Number-Theory Renormalization of Vacuum Energy

Abstract

For QFT on a lattice of dimension $d\geqslant 3$, the vacuum energy (both bosonic and fermionic) is zero if the Hamiltonian is a function of the square of the momentum, and the calculation of the vacuum energy is performed in the ring of residue classes modulo $N$. This fact is related to a problem from number theory about the number of ways to represent a number as a sum of $d$ squares in the ring of residue classes modulo $N$.

Appendix

Appendix A. Quadratic Congruences

1. Quadratic Congruences to Prime Modulus

Simplest quadratic congruence

$$x^2 \equiv 0 (\mathrm{mod} p)$$

in field $\mathbb{Z}(p)$ means that square root of 0 is uniquely defined in $\mathbb{Z}(p)$.

Quadratic congruence

$$ x^2 \equiv a (\mathrm{mod} p), \qquad\text{ in case } {a}\not\equiv0 (\mathrm{mod} p)$$

(A.1)

has solutions for some $a\in \mathbb{Z}(p)$, and such $a$ is said to be a quadratic residue. Note that the trivial case $a=0$ is generally excluded from set of quadratic residues $\text{Res}(p)= \{ 1, r,s,\dots\}$, which are congruent to integers

$$1^2, 2^2, ..., \left( \frac{p-1}{2}\right)^2.$$

The number of quadratic residues is exactly $\frac{p-1}{2}$ because if we take different $x_1$ and $x_2$

$$x_1^2\equiv x_2^2\equiv a\not\equiv0 (\mathrm{mod} p),$$

then, since $\mathbb{Z}(p)$ is field, we have

$$(x_1/x_2)^2\equiv 1 (\mathrm{mod} p)\quad\Rightarrow\quad x_1=\pm x_2.$$

If the congruence (A.1) does not have a solution, then $a$ is said to be a quadratic nonresidue.

The Legendre symbol is defined for integers $a \in{\mathbb{Z}}$:

$$ \chi(a) \stackrel{\text{def}}{=} \quad\left(\frac{a}{p}\right)_L \stackrel{\text{def}}{=} \left\{ \begin{aligned} \, & \quad 0 \quad\text{ if $a$ is divisible by } p,\\ & \quad 1 \quad\text{ if $a$ is quadratic residue modulo } p,\\ & -1 \quad\,\text{if $a$ is quadratic nonresidue}.\\ \end{aligned}\right.$$

(A.2)

Legendre symbol is group theoretical character and we denote it like $\chi(a)$ for $p$ fixed. From Fermat’s and Legendre theorems it follows that

$$ \forall a : \qquad \chi(a) \equiv a^{\frac{p-1}{2}} (\mathrm{mod} p).$$

(A.3)

Using this congruence one can simply evaluate Legendre symbol of $-1$ by formula

$$\chi(-1) = (-1)^{\frac{p-1}{2}}.$$

Also characters’ sum over all elements of $\mathbb{Z}(p)$ is equal to $0$:

$$ \sum_a \chi(a) =0.$$

(A.4)

Hence (in one dimension and $N=p$) the multiplicity (2.7) as number of solutions of congruence (A.1) is of the form

$$c (k)=1+ \chi(k),\quad k\in\mathbb{Z}(p).$$

2. Quadratic Congruence to $p^m$ Modulus

We can consider all solutions $x$ of congruence with parameter $a$

$$ x^2\equiv a (\mathrm{mod} p^m),\quad x,\, a \in \mathbb{Z}(p^m).$$

(A.5)

Any solution of (A.5) will be also the solution of similar congruence with less by 1 power of $p$:

$$ \text{if }\acute{x}:\; \acute{x}^2\equiv a (\mathrm{mod} p^m) \text{ then } \;\Rightarrow\; \acute{x}^2\equiv a (\mathrm{mod} {p^{m-1}}) \;\Rightarrow\;\dots\Rightarrow\; \acute{x}^2\equiv a (\mathrm{mod} {p^1}).$$

(A.6)

Notation of $x$ expansion in powers of $p^\alpha$ and digits $x_\alpha$ looks like:

$$ \begin{aligned} \, x &= x_0 +x_1\, p + x_2\, p^2 + \cdots + x_\alpha p^\alpha + \cdots +x_{m-1}\, p^{m-1},\quad x_\alpha \in \mathbb{Z}(p) ;\\ a &= a_0 +a_1\, p + a_2\, p^2 + \cdots + a_\alpha p^\alpha + \cdots +a_{m-1}\, p^{m-1},\quad a_\alpha \in \mathbb{Z}(p).\\ \end{aligned}$$

(A.7)

First we explain the case when $a$ is relatively prime to $p$.

The greatest common divisor of $a$ and $p$ in this case is 1

$$(a,p)=1.$$

From (A.6) and (A.7) one can immediately find congruence for digit $x_0$

$$ x_0^2 \equiv a_0 (\mathrm{mod} {p^1}),$$

(A.8)

it looks like (A.1). If $a_0$ is a quadratic nonresidue there are no solutions of (A.5). So one-dimensional multiplicity for $N=p^m$

$$c (a)=0 \quad \text{if $(a,p)=1$ and digit $a_0 \not\in\text{Res}(p)$.}$$

We go further with $a_0$ as quadratic residue, so $\; x_0=\pm\sqrt{a_0}\in \mathbb{Z}(p)$. From (A.6) and (A.7) from next $p^2$-modulo congruence $x^2\equiv a (\mathrm{mod} {p^2})\;$ we get

(A.9)

as linear congruence with respect to unknown $x_1$

$$2 x_0 x_1 \; p\equiv a_0 -x_0^2 +a_1 p (\mathrm{mod} {p^2}),$$

one can reduce this right side by $p$ due to (A.8). Then

$$2 x_0 \; x_1 \equiv a_1 + \frac{a_0 -x_0^2}{p} (\mathrm{mod} p).$$

In field $\mathbb{Z}(p)$ since $2x_0\neq0$ we have inverse element $(2 x_0)^{-1}\in \mathbb{Z}(p)$. Digit $x_1$ is uniquely defined:

$$x_1 \equiv \left(a_1 + \frac{a_0 -x_0^2}{p} \right) (2 x_0)^{-1} (\mathrm{mod} p).$$

From next $p^3$-modulo congruence $x^2\equiv a (\mathrm{mod} {p^3})$ we get linear congruence with unknown $x_2$

whose right side has common factor $p^2$ due to (A.9). So at this step one can find $x_2$ uniquely:

$$x_2 \equiv \left( a_2+ \frac{a_0+a_1 p -(x_0 +x_1\, p)^2}{p^2} \right) (2 x_0)^{-1} (\mathrm{mod} p).$$

All digits $x_\alpha$ are found sequentially step by step. Let us consider arbitrary step of finding $x_\alpha$ from $p^{\alpha+1}$-modulo congruence $x^2\equiv a (\mathrm{mod} {p^{\alpha+1}})$.

We have $x_\alpha \in\mathbb{Z}(p)$ uniquely as

$$x_\alpha \equiv \left(a_\alpha+ \frac{[a_0+\cdots] -[x_0 +\cdots]^2}{p^\alpha} \right) (2 x_0)^{-1} (\mathrm{mod} p),$$

when $x_0$ is fixed with its sign, then solution $x$ is calculated uniquely.

Thus multiplicities in one dimension

$$ c_{p^m}(a) =c_{p}(a_0) =1 +\chi(a_0)\quad \text{if digit } a_0\ne0,$$

(A.10)

and we obtain this property of multiplicities for arbitrary $n<m$

$$ c_{p^m}(a)=c_{p^n}\left( a{\mod p^n}\right) \quad \text{if digit }a_0\ne0.$$

(A.11)

From $p^2$-modulo congruence $x^2\equiv a (\mathrm{mod} {p^2})\;$ we get

and anyway no solutions for $x_1$. So if $a_1 \neq0$ than $c_{p^m}(a)=0$, i.e.

$$c_{p^m}(\xi\, p^1)=0, \quad \xi\in\mathbb{Z}(p^{m-1})\setminus\{0\}.$$

We put the lowest non-zero digit in $x$ as $x_\alpha\neq0$. Then congruence (A.5) takes the form

$$(x_\alpha\cdot p^\alpha+\cdots)^2 \equiv (a_{2\lambda+1}+\cdots)p^{2\lambda+1} (\mathrm{mod} {p^m}),$$

$$(x_\alpha+\cdots)^2 p^{2\alpha} \equiv (a_{2\lambda+1}+\cdots)p^{2\lambda+1} (\mathrm{mod} {p^m}).$$

Anyway no solutions exist and multiplicity

$$ c_{p^m}(\xi\, p^{{2\lambda+1}})=0, \quad \text{where } \xi\in \mathbb{Z}(p^{m-2\lambda-1}) \setminus\{0\},\, \lambda\geqslant0.$$

(A.12)

From $p^2$-modulo congruence $x^2\equiv a (\mathrm{mod} {p^2})\;$ we get

Simply $x_0=0$ and from $p^3$-modulo congruence $x^2\equiv a (\mathrm{mod} {p^3})\;$ we get ordinary quadratic congruence for $x_1$

with some solutions if $\sqrt{a_2}$ exists. But now starting equation is not (A.5), because of power descending by 2 units

$$( x_1\, p + x_2\, p^2+\cdots +x_{m-1}\, p^{m-1})^2 \equiv a_2 p^2+\cdots+a_{m-1}\, p^{m-1} (\mathrm{mod} {p^m})$$

thus obviously last digit $x_{m-1}$ can take arbitrary values from $\mathbb{Z}(p)$. Therefore multiplicities as number of solutions of (A.5) are greater by $p$

$$c_{p^m}(a)=\bigl(1+\chi(a_2) \bigr)\, p\quad \text{if digit } a_2\ne0,$$

i.e.

$$c_{p^m}(\xi\, p^2)=\bigl(1+\chi(\xi{\mod p}) \bigr)\, p, \quad \text{where } \xi\in\mathbb{Z}(p^{m-2})\setminus\{0\}.$$

$$(x_\alpha+\cdots)^2 p^{2\alpha} \equiv (a_{2\lambda}+\cdots)p^{2\lambda} (\mathrm{mod} {p^m}), \quad x_\alpha\neq0.$$

The solutions exist only when ${\alpha}=\lambda$. We obtain

(A.13)

and we have the similar case as (A.10) (when $a$ is relatively prime to $p$).

