# Ultra-thin Tunable Optomechanical Metalens

Adeel Afridi  ICFO Institut de Ciencies Fotoniques, Mediterranean Technology
Park, 08860 Castelldefels (Barcelona), Spain  Jan Gieseler  ICFO Institut de
Ciencies Fotoniques, Mediterranean Technology Park, 08860 Castelldefels
(Barcelona), Spain  Nadine Meyer  Nanophotonic Systems Laboratory, Department
of Mechanical and Process Engineering, ETH Zurich, 8092 Zurich, Switzerland
Romain Quidant  Nanophotonic Systems Laboratory, Department of Mechanical and
Process Engineering, ETH Zurich, 8092 Zurich, Switzerland<EMAIL_ADDRESS>
###### keywords:

Optomechanics, Reconfigurable metasurface, Metalens, LaTeX

Nanophotonic Systems Laboratory, Department of Mechanical and Process
Engineering, ETH Zurich, 8092 Zurich, Switzerland Reconfigurable metasurfaces
offer great promises to enhance photonics technology by combining integration
with improved functionalities. Recently, reconfigurability in otherwise static
metasurfaces has been achieved by modifying the electric permittivity of the
meta-atoms themselves or their immediate surrounding. Yet, it remains
challenging to achieve significant and fast tunability without increasing
bulkiness. Here, we demonstrate an ultra-thin tunable metalens whose focal
distance can be changed through optomechanical control with moderate
continuous wave intensities. We achieve fast focal length changes of more than
5% with response time of the order of 10 $\mu$s.

Keywords: Optomechanics, Reconfigurable metasurface, Metalens

A metalens is a two dimensional (2D) metasurface 1, 2, 3 that controls the
amplitude, polarization and phase of the impinging light using engineered sub-
wavelength resonators, also known as meta-atoms (MA), judicially arranged in
periodic or quasi-periodic arrays 4, 2, 5, 6. To date, metalenses are the
center of intense research efforts and have shown great potential to replace
bulky optical elements, sometimes even with superior performance 7. However,
the intrinsic passive nature of metalenses limits their use where active
operation is required, such as adaptive vision and imaging 8, 9, 10, 5, 11,
12, 7, 6. To address this challenge, various approaches have emerged that rely
on changing the optical properties of the meta-atoms themselves or their
surrounding medium 6, 13, 14, 15, 12, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,
26. In particular, promising advances were reported using electro-optic
control 12, 16, 17, 18, temperature-induced effects 19, 20, light-induced
effects 21, 22, 23, 24, 25, phase change media 26 and mechanical actuation 6,
13, 14, 15. Despite this progress, it remains challenging to combine
compactness, fast operation speed and low energy consumption. For instance,
phase change material-based metasurface provides large tunability and
ultrafast switch ON time (up to MHz) but suffer from long switch OFF time 27.
Similarly, electrically controlled liquid crystal (LC) embedded metalenses
operate at low driving voltages ($<$ 10V). However, they require bulky LC
chambers and are polarization dependent due to the birefringence of the LC 28,
12. MEMS-based technologies promise large focal length changes and high
switching speed but require high driving voltages 6, 15. On the other hand,
mechanical stretching requires bulky mechanical arms limiting integration and
response time 14.

Optomechanics 29, 30, 31, 32, 33, 34, where optical forces mediate the
interaction between light and structural mechanics, offers new possibilities
for reconfigurable metasurfaces. When considering suspended meta-atoms
patterned in a thin membrane, the resonantly enhanced electromagnetic forces
at resonance can exceed elastic forces of the material, thereby inducing
mechanical deformation. Already a small deformation alters the phase delay
incurred by the impinging light substantially 35. First proposed theoretically
36, the concept of optomechanical metasurfaces features giant nonlinearity,
optical bistability and asymmetric light transmission. Shortly after,
optomechanically-induced modulation of light transmission through a
metasurface has been experimentally demonstrated 37. Similarly, gold meta-
atoms supported by pairs of free-standing silicon nitride strings exhibit
individual optomechanical plasmonic resonances 38. More recently, an
experimental study of a suspended silicon carbide (SiC) metasurface supporting
multi-mode vibrational resonances was reported 39. These observations suggest
optomechanical control as a promising approach toward compact, fast and low
power tunable metalenses.

In this letter, we design, fabricate and characterize an ultra-thin
optomechanically reconfigurable metalens operating in the near-IR regime. Our
metalens is formed by an ensemble of suspended Huygens’ meta-atoms 40 carved
in a free-standing crystalline silicon membrane, as illustrated in Figure 1.
The metalens is designed to focus probe light at $\lambda_{probe}=1.31\,\mu$m
while being controlled by a pump beam at $\lambda_{pump}=1.55\,\mu$m. Optical
forces experienced by the meta-atoms translate into a mechanical deformation
of the metalens ($\Delta z$) and consequently a change in the focal length
$\Delta f$. Beyond demonstrating the tunability of the focal length by the
pump light, we perform a full characterization of the lens performance
including power dependence, focusing efficiency and time response.

Figure 1: Artistic representation of an optomechanically reconfigurable
metalens. The metalens is designed to focus light at $\lambda_{probe}$. Upon
illumination with a pump beam at $\lambda_{pump}$, the resonantly enhanced
optical forces mechanically deforms the metalens by an amount $\Delta z$
resulting in a change of the focal length ($\Delta f$) at $\lambda_{probe}$.

