}-\xi&\equiv 0\pmod{2\ell+2}.\end{aligned}\right.$

Since $\gcd(2\ell+2,2m+1)=1$, we know that the above system is solvable. So
(5.11) is also satisfied. We conclude by Corollary 5.2 that

$\displaystyle H_{M}\big{(}\mathscr{H}(q)\big{)}=0,$

and therefore confirm (6.62).

###  6.5. Seven families of coefficient functions

All results in Corollary 1.5 are direct consequences of Theorems 1.1–1.4 as
long as we notice that $H_{M}\big{(}G(q)\big{)}=0$ provided that
$H_{M}\big{(}G(q)\cdot F(q^{M})\big{)}=0$ for any given series $F$ in $q^{M}$.

##  7\. Conclusion

The entire project began in an attempt to understand the underlying patterns
in the coefficient-vanishing phenomena associated with theta series. It turns
out that most of the known results on this topic can be covered by several
general relations.

Also of significance is the unified strategy presented in Section 2, which
contains two aspects:

  1. (7.1)

It provides an automatic way to verify if a coefficient-vanishing result, with
or without free parameters, holds true. In fact, if such a result is
discovered experimentally, one may try to fit it into the four models
discussed in Section 6, or other models of a similar nature. The remaining
task then becomes verifying some relations in connection with products of two
theta series under the action of the $\mathbf{H}$-operator. Finally, such a
verification relies on checking certain divisibility criteria presented in
Corollary 5.2 and Theorems 5.3 and 5.4.

  2. (7.2)

It allows us to manually construct specific or generic coefficient-vanishing
results. For instance, the discovery of Theorem 1.4 relies only on one example
or two presented in [32]. Roughly speaking, one may fix some of the free
parameters in a generic model, and then use the divisibility conditions in
Section 5 to constrain the choice of other parameters so that these conditions
remain valid.

The divisibility criteria in Corollary 5.2 and Theorems 5.3 and 5.4 look very
sharp, but they work effectively in practice, at least for most coefficient-
vanishing results we had encountered. So we expect that our approach may light
up a general theory. However, it should be pointed out that two “clouds” need
to be taken into serious consideration.

First, what if one or more of the divisibility conditions fail to hold? For
instance, we observe that (1.13) is still valid even if the latter power
$4m+2$ is replaced by a multiple of $4$. Namely, for $M=4\ell+6m+8$,
$\sigma=-(2\ell+2m+3)k$, $\kappa\in\\{1\\}$, $\lambda\in\\{0,1\\}$ and any $k$
such that $\gcd(k,M)=1$,

$\displaystyle\mathbf{H}_{M}\Bigg{(}q^{\sigma}\cdot
f\big{(}(-1)^{\kappa}q^{2k},(-1)^{\kappa}q^{\mu
M-2k}\big{)}^{2\ell+1}\bigg{(}\frac{f\big{(}{-q^{2k}},-q^{\mu
M-2k}\big{)}}{f\big{(}(-1)^{\lambda}q^{k},(-1)^{\lambda}q^{\mu
M-k}\big{)}}\bigg{)}^{4m+4}\Bigg{)}\overset{?}{=}0.$ (7.1)

For this relation, we find that the corresponding criteria in (5.6) are no
longer true. Thus, a subtle refinement of the results in Sections 4 and 5 is
necessary.

Second, it is easily seen that the original result of Richmond and Szekeres
[24] related to (1.3) cannot be fit into our framework. Although one may
transform (1.3) and its generalization to the summation form by Ramanujan’s
${}_{1}\psi_{1}$ formula as Andrews and Bressoud [2] had done, an obstacle
occurs due to the lack of an expansion formula for the reciprocal of a generic
theta power such as those in Section 3. Also, the coefficient-vanishing
phenomenon appears in series associated with other classical summations such
as the Appell–Lerch sum [10]. These examples suggest one extend the glimmer of
our theory to the endless Galaxy.

### Acknowledgements

The authors would like to acknowledge the anonymous referee for the inspiring
comments. In particular, a couple of typos in an earlier draft of this work
were detected by the referee. Shane Chern was partially supported by a Killam
Postdoctoral Fellowship from the Killam Trusts. Dazhao Tang was partially
supported by the National Natural Science Foundation of China (No. 12201093),
the Natural Science Foundation Project of Chongqing CSTB (No.
CSTB2022NSCQ–MSX0387), the Science and Technology Research Program of
Chongqing Municipal Education Commission (No. KJQN202200509), and the Doctoral
start-up research Foundation (No. 21XLB038) of Chongqing Normal University.

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