In solution $x$ we see that the lowest $\alpha$ digits are zero and the highest $\alpha$ digits one can take arbitrarily. And between them $m-2\alpha$ digits are calculated step by step from linear congruences.

Therefore multiplicities as number of solutions of (A.5) are greater by $p^{\alpha}$

$$ c_{p^m}(a)= \bigl(1+ \chi(a_{2\alpha}) \bigr) p^{\alpha}, \quad \text{if digits $a_0=\cdots =a_{2\alpha-1}=0$, $a_{2\alpha}\neq0$ and } 2\alpha<m.$$

(A.14)

And for arbitrary $n<m$

$$ c_{p^m}(a)=c_{p^n}\left( a{\mod p^n}\right) \quad \text{if digit $a_{2\alpha}\ne0$ and $2\alpha\leqslant n<m$.}$$

(A.15)

In section D we’ll investigate similarity of generating functions in $d=1$ using this relation.

We denote the lowest non-zero digit in $x$ by index $\alpha$.

$$ (x_\alpha p^\alpha+\cdots)^2=(x_\alpha+\cdots+x_{m-1}\, p^{m-1-\alpha})^2 \; p^{2\alpha} \equiv0 (\mathrm{mod} {p^m}).$$

(A.16)

We have solutions here if

$$2\alpha \geqslant m,$$

so the lowest value of index $\alpha$ is ${\left\lfloor\frac{m+1}{2}\right\rfloor}.$ The number of arbitrary digits is $\left\lfloor\frac{m}{2}\right\rfloor$. So multiplicity is $c_{p^m}(0)=p^{\lfloor\frac{m}{2}\rfloor}$. We get all-for-one in formulæ

$$ c_{p^m}(a) =\left\{ \begin{aligned} \, p^{\lfloor\frac{m}{2}\rfloor} \quad &\text{ if }a=0, \\ \bigl(1+ \chi(a_{2\lambda}) \bigr)\, p^{\lambda} \quad &\text{ if $a$ is divisible by $p^{2\lambda}$, i.e. $(a,p^m)=p^{2\lambda}$, and $\lambda\geqslant0$,}\\ 0 \quad &\text{ if }a\text{ is divisible by } p^{2\lambda+1},\text{ i.e. }(a,p^m)=p^{2\lambda+1}, \lambda\geqslant0. \\ \end{aligned}\right.$$

(A.17)

Appendix B. Generating Functions for Multiplicities

We define the polynomial of $\tau$

$$ f^d_N(\tau)=\sum_{k\in\mathbb{Z}(N)} c_{Nd}(k)\,\tau^k$$

(B.1)

as a generating function for multiplicities $c_{N\,d}(k)$ of $E(k)$ on the lattice $\mathbb{Z}^d(N)$. For one-dimensional generating function we skip the upper index $d=1$, i.e. $f_N(\tau)\stackrel{\text{def}}{=}f^1_N(\tau)$. And $f_0(\tau)\stackrel{\text{def}}{=}1$.

We use brackets Footnote

$\left\langle \tau^{k} \mid f_N(\tau) \right\rangle =\frac{1}{N} \sum_{l=1}^N \tau(l)^* f_N(\tau(l))$, where $\tau(l)$ are roots of unity in $\mathbb{C}$ given by $\tau(l)=\mathrm{e}^{2\pi \mathrm{i} \, l/N}$.

to extract multiplicity $c_{N\,d}(k)$ from generating function:

$$ \left\langle \tau^{k} \mid f^d_N(\tau) \right\rangle =c_{N\,d}(k).$$

(B.2)

$\left\langle \tau^{k} \mid f_N(\tau) \right\rangle =\frac{1}{N} \sum_{l=1}^N \tau(l)^* f_N(\tau(l))$, where $\tau(l)$ are roots of unity in $\mathbb{C}$ given by $\tau(l)=\mathrm{e}^{2\pi \mathrm{i} \, l/N}$.

We treat $\tau$ as a formal variable with the equivalence relation:

$$ \tau^N=1,$$

(B.3)

i.e. the powers of $\tau$ in (B.1) can be considered as elements of $\mathbb{Z}(N)$. The multiplication with relation (B.3) we denote as $f\circ g$. Hereafter $\tau^N$ is equivalent to unity in reduction of similar terms so that $\tau^{k+N}=\tau^k$:

$$\left(\sum_{p_1\in\mathbb{Z}(N)} \tau^{p_1^2}\right)^d =\sum_{p_1\in\mathbb{Z}(N)}\cdots \sum_{p_d\in\mathbb{Z}(N)} \tau^{p_1^2}\circ \cdots \circ\tau^{p_d^2} = \sum_{\mathbf{p}\in\mathbb{Z}^d(N)} \tau^{\mathbf{p}^2} ,$$

hence in case $d>1$ the generating function can be represented as a $d$ times product of one-dimensional generating functions:

$$ f^d_N(\tau) =\Bigl(f_N(\tau)\Bigr)^{ d}.$$

(B.4)

The introduced equivalence (B.3) doesn’t violate the commutativity and distributivity of multiplication. Thereby, we reduce our investigation of multiplicities (2.7) to problems of polynomial algebra and arithmetic of quadratic forms [8, 9].

One can assume that variable $\tau$ takes values in the set of complex $N$th roots of unity:

$$\tau\in\{\tau(l)=\mathrm{e}^{2\pi\mathrm{i}\frac{l}{N}},\, l\in\mathbb{Z}(N)\}.$$

In such a representation we find multiplicities $c_{Nd}(k)$ as coefficients of discrete Fourier expansion of the function $f^d_N(\tau(l))$ on the lattice $\mathbb{Z}(N)$.

Polynomial $\phi_N(\tau)$.

We define the polynomial $\phi_N(\tau)$ with all coefficients equal to 1:

$$ \phi_N(\tau)=\sum_{k\in\mathbb{Z}(N)} \tau^k= 1 + \tau + \tau^2 + \cdots + \tau^{N-1}.$$

(B.5)

Here $\phi_N(\tau) \not\equiv \frac{1-\tau^N}{1-\tau}$ because due to (B.3) the value 1 of variable $\tau$ is possible value.

If $\tau$ runs over complex roots of unity then

$$\phi_N(\tau) = \left\{\begin{array}{cc} N &\text{if }\tau=1,\\ 0 &\text{otherwise}. \end{array}\right.$$

From this point of view $\phi_N(\tau)$ is the lattice analogue of the delta function

$$\phi_N\bigr(\tau(l)\bigl)=\left\{\begin{array}{cc} N &\text{if }l=0,\\ 0 &\text{otherwise} . \end{array}\right.$$

For any $k\in\mathbb{Z}(N)$ we have

$$ \tau^k \circ \phi_N \; (\tau)= 1\cdot\phi_N (\tau) ,$$

(B.6)

hence

$$ \phi_N \circ \phi_N = \left( \sum_{k=0}^{N-1} \tau^{k}\right) \circ \phi_N = N\cdot \phi_N$$

(B.7)

and consequently

$$ \phi_N \circ \left(N- \phi_N\right) = 0,$$

(B.8)

$$\begin{aligned} \, \left(N- \phi_N\right)^{\;2} &= N\left(N- \phi_N\right) ,\\ \left(N- \phi_N\right)^{\;m} &= N^{m-1}\left(N- \phi_N\right). \end{aligned}$$

Appendix C. Case $$N=p$$ , $$p\not=2$$

Now $p$ is prime odd number and we put $N=p$ hereafter in section C. During intermediate calculations we sometimes skip the index $p$ (or $N$).

For the one-dimensional case we can derive the explicit expression for the multiplicities $c_{N\,1}(k) =c_{N}(k)$ in terms of quadratic congruences [10].

Polynomial $g_p(\tau)$.

Coefficients $c_{p}(k)$ take values $0$ or $2$, or $1$ (square root of $0$); and the generating function is

$$f_p(\tau)= 1 + 2 \tau+ 2 \tau^r + 2 \tau^s + \cdots =1+\sum_{r\in \text{Res}(p)}2\tau^r,$$

where powers $\tau$ with nonresidues $\tilde{r},\tilde{s},..$ disappeared from $f_p(\tau)$ because of its null coefficients (multiplicities).