Combining light focusing with optomechanical control requires careful
multimode engineering at both pump and probe wavelengths. On the one hand, the
set of meta-atoms must cover phase changes from 0 to 2$\pi$ at
$\lambda_{probe}$. On the other hand, an additional mode at $\lambda_{pump}$
is required to induce optical forces causing a sufficiently large mechanical
deformation. To meet these requirements, we selected a rectangular periodic
arrangement of suspended Huygens’ meta-atoms with periodicity $P_{x}$
($P_{y}$) along the $x$ ($y$) axis, in a 200 nm-thick silicon membrane. Each
meta-atom consists of a disk with radius $r$, attached to a transversal
nanobeam of width $w_{2}$ via a short neck of width $w_{1}$ (Figure 2a).
Design optimization used a commercial finite element method simulator (COMSOL
MULTIPHYSICS). Figure 2b displays the transmission T and reflection R for a
specific meta-atom with r = 0.2 $\mu m$, $w_{1}=w_{2}=0.095\,\mu$m,
$P_{x}=P_{y}=1.275\,\mu$m. Under $x$-polarized light, the disk supports
magnetic and electric dipolar resonances around $\lambda_{probe}=$1.31 $\mu$m.
The interaction between the electric and magnetic dipole resonances,
controlled by the disk radius $r$, introduces a phase delay of the incoming
light 40, 41. Additionally, the disk attached to the supporting beam supports
an electric mode around $\lambda_{pump}$, that we use for optomechanical
control.

Next, we optimized an entire set of meta-atoms to provide 0-2$\pi$ phase
change at $\lambda_{pump}$, by changing the disk radius r and beam width
$w_{2}$ simultaneously. The simulated phase imparted ($\phi_{MA}$) by these
meta-atoms at $\lambda_{probe}$ are shown in Figure 2c as a function of r
(yellow solid line with triangular markers), while the respective $w_{2}$
parameter is given in Table 1 in the supplementary information. The
periodicity and the width were kept constant at $P_{x}=P_{y}=1.275\,\mu$m and
$w_{1}=0.095\,\mu$m, respectively.

Figure 2: Geometry and simulation results of the silicon meta-atom. a) Meta-
atom geometrical parameters. b) Simulated transmission T (blue solid line),
and reflection R (red dashed line) of a meta-atom with disk radius
$r=0.2\,\mu$m. The red (blue) shaded region signifies probe (pump) resonance.
c) Simulated phase due to the meta-atoms at probe wavelength
$\lambda_{pump}=1.31\,\mu$m and calculated force experienced by the meta-atoms
at pump wavelength $\lambda_{probe}=1.55\,\mu$m and pump intensity $I_{p}=1$
$\mu$W/$\mu$m2, as a function of disk radius $r$. d) Numerical simulation of
the deformation $\Delta z$ for meta-atoms in a one dimensional array in the
presence of pump light with $I_{p}=110$ $\mu$W/$\mu$m2.

In addition, we used the time averaged Maxwell’s stress tensors to calculate
the optical forces acting on the meta-atoms, as a function of the disc radius
$r$ (see supplementary information for details). Figure 2c shows that the sub-
unit (see Figure S3a in supplementary information) of the meta-atoms
experience anti-parallel forces along the $z$-direction. The disk experiences
a negative force that pulls against the illumination direction. Conversely,
the beam is pushed by the incoming light. This force imbalance across the
meta-atom introduces a tilt of the disk with respect to the beam, while the
total positive force over a finite periodic array bends the whole array
forward. Furthermore, we numerically simulated the deformation $\Delta z$, for
a maximum pump intensity of $I_{p}=110$ $\mu$W/$\mu$m2 for each meta-atom
arranged equidistantly in a 1D array of 100 $\mu$m 36. Figure 2d shows the
deformation $\Delta z$ as a function of disk radius $r$. We observe a maximum
deformation $\Delta z\approx 380\,$nm for the meta-atom with $r=200$ nm.

To calculate the required phase profile $\phi_{req}(x,y)$ in the $xy$-plane,
that focuses light at focal length $f$, we used the hyperboloidal phase
function:

$\phi_{req}(x,y)=\frac{2\pi}{\lambda_{probe}}(\sqrt{x^{2}+y^{2}+f^{2}}-f)$

We fixed the focal distance and diameter of the metalens at $f=300\,\mu$m and
$D=100\,\mu$m, respectively. We discretized the continuous $\phi_{req}(x,y)$
with a suitable meta-atom such that, at a given $x$ and $y$ position,
$|\phi_{MA}-\phi_{req}(x,y)|$ is minimized.

To fabricate our tunable metalens design, we employed top-down electron beam
(E-beam) lithography. As material substrate we used a commercially available
free standing crystalline (100) silicon membrane from Norcada Inc. We span
coated the membrane sample with the AR-P 6200.04 positive photo-resist with a
thickness of 230 nm followed by baking for 1 minute at 150oC. Afterwards,
E-beam exposure was carried out followed by one and a half minutes development
in AR 600-546 developer at room temperature. We then etched the silicon
membrane using HBr chemistry with an inductively coupled plasma (ICP) etcher.
Finally, we stripped off the photo-resist with an oxygen plasma etcher. Figure
3 shows scanning electron microscopy (SEM) images of the final metalens with a
diameter $D=100\,\mu$m and designed focal length $f=300\,\mu$m.

Figure 3: SEM micrographs of an optomechanically reconfigurable metalens.
Fabricated metalens with diameter $D=100\,\mu$m and designed focal length
$f=300\,\mu$m carved on a free standing silicon membrane (Norcada Inc.) of
thickness 200 nm.

We characterized the fabricated metalens using a home-made two color optical
setup (see Figure S1 of the supplementary information). Collimated probe and
pump beams are linearly polarized. The pump beam passes through a combination
of an electro opto modulator (EOM), a polarizing beam splitter and a half wave
plate to control the power and polarization of the pump beam. Afterwards, both
beams were recombined on a 50:50 beam splitter. A 20x objective lens and NA =
0.4 focuses the probe and pump beam to a focal spot of 100 $\mu$m and 60
$\mu$m onto the metalens. A second identical objective mounted on a
piezoelectric stage collects the light after the metalens. A band pass filter
with central frequency of 1.31 $\mu$m blocks the pump while allowing the probe
to reach a near infrared (NIR) camera and a photo-diode. We acquired an image
stack of 2D intensity maps along the principle optical axis with a step size
of $\sim 1\,\mu$m by moving the collection objective with the piezo stage.
First, we characterized the lens response in the absence of the pump beam.
Figure 4a and b show the 2D intensity map without pump ($I_{p}=0$
$\mu$W/$\mu$m2) in the $xz$-plane (longitudinal) and at the focal plane
(transversal), respectively. The measured focal length $f$ is 328 $\mu$m $\pm$
0.3 $\mu$m. Subsequently, we switched on the pump beam (intensity $I_{p}$ =
110 $\mu W/\mu m^{2}$) and observed a reduction of the focal length by $\Delta
f=-20\,\mu$m. The 2D intensity maps (longitudinal and transversal) are
displayed in Figure 4c and d. For further comparison, we plot the focus
profile ($z$-cut) and 1D focal spot profile with and without pump laser in
Figure 4e and f, respectively. It is noteworthy, that the focusing quality and
the full width half maximum (FWHM) value is well-preserved upon the
introduction of the pump laser. Furthermore, the involved pump intensities are
compatible with LED arrays, thus eliminating the need for a bulky laser
source.