Next we introduce the polynomial $g_p(\tau)$ using the character $\chi(k)$ (defined in section A):

$$ g_p(\tau)= \sum_{k\in\mathbb{Z}(p)} \chi(k)\cdot \tau^k .$$

(C.1)

So, the generating function for multiplicities in the one-dimensional case is the sum of polynomials $g_p(\tau)$ and $\phi_p(\tau)$:

$$ f_p(\tau)= 1+\sum_{r\in \text{Res}(p)}2\tau^r = \phi_p(\tau) +g_p(\tau).$$

(C.2)

And multiplicities in $d=1$ are given by explicit formula:

$$\text{in $d=1$ } \quad c_{p\,1}(k)=1+\chi(k).$$

1. Properties of Polynomials $g_p$ and $\phi_p$ in Multiplication

Due to (B.7) we have

$$ \begin{aligned} \, \phi_p &\circ \phi_p = p\cdot \phi_p \quad\text{ and }\quad \phi_p ^{\; m} = p^{m-1}\cdot \phi_p,\\ (p - \phi_p) &\circ \phi_p = 0, \end{aligned}$$

(C.3)

and we get

$$ \phi_p \circ g_p =0,$$

(C.4)

because of (B.6) and (A.4)

$$\phi_p \circ \left( \sum_{a=0}^{p-1} \chi(a)\,\tau^a \right) =\phi_p \circ \Biggl( \underbrace{\sum_{a=0}^{p-1} \chi(a) }_0 \Biggr) =0.$$

The explicit result for $g_p^{\,2}$ is:

$$ g_p\circ g_p = (-1)^{\frac{p-1}{2}} \left(\vphantom{\frac{a}{a}} p- \phi_p \right).$$

(C.5)

Proof.

Writing out $g_p^{\,2}$ as a double sum and remembering about (B.3), we will make a transition at fixed $a$ from summation variable $b$ to the new summation variable $t$ by simple rule $b\equiv a\cdot t (\mathrm{mod} p)$, since $\mathbb{Z}(p)$ is field. In third line, we split the sum over $t$ into two parts.

$\quad\square$

As a consequence of (C.3), (C.4), (C.5) we have the equation for arbitrary powers $m$ and $n$

$$ \phi_p^{\;m} \circ g_p^{\;n} =0,$$

(C.6)

then we obtain

$$ (\phi_p + g_p) ^{\;m} = \phi_p^{\;m} + g_p^{\;m}.$$

(C.7)

And from (C.5)

$$ g_p\circ g_p \circ g_p = (-1)^{\frac{p-1}{2}} \cdot p\cdot g_p .$$

(C.8)

Arbitrary power $m$:

$$ g_p ^{\; m}= (-1)^{\frac{p-1}{2}\lfloor\frac{m}{2}\rfloor} \begin{cases} p^{\lfloor\frac{m}{2}\rfloor} \cdot g_p &\text{if odd $m$,}\\ p^{\frac{m}{2}-1} \bigl(p - \phi_p) &\text{if even $m$.}\\ \end{cases}$$

(C.9)

2. Multiplicities in $d=2$

From generation function (C.2) we have in two-dimensional case due to (C.7)

$$f^2_p(\tau) = f_p \circ f_p = \phi_p \circ \phi_p + g_p\circ g_p.$$

Now we have mathematical expression for $f^2_p$ for any prime odd $p$:

$$ f^2_p = p\cdot\phi_p + (-1)^{\frac{p-1}{2}} \bigl( p- \phi_p \bigr),$$

(C.10)

in explicit polynomial form we have

$$ f^2_p(\tau)= \begin{cases} (2p-1)\cdot\tau^0 + (p-1)\cdot\tau^1 +\cdots+(p-1)\cdot\tau^{p-1} & \text{if even $\frac{p-1}{2}$,}\\ \qquad 1\cdot\tau^0 \; + \; (p+1)\cdot\tau^1 +\cdots+(p+1)\cdot\tau^{p-1} & \text{if odd $\frac{p-1}{2}$}. \end{cases}$$

(C.11)

So polynomial coefficient from $f_p^2$ is congruent with $\pm1$ and multiplicity is congruent as

$$ \text{in $d=2$ } \quad c_{p\,2}(k)\equiv -\chi(-1) (\mathrm{mod} {p}),\quad k\in\mathbb{Z}(p).$$

(C.12)

3. Multiplicities in $d=3$

Using (C.7) and (C.9) for $f^3_p$ we have

$$f^3_p = f_p \circ f_p \circ f_p = \phi_p^{\;3} + g_p^{\;3} = p^2\,\phi_p +(-1)^{\frac{p-1}{2}}\, p \, g_p.$$

In the expression above one can take the common factor $p$ out of brackets. So each coefficient in polynomial above is divisible by $p$ and we have

$$c_{p\,3}(k)\equiv 0 (\mathrm{mod} p).$$

And for $d>3$ formula

$$c_{p\,d}(k)\equiv 0 (\mathrm{mod} p)$$

is obvious due to factorization (B.4). We proved the proposition of the Theorem in case $N=p$.

Appendix D. Case $$N=p^m$$ , $$p\not=2$$

Cyclotomic Polynomials $\Phi_n(\tau)$.

In this section $N=p^m$, $p$ is prime odd and $m\in\mathbb{N}$, $m>1$. The equivalence relation (B.3) now is

$$\tau^{p^m}=1.$$

In addition to $\phi_N(\tau)$ as sum of all powers in (B.5) we introduce cyclotomic polynomials for $N=p^m$:

$$ \begin{aligned} \, \Phi_p (\tau) &= 1 + \tau + \tau^2 + \cdots + \tau^{p-1}=\phi_p (\tau) , \\ \Phi_{p^2}(\tau) =\Phi_p(\tau^p) &= 1 + \tau^{p} + \tau^{2p}+ \tau^{3p}+ \dots +\tau^{(p-1)p} =\phi_p(\tau^p), \\ \Phi_{p^3}(\tau) =\Phi_p(\tau^{p^2}) &= 1 +\tau^{p^2} +\tau^{2p^2} +\tau^{3p^2} +\cdots+\tau^{(p-1)p^2} =\phi_p(\tau^{p^2}), \\ \vdots\\ \Phi_{p^{m}}(\tau) =\Phi_p(\tau^{p^{m-1}}) &= 1 +\tau^{p^{m-1}} +\tau^{2p^{m-1}} +\tau^{3p^{m-1}} +\cdots+\tau^{(p-1)p^{m-1}} =\phi_p(\tau^{p^{m-1}}) .\\ \end{aligned}$$

(D.1)

One can easily derive that the product of cyclotomic polynomials of successive powers of $p$ is the sum of all powers of $\tau$ in $\phi_N(\tau)$:

$$\phi_{p^{m}} = \Phi_{p}\cdot\Phi_{p^2}\cdots\Phi_{p^{m}}.$$

We have similar to (C.3) equalities

$$ \begin{aligned} \, \tau^{p^{m-1}} \circ \Phi_{p^{m}} = \Phi_{p^{m}}, \\ \Phi_{p^{m}} \circ \Phi_{p^{m}} = p\cdot \Phi_{p^{m}},\\ \left(p- \Phi_{p^{m}} \right) \circ \Phi_{p^{m}} = 0. \end{aligned}$$

(D.2)

And polynomials with modified argument $\tau^{p^{m-1}}$

$$ \Phi_{p^{m}} (\tau) \circ g_p (\tau^{p^{m-1}}) = \phi_p (\tau^{p^{m-1}}) \circ g_p (\tau^{p^{m-1}}) =0,$$

(D.3)

$$ \bigl(\phi_p (\tau^{p^{m-1}}) + g_p (\tau^{p^{m-1}}) \bigr) ^{\;n} = \phi_p (\tau^{p^{m-1}}) ^{\;n} + g_p (\tau^{p^{m-1}}) ^{\;n}.$$

(D.4)

1. Similarity of Successive Generating Functions in $d=1$

The recurrent formula for $f_{p^m}(\tau)=\mathcal{F}\left( f_{p^{m-1}}(\tau)\right)$ is

$$ f_{p^m}(\tau) = \Phi_{p^m}(\tau) \cdot f_{p^{m-1}}(\tau) +(-1)^{\frac{p-1}{2}\lfloor\frac{m}{2}\rfloor} \cdot \bigl( g_p(\tau^{p^{m-1}}) \bigr)^m ,$$

(D.5)

or via (C.9)

$$ f_{p^m}(\tau) = \Phi_{p^m}(\tau)\cdot f_{p^{m-1}}(\tau) + \begin{cases} p^{\lfloor\frac{m}{2}\rfloor} \cdot g_p(\tau^{p^{m-1}}) & \text{if odd } m,\\ p^{\frac{m}{2}-1} \bigl(p - \Phi_{p^m}(\tau)\bigr)& \text{if even } m. \end{cases}$$

(D.6)