Although our metalens was optimized for $x$-polarized pump light, we also
characterized the metalens under $y$-polarized pump light and find a focal
change $\Delta f=-6\,\mu$m for the pump intensity $I_{p}=110\,\mu$W/$\mu$m2
(see Figure S2 in supplementary information). This is consistent with
numerical simulations shown in Figure S3a, which reveal that under
$y$-polarization the effect of the pump on the beam becomes negligible and, in
contrast to the $x$-polarized case, the force on the disk is along the
propagation direction.

Figure 4: Optical characterization of the focal length with pump light ON and
OFF. a) and c) 2D intensity maps in the $xz$-plane with the pump intensities
$I_{p}=0\,\mu$W/$\mu$m2 and $I_{p}=110\,\mu$W/$\mu$m2, respectively. b) and d)
2D intensity profile of the focal spot in its respective focal plane ($xy$)
under the pump intensities $I_{p}=0\,\mu$W/$\mu$m2 and
$I_{p}=110\,\mu$W/$\mu$m2, respectively. e) Axial optical intensity
distribution for the two pump intensities. f) 1D cut of the focal spot across
the $y$-axis for the two pump intensities (blue solid line for
$I_{p}=0\,\mu$W/$\mu$m2 and yellow dashed line for $I_{p}=110\,\mu$W/$\mu$m2.
The FWHM is highlighted by the blue shaded area. All intensity profiles are
normalized to the maximum probe intensity for the given pump intensity.

When sweeping the pump intensity $I_{p}$ from 0 to 110 $\mu$W/$\mu$m2, we
observed a linear power dependence of the focal distance (Figure 5a). Another
important parameter that characterizes the performance of a metalens is the
focusing efficiency (FE), defined as the relative amount of light passing
through an aperture of 3 times the full width half maximum (FWHM) of the focal
spot 42. We experimentally measured the FE for increasing $I_{p}$ and observe
a reduction from 78% (for $I_{p}=0\,\mu$W/$\mu$m2) to 68% (for
$I_{p}=110\,\mu$W/$\mu$m2) as depicted in Figure 5a (blue circular dots).

Finally, we are interested in the switching dynamics of our optomechanically
reconfigurable metalens. To measure its time response, we modulated the pump
beam through the EOM with a square wave of frequency 0.5 Hz and amplitude 110
$\mu$W/$\mu$m2. The switching on rise time (10-90%) and switching off fall
time (90-10%) of the EOM is limited to 14 $\mu$s. Figure 5b and c shows the
time response of both the pump (red solid curve) and the metalens (blue solid
line). We fit a Sigmoid function to the data (black solid line for the
metalens and grey dashed line for the pump) to obtain the rise and fall time.
Our measurements give a rise time $\Delta t_{rise}=17\,\mu$s (10% to 90%) and
fall time $\Delta t_{fall}=18\,\mu$s (90% to 10%) for the metalens highlighted
by yellow shaded area in Figure 5b and c, respectively. From these values, we
rule out that thermal effects dominate in the tunability of our lens. This
conclusion is corroborated by the switching time calculated by Zhang et al.
36.

Figure 5: Focal length reconfigurability and switching dynamics of the
metalens. a) Focal length change and focusing efficiency (FE) as a function of
pump intensity $I_{p}$. Red dots represent experimental data while the black
solid line is a linear fit. The blue circular dots shows the focusing
efficiency (FE). b) Time response (rise time) of the metalens (blue solid
line) switching from $f$ = 320 $\mu$m to 300 $\mu$m under modulation of the
pump beam from 0 to 110 $\mu$W/$\mu$ m2 (red solid line). c) Time response
(fall time) of the metalens (blue solid line) switching from $f$= 300 $\mu$m
to 320 $\mu$m under modulation of the pump beam from 110 to 0 $\mu$W/$\mu$ m2
(red solid line). In b) and c), black solid line and grey dashed lines are
Sigmoid fits to the metalens and pump response, respectively. The 90% to 10%
rise/fall time for the metalens is higlighted by the yellow shaded area.

In conclusion, we proposed and realized an ultra-thin varifocal metalens
actuated by optomechanical control, which combines significant tunability with
10 $\mu$s response time. While the system was optimized to operate in the NIR
region of the spectrum, extension to the visible frequency range could be
achieved through an appropriate adjustment of the meta-atoms parameters and
choice of their constitutive material. We foresee that future development of
optomechanical integrated reconfigurable metalenses and other metasurface-
based functionalities would greatly benefit from further mode engineering,
directly leveraging on the latest advances in dielectric metasurfaces.

## Acknowledgments

Adeel Afridi acknowledges financial support from the Marie Skłodowska-Curie
Co-funding Programme (COFUND-DP, H2020-MSCA-COFUND-2014, GA No. 665884).

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# On multiplicative functions
which are additive on positive cubes

Poo-Sung Park  Department of Mathematics Education, Kyungnam University,
Changwon, 51767, Republic of Korea<EMAIL_ADDRESS>
###### Abstract.

Let $k\geq 3$. If a multiplicative function $f$ satisfies

$f(a_{1}^{3}+a_{2}^{3}+\cdots+a_{k}^{3})=f(a_{1}^{3})+f(a_{2}^{3})+\cdots+f(a_{k}^{3})$

for all $a_{1},a_{2},\ldots,a_{k}\in\mathbb{N}$, then $f$ is the identity
function. The set of positive cubes is said to be a $k$-additive uniqueness
set for multiplicative functions. But, the condition for $k=2$ can be
satisfied by infinitely many multiplicative functions.