For example, if we take $k=1 \in\mathbb{Z}(p^{m})$ then from (A.10) in $d=1$ it follows that $c_{p^{m}}(1) = c_{p\,}(1)$. Using notation with brackets (B.2) we write

$$\left\langle \tau^{1} \mid f_{p^{m}}(\tau) \right\rangle =\left\langle \tau^{1} \mid f_{p}(\tau) \right\rangle = 1+\chi(1) = 2.$$

To generalize this relation we use (A.11) for non divisible by $p$ arbitrary $k\in\mathbb{Z}(p^{m})$ and the equality of $\tau^k$ coefficients $c_{p^{m}}(k)$ in $f_{p^{m}}(\tau)$ and $c_{p^{m-1}}(k {\scriptstyle \mod p^{m-1}} )$ in $f_{p^{m-1}}(\tau)$ is clear. Using cyclotomic polynomials (D.1) we can write this similarity relation

$$ \Bigl\langle \tau^{k} \mid f_{p^{m}}(\tau)\Bigr\rangle = \Bigl\langle \tau^{k} \mid \Phi_{p^{m}}(\tau)\circ f_{p^{m-1}}(\tau)\Bigr\rangle = \Bigl\langle \tau^{k} \mid \Phi_{p^{m}} \cdots\Phi_{p^{2}}\cdot f_p\Bigr\rangle \quad\text{ if } k\not\equiv0 (\mathrm{mod} p).$$

(D.7)

NB. In fact, the largest power of $\Phi_{p^{m}}$ polynomial is $(p-1)p^{m-1}$ and the largest power of $f_{p^{m-1}}$ polynomial is $p^{m-1} -1$ and power of their product is

$$(p-1)p^{m-1} +p^{m-1} -1=p^{m}-1,$$

there is no need to use here $\circ$ product.

We can obtain formula more general than (D.7). Due to (A.15) $c_{p^m}(a)=c_{p^{m-1}}\left( a{\scriptstyle\mod p^{m-1}}\right)$ and then

$$ \Bigl\langle \tau^{k} \mid f_{p^{m}}(\tau)\Bigr\rangle = \Bigl\langle \tau^{k} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle \quad\text{ if } k\not\equiv0 (\mathrm{mod} {p^{m-1}}).$$

(D.8)

We will consider odd $m$ and even separately.

For example, if we take $k=0\in\mathbb{Z}(p^{m})$ then

$$\Bigl\langle \tau^{0} \mid f_{p^{m}}(\tau)\Bigr\rangle = p^\mu \qquad\text{ and }\qquad \Bigl\langle \tau^{0} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle = p^\mu.$$

For $k=p^{2\mu}\in\mathbb{Z}(p^{m})$

$$\begin{aligned} \, \Bigl\langle \tau^{p^{2\mu}} \mid f_{p^{m}}(\tau)\Bigr\rangle &= c_{p^{m}}(1\cdot p^{2\mu} ) = \left( 1+\chi(1) \right) p^\mu, \\ \Bigl\langle \tau^{p^{2\mu}} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle &= \Bigl\langle \tau^{p^{2\mu}} \bigm| (1+{\tau^{p^{2\mu}}}+\cdots)\cdot (p^\mu+2\tau+\cdots) \Bigr\rangle = p^\mu, \end{aligned}$$

so in general case when $k\equiv0 (\mathrm{mod} {p^{m-1}})$

$$ \Bigl\langle \tau^{k} \mid f_{p^{m}}(\tau)\Bigr\rangle = \Bigl\langle \tau^{k} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle + \chi(\kappa)\, p^\mu\, , \qquad\text{ where } k=\kappa\cdot p^{m-1}\in\mathbb{Z}(p^{m}).$$

(D.9)

Relations (D.8) and (D.11) give for odd $m$

$$ f_{p^m}(\tau) = \Phi_{p^m}(\tau) \cdot f_{p^{m-1}}(\tau) +p^\mu \sum_{\kappa=0}^{p-1}\chi(\kappa) \tau^{\kappa\, p^{m-1}} = \Phi_{p^m}(\tau) \cdot f_{p^{m-1}}(\tau) + p^{\lfloor\frac{m}{2}\rfloor} \cdot g_p(\tau^{p^{m-1}}).$$

(D.10)

If we take $k=0\in\mathbb{Z}(p^{m})$ then due to (A.17)

$$\Bigl\langle \tau^{0} \mid f_{p^{m}}(\tau)\Bigr\rangle = p^\mu \qquad\text{ and }\qquad \Bigl\langle \tau^{0} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle = p^{\mu-1}.$$

For $k=p^{2\mu}\in\mathbb{Z}(p^{m})$

$$\begin{aligned} \, \Bigl\langle \tau^{p^{m-1}} \mid f_{p^{m}}(\tau)\Bigr\rangle &= 0, \\ \Bigl\langle \tau^{p^{m-1}} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle &= \Bigl\langle \tau^{p^{2\mu-1}} \bigm| (1+{\tau^{p^{2\mu-1}}}+\cdots)\cdot (p^{\mu-1}+2\tau+\cdots) \Bigr\rangle = p^{\mu-1}, \end{aligned}$$

so in general case when $k\equiv0 (\mathrm{mod} {p^{m-1}})$

$$ \Bigl\langle \tau^{k} \mid f_{p^{m}}(\tau)\Bigr\rangle = \Bigl\langle \tau^{k} \mid \Phi_{p^{m}}(\tau)\cdot f_{p^{m-1}}(\tau)\Bigr\rangle + p^\mu\cdot\delta_{\kappa 0} - p^{\mu-1}\cdot1 , \quad\text{ where } k=\kappa\cdot p^{m-1}\in\mathbb{Z}(p^{m}).$$

(D.11)

We obtain

$$ \text{for } m=2\mu :\quad f_{p^m}(\tau) = \Phi_{p^m}(\tau) \cdot f_{p^{m-1}}(\tau) + p^{\frac{m}{2}-1} \bigl(p - \Phi_{p^m}(\tau)\bigr).$$

(D.12)

2. Multiplicities in $d=2$

In two-dimensional case due to (D.4) from one-dimensional (D.5) we have

$$ \begin{aligned} \, f^2_{p^m} &= f_{p^m} \circ f_{p^m} = \left( \Phi_{p^m}\circ f_{p^{m-1}}\right) ^2 \, + \Bigl( g_p(\tau^{p^{m-1}}) \Bigr)^{2m} \\ &= p\cdot \Phi_{p^m}\cdot f^2_{p^{m-1}} \; + \left( (-1)^{\frac{p-1}2} \cdot\left(p -\Phi_{p^m}\right)\right) ^{m} \\ &= p\cdot \Phi_{p^m}\cdot f_{p^{m-1}}^2 \; + (-1)^{\frac{p-1}{2}\,{m}} \cdot p^{m-1} \cdot\left(p -\Phi_{p^m}\right). \end{aligned}$$

(D.13)

3. Multiplicities in $d=3$

In three-dimensional case

$$ \begin{aligned} \, f^3_{p^m} = f_{p^m} \circ f_{p^m} \circ f_{p^m} &= \left( \Phi_{p^m}\circ f_{p^{m-1}}\right) ^3\, + (-1)^{\frac{p-1}{2}\,\lfloor\frac{m}{2}\rfloor\, m} \Bigl( g_p(\tau^{p^{m-1}}) \Bigr)^{3m} \\ &= p^2\cdot \Phi_{p^m}\cdot f^3_{p^{m-1}} \, + (-1)^{\frac{p-1}{2}\lfloor\frac{m}{2}\rfloor m} \Bigl( (-1)^{\frac{p-1}2}\, p\cdot g_p(\tau^{p^{m-1}}) \Bigr)^{m} \\ &= p^2\cdot \Phi_{p^m}\cdot f^3_{p^{m-1}} \, + (\pm)\, p^{m} \Bigl( g_p(\tau^{p^{m-1}})\Bigr)^{m} ,\\ \end{aligned}$$

(D.14)

where the sign $\pm$ equals $(-1)$ if $\quad -p\equiv m\equiv1 (\mathrm{mod} 4)\quad$ and $(+1)$ otherwise.

It is clear that if $f^3_{p^{m-1}}$ is divisible by $p^{m-1}$ than $f^3_{p^{m}}$ is also divisible by $p^{m}$. In section C the Theorem is proved for $N=p^1$ and hence by induction we proved the Theorem for $N=p^m$.

Appendix E. Definitions for Proof-2

1. $d$-Dimensional Integer Space

Let us consider the $d$-dimensional space $\mathbb{Z}^d$ of integer vectors

$$\boldsymbol{\xi}=(\xi_1, \xi_2, ..., \xi_d),\quad \xi_i \in \mathbb{Z}.$$

In the figures, the elements of space $\mathbb{Z}^d$ is represented by cells of unit size in all coordinates.

Let $\mathrm L$ be some finite subset of $\mathbb{Z}^d$

$$\mathrm L \subset \mathbb{Z}^d,\quad | \mathrm L | \in\mathbb{N},$$

where $| \mathrm L |$ is the number of elements in $\mathrm L$.

Notation

$$n \cdot \mathrm L + \mathbf{a}=\{n\cdot\boldsymbol{\xi}+\mathbf{a}|\boldsymbol{\xi}\in\mathrm L\},$$

where $n \in \mathbb{N}$, $\mathbf{a} \in \mathbb{Z}^d$, is used for the set $\mathrm L$ stretched by factor $n$ and shifted by the vector $\mathbf{a}$. That is, the points of this set are obtained from the points of the set $\mathrm L$ by multiplying all coordinates by $n$ and shifting by vectors $\mathbf{a}$:

$$\xi_i \rightarrow n\xi_i + a_i.$$

The notation $\mathrm L + \mathbf{a}$ is equivalent to $1 \cdot \mathrm L + \mathbf{a}$.

The notation $n \cdot \mathrm L$ is equivalent to $n \cdot \mathrm L + \mathbf{0}$, where $\mathbf{0}$ is the null vector in $\mathbb{Z}^d$.

2. Sets of Integers

Definition E.1.

$\mathbb{Z}(\alpha)$ is the set of integers from $0$ to $\alpha-1$, i.e.

$$\mathbb{Z}(\alpha) = \left\{0, 1, 2, \ldots, \alpha-1 \right\}.$$

Definition E.2.

$n\mathbb{Z}(\alpha) + a$ is the set of all numbers from $\mathbb{Z}(\alpha)$, multiplied by $n$ and shifted by $a$, i.e.

$$n\mathbb{Z}(\alpha)+a=\left\{a, n+a, 2n+a, \ldots, (\alpha-1)n+a\right\}.$$

The notation $\mathbb{Z}(\alpha)+a$ is equivalent to $1\mathbb{Z}(\alpha)+a$.

The notation $n\mathbb{Z}(\alpha)$ is equivalent to $n\mathbb{Z}(\alpha)+0$.

3. Lattices

Let us introduce the notation for the $b$-dimensional lattice $\underbrace{N \times N \times \ldots \times N}_b$.

Definition E.3.

$\operatorname{lattice}^b(N)$ for $a\leqslant d$ is the set of $\boldsymbol{\xi}\in\mathbb{Z}^d$ such that $\xi_1,\dots,\xi_b\in\mathbb{Z}(N)$, and $\xi_{b+1}=\dots=\xi_d=0$.

Definition E.4.

$\operatorname{lattice}_{i_1\dots i_b}^b(N)$ for $b\leqslant d$ is the set of $\boldsymbol{\xi}\in\mathbb{Z}^d$ such that $\xi_{i_1},\dots,\xi_{i_b}\in\mathbb{Z}(N)$, and $\xi_j=0$ if $j\not\in\{i_1,\dots,i_b\}$.

For example, if $d=6$, then $\operatorname{lattice}^3_{2,3,5}(N)$ is the set of $\boldsymbol{\xi}\in\mathbb{Z}^6$ such that $\xi_2,\xi_3,\xi_5\in\mathbb{Z}(\alpha)$, and $\xi_1=\xi_4=\xi_6=0$.

4. Functions

Definition E.5.

$\operatorname{ncells}(\mathrm L, \alpha, k)$ (”the number of cells”) is the number of cells $\boldsymbol{\xi}\in\mathrm L\subset\mathbb{Z}^d$, such that

$$\sum_{i=1}^d \xi_i^2\equiv k (\mathrm{mod} \alpha).$$

In terms of $\operatorname{ncells}$ the multiplicity function $c_{Nd}$ (2.7) has the following form

$$c_{Nd}(k)=\operatorname{ncells}(\operatorname{lattice}^d(N),N,k).$$

Definition E.6.

The logical function $\operatorname{zrf}(\mathrm L, \alpha, \beta)$ (”zerofy”) is equal to “$true$” if

$$\forall k\in\mathbb{Z}(\alpha)\quad \operatorname{ncells}(\mathrm L, \alpha, k)\equiv 0 (\mathrm{mod} \beta),$$

otherwise it is “$false$”.

The notation $\operatorname{zrf}(\mathrm L, \alpha, \beta)$ is equivalent to $\operatorname{zrf}(\mathrm L, \alpha, \beta) = true$.

The parameter $\alpha$ of the functions $\operatorname{ncells}$ and $\operatorname{zrf}$ is called the momentum ring parameter, and $\beta$ is called the energy ring parameter.

In terms of fumction $\operatorname{zrf}$ the theorem (2.8) has the following form

$$d\geqslant3\text{ or }(d\geqslant2,N=2^n) \Rightarrow \operatorname{zrf}(\operatorname{lattice}^d(N),N,N).$$

Appendix F. Lemmas

To simplify the proof we introduce several simple lemmas. For obvious lemmas, the proofs are skipped.

Lemma F.1 (on increasing of the momentum ring parameter).

$$ \operatorname{ncells}(\mathrm L, \alpha, k) = \sum\limits_{j=0}^{n-1} \operatorname{ncells}(\mathrm L, n\alpha, k + j\alpha).$$

(F.1)

Proof.

The number $k \in \mathbb{Z}(\alpha)$ corresponds to the numbers $k$, $k+\alpha$, $k+2\alpha$, $k+3\alpha$,... , $k+( n-1)\alpha$ in $\mathbb{Z}(n\alpha)$. To go from modulus $n\alpha$ to modulus $\alpha$, one needs to sum the corresponding numbers of cells. $\quad\square$

Lemma F.2 (on decreasing of the momentum ring parameter).

$$\operatorname{zrf}(\mathrm L, n\alpha, \beta) \Rightarrow \operatorname{zrf}(\mathrm L, \alpha, \beta).$$

Lemma F.3 (on decreasing of the energy ring parameter).

$$\operatorname{zrf}(\mathrm L, \alpha, n\beta) \Rightarrow \operatorname{zrf}(\mathrm L, \alpha, \beta).$$

Lemma F.4 (on lattice stretching).

$$\operatorname{zrf}(\mathrm L, \alpha, \beta) \Rightarrow {\operatorname{zrf}(n \cdot \mathrm L, n^2\alpha, \beta)}.$$

Lemma F.5 (on lattice shift).

$$\operatorname{ncells}(n\cdot\mathrm L+n\alpha\mathbf{a},n^2\alpha,k) = \operatorname{ncells}(n\cdot\mathrm L, n^2\alpha,k).$$

Proof.

Let $\xi_i$ be the coordinates of a cell in the set $\mathrm L$. $n\xi_i$ are the coordinates of a cell in the set $n\cdot \mathrm L$. When shifted by the vector $n\alpha\mathbf{a}$, the coordinates of the cells of the set $n\cdot\mathrm L$ are shifted by the corresponding components of this vector:

$$(n\xi_1,n\xi_2,n\xi_3,\ldots)\rightarrow(n\xi_1+n\alpha a_1,n\xi_2+n\alpha a_2,n\xi_3+n\alpha a_3,\ldots).$$

The sum of the squares of the coordinates changes as follows:

$$(n\xi_1)^2+(n\xi_2)^2+(n\xi_3)^2+\ldots\rightarrow (n\xi_1+n\alpha a_1)^2+(n\xi_2+n\alpha a_2)^2+(n\xi_3+n\alpha a_3)^2+\ldots.$$

Increment of the sum of squares:

$$\begin{aligned} \, (n\xi_1+n\alpha a_1)^2+(n\xi_2+n\alpha a_2)^2+(n\xi_3+n\alpha a_3)^2+\ldots -(n\xi_1)^2-(n\xi_2)^2-(n\xi_3)^2-\ldots \\ =2n^2\alpha a_1\xi_1+n^2\alpha^2 a_1^2+2n^2\alpha a_2\xi_2+n^2\alpha^2 a_2^2+2n^2\alpha a_3\xi_3 +n^2 \alpha^2 a_3^2+\ldots \\ =n^2\alpha(2a_1\xi_1+\alpha a_1^2+2a_2\xi_2+\alpha a_2^2+2a_3\xi_3+\alpha a_3^2+\ldots) \equiv 0 (\mathrm{mod} {n^2\alpha}). \end{aligned}$$

Therefore

$$\operatorname{ncells}(n\cdot \mathrm L+n\alpha\mathbf{a},n^2\alpha,k) = \operatorname{ncells}(n\cdot\mathrm L,n^2\alpha,k).$$

$\quad\square$

Lemma F.6 (on sum of lattices).

$$\mathrm L_1 \cap \mathrm L_2 = \varnothing, \operatorname{zrf}(\mathrm L_1, \alpha, \beta), \operatorname{zrf}(\mathrm L_2, \alpha, \beta) \Rightarrow \operatorname{zrf}(\mathrm L_1 \cup \mathrm L_2, \alpha, \beta).$$

Lemma F.7 (on subtraction of lattices).

$$\mathrm L_1 \cap \mathrm L_2 = \varnothing, \operatorname{zrf}(\mathrm L_1 \cup \mathrm L_2, \alpha, \beta), \operatorname{zrf}(\mathrm L_2, \alpha, \beta) \Rightarrow \operatorname{zrf}(\mathrm L_1, \alpha, \beta).$$

Lemma F.8 (on multiple lattices).

Let $\mathrm L_i \cap \mathrm L_j = \varnothing$ if $i \neq j$, and $\operatorname{ncells}(\mathrm L_i,\alpha,k) = \operatorname{ncells}(\mathrm L_j,\alpha,k)$. Then

$$\operatorname{zrf}(\mathrm L_1, \alpha, \beta) \Rightarrow \operatorname{zrf}(\bigcup\limits_{j=1}^{n} \mathrm L_j, \alpha, n\beta).$$

Lemma F.9 (on periodicity).