Besides, if $k\geq 3$ and a multiplicative function $g$ satisfies

$g(a_{1}^{3}+a_{2}^{3}+\cdots+a_{k}^{3})=g(a_{1})^{3}+g(a_{2})^{3}+\cdots+g(a_{k})^{3}$

for all $a_{1},a_{2},\ldots,a_{k}\in\mathbb{N}$, then $g$ is the identity
function. However, when $k=2$, there exist three different types of
multiplicative functions.

###### Key words and phrases:

$k$-additive uniqueness, multiplicative function, positive cube

###### 2020 Mathematics Subject Classification:

Primary 11A25; Secondary 11N64

This research was supported by Basic Science Research Program through the
National Research Foundation of Korea (NRF) funded by the Ministry of
Education (NRF-2021R1A2C1092930).

##  1\. Introduction

The additive uniqueness, briefly AU, was initiated by Spiro in 1992 [12]. She
call a set $E\subset\mathbb{N}$ a AU set for a set $S$ of arithmetic functions
if the condition

$f\in S\text{ and }f(a+b)=f(a)+f(b)\text{ for all }a,b\in E$

determines $f$ uniquely. She showed that the set of prime numbers is AU for
multiplicative functions $f$ with $f(p_{0})\neq 0$ for some $p_{0}$ prime.

Two decades later, Fang [7] and Dubickas and Šarka [6] showed that the set of
prime numbers with condition

$f(p_{1}+p_{2}+\cdots+p_{k})=f(p_{1})+f(p_{2})+\cdots+f(p_{k})\text{ for fixed
}k\geq 3$

also determines the multiplicative function $f$ uniquely if $f(p_{0})\neq 0$
for some $p_{0}$ prime. Henceforth, let us call such a set $k$-AU.

In 1996 Chung [4] studied condition

$f(m^{2}+n^{2})=f(m^{2})+f(n^{2})\text{ for all }m,n\in\mathbb{N}$

and showed that the set of positive squares is not AU for multiplicative
functions.

The author, two decades later, showed that the set of positive squares is
$k$-AU for $k\geq 3$ [8]. This is the first case for $k$-AU but not $\ell$-AU
with $\ell<k$.

In this paper we show that the set of positive cubes is a $k$-additive
uniqueness set, like the set of positive squares, for multiplicative functions
when $k\geq 3$.

######  Theorem 1.1.

The set of positive cubes is $k$-AU for $k\geq 3$. That is, if a
multiplicative function $f_{k}$ satisfies the condition

$f_{k}(a_{1}^{3}+a_{2}^{3}+\cdots+a_{k}^{3})=f_{k}(a_{1}^{3})+f_{k}(a_{2}^{3})+\cdots+f_{k}(a_{k}^{3})$

for all $a_{1},a_{2},\ldots,a_{k}\in\mathbb{N}$, then $f_{k}$ is the identity
function.

But, $f_{2}$ is not determined uniquely. So, we proceed as follows:

  * §2

Classify the multiplicative functions $f_{2}$.

  * §3

Show that $f_{3}$ is the identity function.

  * §4

Show that $f_{4}$ is the identity function.

  * §5

Show that $f_{k}$ is the identity function with $k\geq 5$.

In §6 we consider a variant problem: For fixed $k\geq 2$, if a multiplicative
function $g_{k}$ satisfies

$g_{k}(a_{1}^{3}+a_{2}^{3}+\cdots+a_{k}^{3})=g_{k}(a_{1})^{3}+g_{k}(a_{2})^{3}+\cdots+g_{k}(a_{k})^{3}$

for all $a_{1},a_{2},\ldots,a_{k}\in\mathbb{N}$, then is $g_{k}$ determined
uniquely? If we set $K(x_{1},\ldots,x_{k})=x_{1}^{3}+\cdots+x_{k}^{3}$, this
problem is to find a multiplicative function $g_{k}$ commutable with $K$.

In 2014 Bašić [1] showed that various multiplicative function $g$ satisfies

$g(a^{2}+b^{2})=g(a)^{2}+g(b)^{2}$

and the author, a couple of years later, showed that a multiplicative function
$g$ is the identity function if $k\geq 3$ is fixed and

$g(a_{1}^{2}+a_{2}+\cdots+a_{k}^{2})=g(a_{1})^{2}+g(a_{2})^{2}+\cdots+g(a_{k})^{2}$

holds for all $a_{1},a_{2},\ldots,a_{k}\in\mathbb{N}$ [9].

In this section, we show that $g_{k}$ is uniquely determined for $k\geq 3$ and
$g_{2}$ is one of three functions.

##  2\. non $2$-additive uniqueness

The set of positive cubes is not (2-)AU for multiplicative functions. That is,
although a multiplicative function $f_{2}$ satisfies

$f_{2}(a^{3}+b^{3})=f_{2}(a^{3})+f_{2}(b^{3})$

for all $a,b\in\mathbb{N}$, $f_{2}$ may not be the identity function.

Since $n^{3}\equiv 0,1,-1\pmod{9}$, the sum of two cubes cannot be congruent
to $\pm 3$ modulo $9$. If we want to determine $f_{2}(3)$, we should use
$f_{2}(3m)$ with $3\nmid m$. But, since $3m\equiv\pm 3\pmod{9}$, $3m$ cannot
be written as a sum of two positive cubes. So, for example, if we define a
multiplicative function $f:\mathbb{N}\to\mathbb{C}$ by

$f(n)=\begin{cases}0&\text{ if }3\mid n\text{ and }3^{2}\nmid n,\\\ n&\text{
otherwise,}\end{cases}$

this function satisfies the additive condition but is not the identity
function.

In general, we can characterize the function $f_{2}$ completely.

######  Theorem 2.1.

If a multiplicative function $f_{2}$ satisfies

$f_{2}(a^{3}+b^{3})=f_{2}(a^{3})+f_{2}(b^{3})$

for $a,b\in\mathbb{N}$, then $f_{2}(n)=n$ unless $3\,|\,n$ and $3^{2}\nmid n$.
We can set the value $f_{2}(3)$ arbitrarily.