Let

$$\forall k \in \mathbb{Z}(P\alpha) {\operatorname{ncells}(\mathrm L, P\alpha, k \oplus P) = \operatorname{ncells}(\mathrm L, P\alpha, k)},$$

where $\oplus$ is addition in the ring $\mathbb{Z}(P\alpha)$. Then

$$\operatorname{zrf}(\mathrm L, P, \alpha \beta) \Rightarrow \operatorname{zrf}(\mathrm L, P\alpha, \beta).$$

Proof.

By Lemma F.1 (on increasing of the momentum ring parameter) and taking into account the periodicity of $\operatorname{ncells}(\mathrm L, P\alpha, k)$,

$$\operatorname{ncells}(\mathrm L, P,k) = \sum\limits_{j=0}^{\alpha-1} \operatorname{ncells}(\mathrm L, P\alpha, k + jP) = \alpha \cdot \operatorname{ncells}(\mathrm L, P\alpha, k),$$

$$\frac{\operatorname{ncells}(\mathrm L, P, k)}{\alpha} = \operatorname{ncells}(\mathrm L, P\alpha, k).$$

If $\operatorname{ncells}(\mathrm L, P, k)$ is a multiple of $\alpha\beta$, then $\operatorname{ncells}(\mathrm L, P\alpha, k)$ is a multiple of $\beta$. $\quad\square$

Lemma F.10 (on permutation of remainders).

Let $\alpha$ and $n$ be coprime, $A$ be the set of the remainders of the division of elements $n\mathbb{Z}(\alpha)+a$ by $\alpha$. Then $A = \mathbb{Z}(\alpha)$.

Appendix G. Proof-2

Theorem on number-theory renormalization (2.8) formulated in terms of the function $\operatorname{zrf}$: for $d\geqslant3$ and any natural $N$, as well as for $d=2$ and $N=2^ m, m \in \mathbb{N}$

$$ \operatorname{zrf}(\operatorname{lattice}^d(N), N, N).$$

(G.1)

The theorem is obvious in the trivial case $N=1$, easily verified for $N=2$, and proved in the Appendix C for odd primes $N$. It remains to prove the theorem for composite $N$.

We will prove the theorem separately for each of the following $3$ cases of composite $N$:

• • $N$ is a power of an odd prime $p$,

• • $N$ is a power of $2$,

• • $N$ is not a prime power.

1. Powers of Odd Primes

Let us prove the theorem (G.1) for the $m$-th power of an odd prime $p$:

$$ N = p^m.$$

(G.2)

Unless otherwise specifically stated, we will assume that $d=3$.

We will prove by induction, assuming that the hypothesis is true for $p^{m-1}$ and for $p^{m-2}$:

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-1}), p^{m-1}, p^{m-1}),$$

(G.3)

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-2}), p^{m-2}, p^{m-2}).$$

(G.4)

The basis of the induction is the proven above validity of the hypothesis for $d=3$ for ${m=0,1}$, i.e. for $N=1$ and $N=p$:

$$ \operatorname{zrf}(\operatorname{lattice}^d(1), 1, 1),$$

(G.5)

$$ \operatorname{zrf}(\operatorname{lattice}^d(p), p, p).$$

(G.6)

Divide the $\operatorname{lattice}^d(N)$ (Fig. 1) into 2 parts: $p \cdot \operatorname{lattice}^d(p^{m-1})$ and $\operatorname{lattice}^d(N ) - p \cdot \operatorname{lattice}^d(p^{m-1})$.

The first part ($p \cdot \operatorname{lattice}^d(p^{m-1})$) consists of all cells of $\operatorname{lattice}^d(N)$ such that each of their coordinates $\xi_i$ is a multiple of $p$ (Fig. 2).

The second part ($\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$) consists of all other cells of $\operatorname{lattice}^d(N)$, i.e. those with at least one of their coordinates $\xi_i$ is not a multiple of $p$ (Fig. 3).

Similarly, we divide the $\operatorname{lattice}^d(p^{m-1})$ into 2 parts: $p \cdot \operatorname{lattice}^d(p^{m-2})$ and $\operatorname{lattice}^d(p^{m-1 }) - p \cdot \operatorname{lattice}^d(p^{m-2})$. (Fig. 1)

We will prove zerofying for each of the 2 parts of $\operatorname{lattice}^d(N)$ separately:

$$ \operatorname{zrf}(p \cdot \operatorname{lattice}^d(p^{m-1}), N, N),$$

(G.7)

$$ \operatorname{zrf}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), N, N),$$

(G.8)

and then, by Lemma F.6 (on sum of lattices), zerofying will be proved for the entire $\operatorname{lattice}^d(N)$ (G.1).

Since the set $p \cdot \operatorname{lattice}^d(p^{m-2})$ is $\operatorname{lattice}^d(p^{m-2})$ stretched by $p$ times, by virtue of the hypothesis for $p^{m-2}$ (G.4), Lemma F.4 (on lattice stretching) and the fact that $N=p^m$ (G.2)

$$ \operatorname{zrf}(p \cdot \operatorname{lattice}^d(p^{m-2}), N, p^{m-2}).$$

(G.9)

Note that the set $p \cdot \operatorname{lattice}^d(p^{m-1})$ can be divided into $p^d$ subsets $p \cdot \operatorname{lattice}^d(p^{m-2}) + p^{ m-1}\mathbf{a}$, where $\mathbf{a} \in \operatorname{lattice}^d(p)$. One of these subsets is $p \cdot \operatorname{lattice}^d(p^{m-2})$, and the rest are obtained from $p \cdot \operatorname{lattice}^d(p^{m-2})$ by such a shift of all its cells, that each coordinate difference is a multiple of $p^{m-1}$ (Fig. 2).

Since $p^{m-1}$ and the coordinates of all cells from $p \cdot \operatorname{lattice}^d(p^{m-2})$ are divisible by $p$, by Lemma F.5 (on lattice shift) for each set ${p \cdot \operatorname{lattice}^d(p^{m-2}) + p^{m-1}\mathbf{a}}$ the value of $\operatorname{ncells}(p \cdot \operatorname{lattice}^d(p^ {m-2}) + p^{m-1}\mathbf{a}, N, k)$ does not depend on $\mathbf{a}$. Therefore, since one of the sets ${p \cdot \operatorname{lattice}^d(p^{m-2}) + p^{m-1}\mathbf{a}}$ is the set $p \cdot \operatorname{lattice}^d(p ^{m-2})$, and there are $p^d$ such sets, due to (G.9) and Lemma F.8 (on multiple lattices)

$$ \operatorname{zrf}(p \cdot \operatorname{lattice}^d(p^{m-1}), N, p^{m+d-2}).$$

(G.10)

Hence, taking into account that $N=p^m$ (G.2), if $d \geqslant 2$, by Lemma F.3 (on decreasing of the energy ring parameter)

$$ \operatorname{zrf}(p \cdot \operatorname{lattice}^d(p^{m-1}), N, N).$$

(G.11)

From (G.9) and by Lemma F.2 (on decreasing of the momentum ring parameter) it follows that

$$ \operatorname{zrf}(p \cdot \operatorname{lattice}^d(p^{m-2}), p^{m-1}, p^{m-2}).$$

(G.12)

From the validity of the hypothesis for $p^{m-1}$ (G.3) and by virtue of the Lemma F.3 (on decreasing of the energy ring parameter), it follows that

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-1}), p^{m-1}, p^{m-2}).$$

(G.13)

Due to (G.12), (G.13) and Lemma F.7 (on subtraction of lattices)

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2}), p^{m-1}, p^{m-2}).$$

(G.14)

Now we divide the set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ into subsets $[\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2})] + p^{m-1}\mathbf{a}$, where $\mathbf {a} \in \operatorname{lattice}^d(p)$ (Fig. 3) in the same way as the set $p \cdot \operatorname{lattice}^d(p^{m-1})$ was divided into sets $p \cdot \operatorname{lattice}^d(p^{m-2}) + p^{m-1}\mathbf{a}$, where $\mathbf{a} \in \operatorname{lattice}^d(p)$.

Similarly, by Lemma F.5 (on lattice shift), for each of these sets the values of $\operatorname{ncells}([\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m- 2})] + p^{m-1}\mathbf{a}, p^{m-1}, k)$ does not depend on $\mathbf{a}$. Therefore, since one of these sets is ${\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2})}$, and there are $p^d$ such sets, due to (G.14) and Lemma F.8 (on multiple lattices)

$$ \operatorname{zrf}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), p^{m-1}, p^{m+d-2}).$$

(G.15)

For $d \geqslant 3$, if we succeed in applying Lemma F.9 (on periodicity), then, taking into account that $N=p^m$ (G.2), zeroing for the second part of the lattice (G.8) will be proven. For $d = 2$ a separate consideration is required.