We will prove through the following process:

  1. (1)

Determine $f_{2}(n)$ for some $n$’s.

  2. (2)

Show that $f_{2}$ fixes $2^{r}$, $3^{r}$, and $7^{r}$ except for $3$.

  3. (3)

Assume that $f_{2}$ fixes $q^{r}$ for all primes $q<p$ except for $f_{2}(3)$
to use induction on prime $p$.

  4. (4)

Show that $f_{2}$ fixes $p$, $p^{2}-p+1$, $p^{3}$, $p^{2}+3$, $(p^{2}+1)^{3}$,
$p^{4}+3$.

  5. (5)

Show that $f_{2}(p^{r})=p^{r}$.

######  Lemma 2.2.

$f_{2}(n)=n$ for $n=1,2,2^{2},2^{3},3^{2},3^{3},5,5^{3},7,7^{3},13,19$.

###### Proof.

The following Mathematica code yields the required results except for $n=4$,
where $x_{n}=f_{2}(n)$.

    Solve[x1 == 1 && x2 == x1 + x1 && x9 == x8 + x1
     && x5 x7 == x27 + x8 && x5 x13 == x64 + x1
     && x8 x9 == x64 + x8 && x7 x13 == x64 + x27
     && x2 x9 x7 == x125 + x1 && x8 x19 == x125 + x27
     && x27 x7 == x125 + x64 && x8 x43 == x343 + x1
     && x27 x13 == x343 + x8 && x27 x19 == x512 + x1
     && x13 x43 == x343 + x8 x27 && x9 x5 x19 == x512 + x343,
    {x1, x2, x5, x7, x8, x9, x13, x19, x27, x43, x64, x125, x343, x512}]

Then, $f_{2}(4)=4$ can be derived from
$f_{2}(4)\,f_{2}(7)=f_{2}(1^{3}+3^{3})=1+f_{2}(27)$. ∎

######  Lemma 2.3.

$f_{2}(p^{r})=p^{r}$ for $p\in\\{2,3,7\\}$ and $r\geq 1$ except for
$f_{2}(3)$.

###### Proof.

We use induction on $r$. Assume that $2^{r}$ is fixed by $f_{2}$ for all
$r\leq 3m$. It is clear that $f_{2}(2^{3m+1})=2^{3m+1}$ by

$f_{2}(2^{3m+1})=f(2^{3m}+2^{3m})=f(2^{3m})+f(2^{3m}).$

We obtain $f_{2}(2^{3m+2})=2^{3m+2}$ by calculating $f_{2}(7\cdot 2^{3m+2})$
in two ways:

$f_{2}(7)\,f_{2}(2^{3m+2})=7f_{2}(2^{3m+2})\\\ $

and

$\displaystyle f_{2}(7\cdot 2^{2}\cdot 2^{3m})$
$\displaystyle=f_{2}\bigl{(}(1^{3}+3^{3})\,2^{3m}\bigr{)}=f_{2}(2^{3m})+f_{2}(3^{3})\,f_{2}(2^{3m})$
$\displaystyle=2^{3m}+3^{3}\cdot 2^{3m}=7\cdot 2^{3m+2}.$

We can also obtain that $f_{2}(2^{3(m+1)})=2^{3(m+1)}$ by calculating
$f_{2}(9\cdot 2^{3m})$ in two ways:

$f_{2}(9)\,f_{2}(2^{3m})=9\cdot 2^{3m}$

and

$f_{2}\bigl{(}(1^{3}+2^{3})\,2^{3m}\bigr{)}=f_{2}(2^{3m})+f_{2}(2^{3(m+1)})=2^{3m}+f_{2}(2^{3(m+1)}).$

Assume that $3^{r}$ is fixed by $f_{2}$ for $2\leq r\leq 3m$. Note that
$f_{2}(11^{3})=11^{3}$ by

$f_{2}(11^{3}+5^{3})=f_{2}(11^{3})+f_{2}(5^{3})=f(2^{4})\,f(7)\,f(13).$

Now, $f_{2}(3^{3m+1})=3^{3m+1}$ from

$\displaystyle f_{2}(7\cdot 19\cdot 3^{3m+1})$ $\displaystyle=f_{2}(7\cdot
19\cdot 3^{4}\cdot
3^{3(m-1)})=f_{2}\bigl{(}(5^{3}+22^{3})\,3^{3(m-1)}\bigr{)}.$

It is easy to show that $f_{2}(3^{3m+2})=3^{3m+2}$ by

$\displaystyle f_{2}(3^{3m+2})$ $\displaystyle=f_{2}(9\cdot
3^{3m})=f_{2}\bigl{(}(1^{3}+2^{3})\,3^{3m}\bigr{)}.$

We can deduce that $f_{2}(3^{3(m+1)})=3^{3(m+1)}$ by calculating

$\displaystyle f_{2}(7\cdot 3^{3(m+1)})$ $\displaystyle=f_{2}(7\cdot
3^{3}\cdot 3^{3m})=f_{2}\bigl{(}(4^{3}+5^{3})\,3^{3m}\bigr{)}.$

Now, we can show that $f_{2}$ fixes $7^{3m+1}$, $7^{3m+2}$, and $7^{3(m+1)}$
with $m\geq 0$ by calculating

$\displaystyle f_{2}(2^{2}\cdot 7^{3m+1})$ $\displaystyle=f_{2}(2^{2}\cdot
7\cdot 7^{3m})=f_{2}\bigl{(}(1^{3}+3^{3})\,7^{3m}\bigr{)},$ $\displaystyle
f_{2}(11\cdot 7^{3m+2})$ $\displaystyle=f_{2}(11\cdot 7^{2}\cdot
7^{3m})=f_{2}\bigl{(}(3^{3}+8^{3})\,7^{3m}\bigr{)},$ $\displaystyle
f_{2}(3^{3}\cdot 13\cdot 7^{3m})$
$\displaystyle=f_{2}\bigl{(}(2^{3}+7^{3})7^{3m}\bigr{)}.$

This fact about $7^{r}$ will be used in Lemma 2.6. ∎

Now, we use induction on prime $p\geq 5$ under the assumption

$f_{2}(q^{r})=q^{r}\text{ for all primes }q<p\text{ except for }f_{2}(3).$

######  Lemma 2.4.

Under the assumption (2), $f_{2}(p)=p$ for all prime $p$.

###### Proof.

Let $p=2r+1\geq 5$. Note that $f_{2}$ fixes $(r+1)^{3}$ and $r+2$, since their
all prime divisors are less than $p$ and $3\nmid(r+2)$. Now we can say that

$\displaystyle f_{2}\bigl{(}(r+1)^{3}+1^{3}\bigr{)}$
$\displaystyle=f_{2}\bigl{(}(r+1)^{3}\bigr{)}+f_{2}(1^{3})=(r+1)^{3}+1$
$\displaystyle=f_{2}\\!\bigl{(}(r+2)(r^{2}+r+1)\bigr{)}=f_{2}(r+2)\,f_{2}(r^{2}+r+1)$
$\displaystyle=(r+2)\,f_{2}(r^{2}+r+1),$

since $\gcd(r+2,r^{2}+r+1)=\gcd(r+2,3)=1$. Hence, $f_{2}$ fixes $r^{2}+r+1$.

On the other hand,

$\displaystyle f_{2}\bigl{(}(r+1)^{3}+r^{3}\bigr{)}$
$\displaystyle=f_{2}\bigl{(}(r+1)^{3}+r^{3}\bigr{)}+f_{2}(r^{3})=(r+1)^{3}+r^{3}$
$\displaystyle=f_{2}\bigl{(}(2r+1)(r^{2}+r+1)\bigr{)}=f_{2}(2r+1)\,f_{2}(r^{2}+r+1)$
$\displaystyle=f_{2}(p)\,(r^{2}+r+1)$

holds, since all prime divisors of $r^{3}$ is less than $p$ and
$\gcd(2r+1,r^{2}+r+1)=1$. Hence, $f_{2}(p)=p$. ∎

######  Lemma 2.5.

Under the assumption (2), $f_{2}$ fixes $p^{3}$ and $p^{2}-p+1$ or
$\frac{p^{2}-p+1}{3}$.

###### Proof.

Let $p=3^{r}s-1$ with $r\geq 1$ and $3\nmid s$. Note that

$\displaystyle f_{2}(p^{3}+1^{3})$
$\displaystyle=f_{2}(3^{r+1})\,f_{2}\\!\left(\frac{p+1}{3^{r}}\right)f_{2}\\!\left(\frac{p^{2}-p+1}{3}\right),$
$\displaystyle f_{2}\bigl{(}p^{3}+(p-1)^{3}\bigr{)}$
$\displaystyle=f_{2}(3^{2})\,f_{2}\\!\left(\frac{2p-1}{3}\right)f_{2}\\!\left(\frac{p^{2}-p+1}{3}\right),$

since $\gcd(p+1,p^{2}-p+1)=\gcd(2p-1,p^{2}-p+1)=3$ and $\frac{p+1}{3^{r}}$,
$\frac{p^{2}-p+1}{3}$, $\frac{2p-1}{3}$ are indivisible by $3$.

By induction hypothesis, we can say $(p-1)^{3}$, $\frac{p+1}{3^{r}}$ and
$\frac{2p-1}{3}$ are fixed by $f_{2}$ since their prime divisors are less than
$p$. Then, we can deduce that

$f_{2}(p^{3})=p^{3}\qquad\text{ and }\qquad
f_{2}\\!\left(\frac{p^{2}-p+1}{3}\right)=\frac{p^{2}-p+1}{3}.$

If $p=3^{r}s+1$ with $r\geq 1$ and $3\nmid s$, then consider

$\displaystyle f_{2}\bigl{(}(p+1)^{3}+1^{3}\bigr{)}$
$\displaystyle=f_{2}(3^{2})\,f_{2}\\!\left(\frac{2p+1}{3}\right)f_{2}\\!\left(\frac{p^{2}+p+1}{3}\right),$
$\displaystyle f_{2}\bigl{(}(p+1)^{3}+p^{3}\bigr{)}$
$\displaystyle=f_{2}(3^{2})\,f_{2}\\!\left(\frac{p+2}{3}\right)f_{2}\\!\left(\frac{p^{2}+p+1}{3}\right),$

where $\gcd(2p+1,p^{2}+p+1)=\gcd(p+2,p^{2}+p+1)=3$ and $\frac{2p+1}{3}$,
$\frac{p^{2}+p+1}{3}$, $\frac{p+2}{3}$ are indivisible by $3$.

Since $p+1$ is even and $\frac{p+1}{2}<p$, we can say that $(p+1)^{3}$ is
fixed by $f_{2}$. Then,

$f_{2}(p^{3})=p^{3}\qquad\text{ and }\qquad
f_{2}\\!\left(\frac{p^{2}+p+1}{3}\right)=\frac{p^{2}+p+1}{3}$

by induction hypothesis. This yields $f_{2}(p^{2}-p+1)=p^{2}-p+1$, since
$\gcd(p+1,p^{2}-p+1)=1$ and

$\displaystyle f_{2}(p^{3}+1^{3})$
$\displaystyle=f_{2}(p+1)\,f_{2}(p^{2}-p+1).$

We can conclude that $f_{2}$ fixes $\frac{p^{2}-p+1}{3}$ if
$p\equiv-1\pmod{3}$ and fixes $p^{2}-p+1$ otherwise. ∎

######  Lemma 2.6.

Under the assumption (2), $f_{2}$ fixes $p^{2}+3$, $(p^{2}+1)^{3}$, and
$p^{4}+3$.