Now we divide the set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ into disjoint sets of the form

$$\begin{aligned} \, p^{m-1} \cdot lattice(p) + (\nu_1, \xi_2, \xi_3, \ldots),\\ p^{m-1} \cdot lattice_2(p) + (\xi_1, \nu_2, \xi_3, \ldots),\\ p^{m-1} \cdot lattice_3(p) + (\xi_1, \xi_2, \nu_3, \ldots),\\ \ldots\qquad\qquad\qquad \end{aligned}$$

$$\nu_i \in \mathbb{Z}({p^{m-1}}),\quad \nu_i \mod{p} \neq 0.$$

with $p$ cells in each set (Fig. 4) as follows. Let’s take an arbitrary cell $\boldsymbol{\xi}$ from $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$, take from its $d$ coordinates the first not multiple of $p$ (according to the definition of this set, at least one of the coordinates is not a multiple of $p$), and then we will shift this coordinate by multiples of $p^{m-1}$. Collect the resulting $p$ cells into a set. Regardless of which point of this set we start from, we will collect the same set. So, the partition of $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ into subsets is unique.

Suppose we have selected a cell $\boldsymbol{\xi}$, and $\xi_1$ is the first coordinate not a multiple of $p$. We will shift $\xi_1$ by multiples of $p^{m-1}$. We get the set $p^{m-1} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots)$. The elements of the set has the form

$$ (\xi_1, \xi_2, \xi_3, \ldots)= (\nu_1 + kp^{m-1}, \xi_2, \xi_3, \ldots),$$

(G.16)

where $k \in \mathbb{Z}(p)$, $\nu_1 \in \mathbb{Z}({p^{m-1}})$, $\nu_i \mod{p}\ne0$.

The sum of squares of the coordinates (G.16) of cells from this set is

$$ (\nu_1 + kp^{m-1})^2 + \xi_2^2 + \xi_3^2 + \ldots= k^2p^{2m-2} + 2kp^{m-1}\nu_1 + \nu_1^2 + \xi_2^2 + \xi_3^2 + \ldots.$$

(G.17)

Consider the sum (G.17) modulo $N$. If $m \geqslant 2$, then $p^{2m-2}$ is a multiple of $N=p^m$, so the term $k^2p^{2m-2}$ can be dropped:

$$ (2kp^{m-1}\nu_1 + \nu_1^2 + \xi_2^2 + \xi_3^2 + \ldots) \mod{N}.$$

(G.18)

Consider the the remainder of the division of the term $2kp^{m-1}\nu_1$ by $N$ (the other terms of (G.18) are constant):

$$ 2kp^{m-1}\nu_1 \mod{N} = p^{m-1}(2k\nu_1 \mod{p}).$$

(G.19)

Since $\nu_i \mod{p} \neq 0$ and $p$ are odd primes, $2\nu_1$ and $p$ are coprime. Therefore, by Lemma F.10 (on permutation of remainders), the set of possible values of $2k\nu_1 \mod{p}$ coincides with $\mathbb{Z}(p)$. Therefore, the set of possible values (G.19) is the same as $p^{m-1}\mathbb{Z}(p)$.

Therefore, for the set $p^{m-1} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots)$ the function $\operatorname{ncells}(p^{m-1} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots), N,k)$ is periodic in $k$ with period $p^{m-1}$:

$$\begin{gathered} \, \operatorname{ncells}(p^{m-1} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots), N, k\oplus p^{m-1}) \\ = \operatorname{ncells}(p^{m-1} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots), N, k). \end{gathered}$$

(G.20)

The function (G.20) takes the values $0$ and $1$, and it takes the values $1$ only for $k-(\nu_1^2+\xi_2^2+\xi_3^2+\dots)\in p^{m-1}\mathbb{Z}(p)$.

Since the set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ consists entirely of such disjoint sets with property (G.20), and for union of disjoint sets, these functions add up, then for the entire set ${\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})}$ this function is periodic in $k$ with period $p ^{m-1}$, that is

$$\begin{gathered} \, \operatorname{ncells}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), N, k \oplus p^{m-1}) \\ = \operatorname{ncells}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), N, k). \end{gathered}$$

(G.21)

Hence, by virtue of (G.15) and Lemma F.9 (on periodicity)

$$ \operatorname{zrf}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), N, p^{m+d-3}).$$

(G.22)

If $d \geqslant 3$, then, taking into account that $N=p^m$ (G.2), by virtue of Lemma F.3 (on decreasing of the energy ring parameter) it follows from (G.22) that

$$ \operatorname{zrf}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), N, N).$$

(G.23)

Thus, applying Lemma F.6 (on sum of lattices) to (G.11) and (G.23) we have proved the theorem for $N=p^m$, $d\geqslant3$

$$\operatorname{zrf}(\operatorname{lattice}^d(N), N, N).$$

2. Proof for $d\geqslant2$, $N=2^m$

If we apply the proof for powers of primes for the case $d=2$, $p=2$, then it turns out that the proof for the first part of the lattice works, but the proof for the second part of the lattice is valid up to the formula (G.15) before the assumptions $p\not=2$ and $d\geqslant3$ are used. A closer look shows that for $d=p=2$ we need to retreat to the formula (G.10).

Now, as an induction hypothesis, we take the validity of the hypothesis not for the previous two powers ($m-1$, $m-2$), but for three ($m-1$, $m-2$, $m-3$):

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-1}), p^{m-1}, p^{m-1}),$$

(G.24)

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-2}), p^{m-2}, p^{m-2}),$$

(G.25)

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-3}), p^{m-3}, p^{m-3}).$$

(G.26)

Here $d=p=2$, but we do not substitute $2$ for these parameters for the convenience of comparison with the previously analyzed cases.

Since we now use the hypothesis for the previous three powers as the induction hypothesis, we can use the proof using the previous two powers by substituting $p^{m-1}$ instead of $N$ (that is, lowering the required powers of $p$ by $1$). For $d\geqslant2$ from (G.10) by reducing the powers of $p$ by $1$ it follows

$$ \operatorname{zrf}(p \cdot \operatorname{lattice}^d(p^{m-2}), p^{m-1}, p^{m-1}).$$

(G.27)

From the validity of the hypothesis for $p^{m-1}$ (G.24), due to (G.27) and the Lemma F.7 (on subtraction of lattices)

$$ \operatorname{zrf}(\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2}), p^{m-1}, p^{m-1}).$$

(G.28)

In the same way as (G.15) was obtained from (G.14), we obtain (G.29) from (G.24).

Divide the set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ into subsets $[\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2})] + p^{m-1}\mathbf{a}$, where $\mathbf{a} \in \operatorname{lattice}^d(p)$ (Fig. 3).

Similarly, by Lemma F.5 (on lattice shift), for each of these sets the values of $\operatorname{ncells}([\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m- 2})] + p^{m-1}\mathbf{a}, p^{m-1}, k)$ does not depend on $\mathbf{a}$. Therefore, since one of these sets is ${\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2})}$, and there are $p^d$ such sets, due to (G.28) and Lemma F.8 (on multiple lattices)

$$ \operatorname{zrf}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), p^{m-1}, p^{m+d-1}).$$

(G.29)

Now (similarly to derivation of (G.15)) we need to apply the F.9 Lemma (on periodicity) to increase the momentum ring parameter by decrease the energy ring parameter.

For $p=2$, the set of possible values (G.19) no longer coincides with $p^{m-1}\mathbb{Z}(p)$, which was based on the fact that $2\nu_1$ and $p$ are coprime, but this is not true for the case $p=2$.

Therefore, for $p=2$ we will divide into sets of $p$ cells not the set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ as a whole, but each of the $p^d$ subsets $[\operatorname{lattice}^d(p^{m-1})-p\cdot \operatorname{lattice}^d(p^{m-2})]+p^{m-1}\mathbf{a}$, where $\mathbf{a} \in \operatorname{lattice}^d(p)$. So, the coordinate in each of the resulting sets will be shifted by a value that is a multiple of $p^{m-2}$, not $p^{m-1}$.

The set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ is divided into disjoint sets of the form

$$p^{m-2} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots),$$

$$p^{m-2} \cdot \operatorname{lattice}_2(p) + (\xi_1, \nu_2, \xi_3, \ldots),$$

$$p^{m-2} \cdot \operatorname{lattice}_3(p) + (\xi_1, \xi_2, \nu_3, \ldots),$$

$$\ldots$$

$$\nu_i \in \mathbb{Z}({p^{m-2}})+p^{m-1}\mathbb{Z}(p),\qquad \nu_i \mod{p}\ne0.$$

with $p=2$ cells in each.

This partition is performed as follows. Take an arbitrary cell $\boldsymbol{\xi}$ from $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$, select from its $d$ coordinates the first non-multiple of $p$ (according to the definition of this set, at least one of the coordinates is not a multiple of $p$), and then we will shift this coordinate by multiples of $p^{m-2}$, so as not to go beyond a specific set $[\operatorname{lattice}^d(p^{m-1}) - p \cdot \operatorname{lattice}^d(p^{m-2})] + p^{m-1}\mathbf{a}$ (Fig. 5). Collect the resulting $p$ cells into a set. Regardless of which point of this set we start from, we will collect the same set, that is, such a partition of $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ into subsets is unique.

Suppose we have selected a cell $\boldsymbol{\xi}$, and the first coordinate not a multiple of $p$ is $\xi_1$. We will shift $\xi_1$ by multiples of $p^{m-2}$. We get the set $p^{m-2} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots)$ from the cells

$$ (\xi_1, \xi_2, \xi_3, \ldots)=(\nu_1 + kp^{m-2}, \xi_2, \xi_3, \ldots),$$

(G.30)

where $k \in \mathbb{Z}(p)$, $\nu_1 \in \mathbb{Z}({p^{m-2}})+p^{m-1}\mathbb{Z}(p),$ $\nu_1 \mod{p}\ne0$.