###### Proof.

Note that $\left(\frac{p+1}{2}\right)^{3}$ and
$\left(\frac{p-1}{2}\right)^{3}$ are fixed by $f_{2}$ since $\frac{p+1}{2}$
and $\frac{p-1}{2}$ are less than $p$. Thus, we can conclude that
$f_{2}(p^{2}+3)=p^{2}+3$ from

$f_{2}\\!\left(\left(\frac{p+1}{2}\right)^{3}+\left(\frac{p-1}{2}\right)^{3}\right)=f_{2}(p)\,f_{2}\\!\left(\frac{p^{2}+3}{4}\right)=p\,f_{2}\\!\left(\frac{p^{2}+3}{4}\right).$

Next, consider

$\left(\frac{p^{2}+1}{2}\right)^{3}+\left(\frac{p-1}{2}\right)^{3}=p\cdot\frac{p+1}{2}\cdot(p^{2}-p+1)\cdot\frac{p^{2}+3}{4},$

where

$\displaystyle\gcd\\!\left(p,\frac{p+1}{2}\right)=\gcd\left(p,p^{2}-p+1\right)=\gcd\\!\left(p,\frac{p^{2}+3}{4}\right)=1,$
$\displaystyle\gcd\\!\left(\frac{p+1}{2},p^{2}-p+1\right)=\begin{cases}1&\text{if
}p\equiv 1\pmod{3},\\\ 3&\text{if }p\equiv-1\pmod{3},\end{cases}$
$\displaystyle\gcd\\!\left(\frac{p+1}{2},\frac{p^{2}+3}{4}\right)=1,$
$\displaystyle\gcd\\!\left(p^{2}-p+1,\frac{p^{2}+3}{4}\right)=\begin{cases}1&\text{if
}p\not\equiv 5\pmod{7},\\\ 7&\text{if }p\equiv 5\pmod{7}.\end{cases}$

We can verify that $(p^{2}+1)^{3}$ is fixed by $f_{2}$ for each case.

For example, assume $p=3^{r}s-1$ with $3\nmid s$ and $p\equiv 5\pmod{7}$.
Then, since $3\mid(p^{2}-p+1)$ but $3^{2}\nmid(p^{2}-p+1)$,

$\displaystyle
f_{2}\\!\left(\left(\frac{p^{2}+1}{2}\right)^{3}+\left(\frac{p-1}{2}\right)^{3}\right)=f_{2}\\!\left(\left(\frac{p^{2}+1}{2}\right)^{3}\right)+\left(\frac{p-1}{2}\right)^{3}$
$\displaystyle=f_{2}(3^{r+1})\,f_{2}(7^{a+b})\,f_{2}(p)\,f_{2}\\!\left(\frac{p+1}{2\cdot
3^{r}}\right)\,f_{2}\\!\left(\frac{p^{2}-p+1}{3\cdot
7^{a}}\right)\,f_{2}\\!\left(\frac{p^{2}+3}{4\cdot 7^{b}}\right)$

with $a,b\geq 1$ and all terms on RHS are fixed by $f_{2}$. Thus,
$f_{2}\bigl{(}(p^{2}+1)^{3}\bigr{)}=(p^{2}+1)^{3}$.

Finally, note that

$\displaystyle f_{2}\\!\left(\left(\frac{p^{2}+1}{2}\right)^{3}+1^{3}\right)$
$\displaystyle=\left(\frac{p^{2}+1}{2}\right)^{3}+1^{3}.$

On the other hand,

$\displaystyle f_{2}\\!\left(\left(\frac{p^{2}+1}{2}\right)^{3}+1^{3}\right)$
$\displaystyle=f_{2}(2)\,f_{2}\\!\left(\frac{p^{2}+3}{4}\right)f_{2}\\!\left(\frac{p^{4}+3}{4}\right)$

since $\gcd\left(\frac{p^{2}+3}{4},\frac{p^{4}+3}{4}\right)=1$.

Then, we can say that $f_{2}$ fixes $p^{4}+3$. ∎

Now we are ready to prove Theorem 2.1. It is enough to show that
$f_{2}(p^{r})=p^{r}$ for all prime $p\geq 5$ and $r\in\mathbb{N}$. We use
induction on $p$.

It is proved in Lemmas 2.4 and 2.5 that $f_{2}(p)=p$ and $f_{2}(p^{3})=p^{3}$.
Suppose that $f_{2}(p^{3m})=p^{3m}$ for some $m\geq 1$. To show
$f_{2}(p^{3m+1})=p^{3m+1}$ we use the equality

$\displaystyle
f_{2}\\!\left(p^{3m}\left(\frac{p+1}{2}\right)^{3}+p^{3m}\left(\frac{p-1}{2}\right)^{3}\right)=f_{2}(p^{3m+1})\,f_{2}\\!\left(\frac{p^{2}+3}{4}\right),$

which was slightly modified from that used in Lemma 2.6. Since $(p\pm 1)^{3}$
and $p^{2}+3$ are fixed, we obtain $f_{2}(p^{3m+1})=p^{3m+1}$.

Similarly, $f_{2}(p^{3m+2})=p^{3m+2}$ from

$\displaystyle
f_{2}\\!\left(p^{3m}\left(\frac{p^{2}-1}{2}\right)^{3}+p^{3m}\left(\frac{p^{2}+1}{2}\right)^{3}\right)=f_{2}(p^{3m+2})\,f_{2}\\!\left(\frac{p^{4}+3}{4}\right),$

where $\frac{p-1}{2}<\frac{p+1}{2}<p$ and $f_{2}$ fixes $(p^{2}+1)^{3}$ and
$p^{4}+3$. This also works for $m=0$. Thus, $f_{2}(p^{2})=p^{2}$.

It can be verified that $f_{2}(p^{3(m+1)})=p^{3(m+1)}$ from

$\displaystyle f_{2}\bigl{(}p^{3m}(p^{3}+1)\bigr{)}$
$\displaystyle=f_{2}(p^{3(m+1)}+p^{3m})$
$\displaystyle=f_{2}(p^{3(m+1)})+f_{2}(p^{3m})=f_{2}(p^{3(m+1)})+p^{3m}$
$\displaystyle=f_{2}(p^{3m})\,f_{2}(p^{3}+1)=p^{3m}(p^{3}+1).$

Thus, proof is completed.