The sum of squares of the coordinates (G.30) of cells from this set is

$$(\nu_1 + kp^{m-2})^2 + \xi_2^2 + \xi_3^2 + \ldots =k^2p^{2m-4} + 2kp^{m-2}\nu_1 + \nu_1^2 + \xi_2^2 + \xi_3^2 + \ldots.$$

(G.31)

Consider this sum of squares modulo $N$. If $m \geqslant 4$, then $p^{2m-4}$ is a multiple of $N$, and the term $k^2p^{2m-4}$ can be dropped:

$$ (2kp^{m-2}\nu_1 + \nu_1^2 + \xi_2^2 + \xi_3^2 + \ldots) \mod{N}.$$

(G.32)

Due to the fact that it is possible to pass from the expression (G.31) to the expression (G.32) only when $m \geqslant 4$, for $p=2$ we have to choose as induction basis not $m=0,1,2$, but $m=1,2,3$.

Taking into account that $p=2$, so $\nu_1=1$ (since $\nu_1 \mod{p}\ne0$), we rewrite (G.32) as

$$ (kp^{m-1} + 1 + \xi_2^2 + \xi_3^2 + \ldots) \mod{N}.$$

(G.33)

Consider the remainder of the division of the term $2kp^{m-1}\nu_1$ by $N$ (the other terms of (G.33) are constant):

$$ kp^{m-1}\in p^{m-1}\mathbb{Z}(p).$$

(G.34)

Therefore, for the set $p^{m-2} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots)$ the function $\operatorname{ncells}(p^{m-2} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots), N, k)$ is periodic in $k$ with period $p^{m-1}$:

$$\begin{gathered} \, \operatorname{ncells}(p^{m-2} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots),N, k \oplus p^{m-1}) \\ = \operatorname{ncells}(p^{m-2} \cdot \operatorname{lattice}(p) + (\nu_1, \xi_2, \xi_3, \ldots), N, k). \end{gathered}$$

(G.35)

Since the set $\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})$ consists entirely of such disjoint sets, for each of which this function is periodic with period $p^{m-1 }$, and for union of disjoint sets, these functions add up, then for the entire set ${\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1})}$ this function is periodic in $k$ with period $p ^{m-1}$, that is

$$\begin{gathered} \, \operatorname{ncells}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}),N,k\oplus p^{m-1}) \\ =\operatorname{ncells}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}),N,k). \end{gathered}$$

(G.36)

Further, similarly to the statement (G.22) we obtain

$$ \operatorname{zrf}(\operatorname{lattice}^d(N) - p \cdot \operatorname{lattice}^d(p^{m-1}), N, N).$$

(G.37)

The induction base ($N=2,4,8$) is verified numerically.

3. Composite Numbers that are not Prime Powers

A composite number $N$ can be represented as a product of powers of unequal primes $p_i$ (if $i \neq j$, then $p_i \neq p_j$):

$$ N=\prod\limits_{i=1}^n p_i^{m_i}.$$

(G.38)

Denote

$$ M=\prod\limits_{i=1}^{n-1} p_i^{m_i},$$

(G.39)

$$ K=p_n^{m_n}.$$

(G.40)

That is

$$ N=MK.$$

(G.41)

$M$ and $K$ are coprime.

We will prove by induction. As a step of induction, we deduce the validity of the hypothesis for $N$ from its validity for $M$ and $K$:

$$ \operatorname{zrf}(\operatorname{lattice}^d(M), M, M), \operatorname{zrf}(\operatorname{lattice}^d(K), K, K) \Rightarrow \operatorname{zrf}(\operatorname{lattice}^d(N), N, N).$$

(G.42)

As the basis of the induction, we take the validity of the hypothesis for $p_1^{m_1}$:

$$ \operatorname{zrf}(\operatorname{lattice}^d(p_1^{m_1}), p_1^{m_1}, p_1^{m_1}).$$

(G.43)

Since $p_1^{m_1}$ and $K$ are powers of primes, for $d=3$ the base of induction has already been proved and $\operatorname{zrf}(\operatorname{lattice}^d(K), K, K)$ is proved.

Represent the lattice $\operatorname{lattice}^d(N)$ as $K^d$ pairwise disjoint lattices $K \cdot \operatorname{lattice}^d(M) + \mathbf{a}$, where $\mathbf{a} \in \operatorname{lattice}^d (K)$ (Fig. 6).

We prove that for each lattice $K \cdot \operatorname{lattice}^d(M) + \mathbf{a}$

$$ \operatorname{zrf}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a},N,N)$$

(G.44)

after which, by Lemma F.6 (on sum of lattices), it will be proved that

$$ \operatorname{zrf}(\operatorname{lattice}^d(N),N,N).$$

(G.45)

Each lattice coordinate $\operatorname{lattice}^d(M)$ runs through all values in $\mathbb{Z}(M)$. By Lemma F.10 (on permutation of remainders), each coordinate of the lattice $K \cdot \operatorname{lattice}^d(M) + \mathbf{a}$ modulo $M$ also runs through all values in $\mathbb{ Z}(M)$. That is, the lattices $\operatorname{lattice}^d(M)$ and $K \cdot \operatorname{lattice}^d(M) + \mathbf{a}$ consist of the same cells if their coordinates are taken modulo $M$. Therefore, the validity of the hypothesis for $M$ (G.42) implies

$$ \operatorname{zrf}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, M, M).$$

(G.46)

Consider two cells in $K \cdot \operatorname{lattice}^d(M) + \mathbf{a}$:

$$K\boldsymbol\xi + \mathbf{a},$$

$$K\boldsymbol{\xi^\prime} + \mathbf{a},$$

where $\boldsymbol\xi, \boldsymbol{\xi^\prime} \in \operatorname{lattice}^d(M)$. The difference of the sums of squares of their coordinates:

$$\begin{aligned} \, (K\xi_1+a_1)^2 + (K\xi_2+a_2)^2 + \ldots - (K\xi_1^\prime+a_1)^2 - (K\xi_2^\prime+a_2)^2 - \ldots\\ \nonumber =K^2(\xi_1^2 + \xi_2^2 + \ldots - {\xi_1^\prime}^2 - {\xi_2^\prime}^2 - \ldots) + 2K((\xi_1-\xi_1^\prime)a_1 + (\xi_2-\xi_2^\prime)a_2 + \ldots). \end{aligned}$$

(G.47)

So, the difference between the sums of squared coordinates of two cells from ${K \cdot \operatorname{lattice}^d(M) + \mathbf{a}}$ is a multiple of $K$.

Compare the functions $\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, M, z)$ and $\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, N, y)$.

Each $z\in\mathbb{Z}(M)$ corresponds to $K$ different values of $y\in\mathbb{Z}(N)$ such that $z= y \mod{M}.$

Let $\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, M, z_0)=0$, then $\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, N, y_0)=0$ for all $z_0= y_0 \mod{M}.$

Let $\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, M, z_1)\ne0$, then there exists a non-empty subset of the lattice $K \cdot \operatorname{lattice}^d(M) + \mathbf{a}$ such that for each of its cells the sum of squared coordinates modulo $M$ is equal to $k_1$. That is, in this subset, the sums of squares of cell coordinates differ from each other by multiples of $M$. But we came to the conclusion that these quantities are multiples of $K$ as well. Therefore, they are multiples of the least common multiple of $M$ and $K$. Since $M$ and $K$ are coprime, these quantities are multiples of $N = MK$ (G.41). Since $y_1\in\mathbb{Z}(N)$, among all $K$ distinct $y_1\equiv z_1 \mod{M}$ there is one value $\tilde y_1$ for which

$$ \operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, N, \tilde y_1) =\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, M, z_1),$$

(G.48)

for the remaining $y_1\ne\tilde y_1$ we get $\operatorname{ncells}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, N, y_1)=0$.

Therefore, due to (G.46)

$$ \operatorname{zrf}(K \cdot \operatorname{lattice}^d(M) + \mathbf{a}, N, M),$$

(G.49)

using the lemma F.6 (on sum of lattices) we obtain

$$ \operatorname{zrf}(\operatorname{lattice}^d(N), N, M).$$

(G.50)

By permutation of $M$ and $K$, we similarly prove that

$$ \operatorname{zrf}(M \cdot \operatorname{lattice}^d(K) + \mathbf{a}, N, K),\quad \mathbf{a}\in \operatorname{lattice}^d(M),$$

(G.51)

using the lemma F.6 (on sum of lattices) we obtain

$$ \operatorname{zrf}(lattice^d(N), N, K).$$

(G.52)

Since $M$ and $K$ are coprime and $N = MK$ (G.41), their least common multiple is equal to $N$, so by (G.50) and (G.52) lattice $\operatorname{lattice}^d(N)$ is zerofied modulo $N$:

$$ \operatorname{zrf}(\operatorname{lattice}^d(N),N,N).$$

(G.53)

Conflict of Interest

The authors of this work declare that they have no conflicts of interest.