##  3\. $3$-additive uniqueness

In the previous section, the set of positive cubes is not 2-AU. But, in this
section, we show that the set of positive cubes is 3-AU. The key idea is
algebraic representations of rational numbers as sums of three cubes of
rational numbers.

Let $f_{3}$ be a multiplicative function satisfying the condition

$f_{3}(a^{3}+b^{3}+c^{3})=f_{3}(a^{3})+f_{3}(b^{3})+f_{3}(c^{3}).$

######  Lemma 3.1.

$f_{3}(n)=n$ for $n=1,2,2^{3},3,3^{2},3^{3},5,5^{3},7,7^{3},11,17,23$.

###### Proof.

It is clear that $f_{3}(3)=3$.

If we set $y_{n}=f_{3}(n)$, then the following Mathematica code yields the
required results.

    Solve[y1==1 && y3==3
     && y2 y5 == y1 + y1 + y8 && y17 == y1 + y8 + y8
     && y4 y9 = y8 + y8 + y27 && y43 = y8 + y27 + y27
     && y5 y11 == y1 + y27 + y27 && y2 y31 == y8 + y27 + y27
     && y2 y3 y11 == y1 + y1 + y64 && y4 y23 == y1 + y27 + y64
     && y9 y11 == y8 + y27 + y64 && y3 y43 == y1 + y64 + y64
     && y8 y17 == y8 + y64 + y64 && y9 y17 == y8 + y64 + y64
     && y5 y31 == y27 + y64 + y64 && y8 y27 == y27 + y64 + y125
     && y3 y5 y23 == y1 + y1 + y343 && y2 y27 y7 == y27 + y64 + y125,
    {y1, y2, y3, y5, y7, y8, y9, y11, y17, y23, y27, y31, y43, y64,
     y125, y343}]

∎

######  Lemma 3.2.

$f_{3}$ fixes $2^{r}$ and $3^{r}$ with $r\geq 1$.

###### Proof.

We use induction. Assume that $f_{2}$ fixes $2^{r}$ for all $r<3m$.

We can say that $f_{3}(2^{3m+1})=2^{3m+1}$ by

$\displaystyle f_{3}(23\cdot 2^{3m+1})$
$\displaystyle=f_{3}(23)\,f_{3}(2^{3m+1})=23f_{3}(2^{3m+1})$
$\displaystyle=f_{3}(23\cdot 16\cdot
2^{3(m-1)})=f_{3}\bigl{(}(3^{3}+5^{3}+6^{3})2^{3(m-1)}\bigr{)}$
$\displaystyle=f_{3}(3^{3})\,f_{3}(2^{3(m-1)})+f_{3}(5^{3})\,f_{3}(2^{3(m-1)})+f_{3}(3^{3})\,f_{3}(2^{3m}).$

Next, $f_{3}(2^{3m+2})=2^{3m+2}$ by

$\displaystyle f_{3}(5\cdot 2^{3m+2})$
$\displaystyle=f_{3}(5)\,f_{3}(2^{3m+2})=5f_{3}(2^{3m+2})$
$\displaystyle=f_{3}(5\cdot 2^{5}\cdot
2^{3(m-1)})=f_{3}\bigl{(}(2^{3}+3^{3}+5^{3})2^{3(m-1)}\bigr{)}$
$\displaystyle=f_{3}(2^{3m})+f_{3}(3^{3})\,f_{3}(2^{3(m-1)})+f_{3}(5^{3})\,f_{3}(2^{3(m-1)}).$

Finally, $f_{3}(2^{3(m+1)})=2^{3(m+1)}$ from

$\displaystyle f_{3}(5\cdot 2^{3m+1})$
$\displaystyle=f_{3}(5)\,f_{3}(2^{3m+1})=5f_{3}(2^{3m+1})$
$\displaystyle=f_{3}(10\cdot
2^{3m})=f_{3}\bigl{(}(1^{3}+1^{3}+2^{3})2^{3m}\bigr{)}$
$\displaystyle=f_{3}(2^{3m})+f_{3}(2^{3m})+f_{3}(2^{3(m+1)}).$

Now, assume that $f_{3}$ fixes upto $3^{3m}$. We can say that
$f_{3}(3^{3m+1})=3^{3m+1}$ by

$\displaystyle f_{3}(3^{3m+1})$
$\displaystyle=f_{3}(3^{3m}+3^{3m}+3^{3m})=3f_{3}(3^{3m}).$

Let us consider $f_{3}(3^{3m+2})$. Note that $f_{3}$ fixes $2^{6}$ and
$5^{3}$. Then, $f_{3}(3^{3m+2})=3^{3m+2}$ by

$\displaystyle f_{3}(2^{4}\cdot 3^{3m+2})$
$\displaystyle=f_{3}(2^{4})\,f_{3}(3^{3m+2})=2^{4}\cdot f_{3}(3^{3m+2})$
$\displaystyle=f_{3}(2^{4}\cdot 3^{5}\cdot
3^{3(m-1)})=f_{3}\bigl{(}(1^{3}+8^{3}+15^{3})\,3^{3(m-1)}\bigr{)}$
$\displaystyle=f_{3}(3^{3(m-1)})+f_{3}(2^{6})\,f_{3}(3^{3(m-1)})+f_{3}(5^{3})\,f_{3}(3^{3m}).$

Finally, $f_{3}(3^{3(m+1)})=3^{3(m+1)}$ by

$\displaystyle f_{3}(4\cdot 3^{3m+2})$
$\displaystyle=f_{3}(4)\,f_{3}(3^{3m+2})=4f_{3}(3^{3m+2})$
$\displaystyle=f_{3}(4\cdot 9\cdot
3^{3m})=f_{3}\bigl{(}(1^{3}+2^{3}+3^{3})\,3^{3m}\bigr{)}$
$\displaystyle=f_{3}(3^{3m})+f_{3}(2^{3})\,f_{3}(3^{3m})+f_{3}(3^{3(m+1)}).$

∎

######  Lemma 3.3.

$f_{3}$ fixes $n^{3}$.

###### Proof.

It is enough to show that $f_{3}(p^{3r})=p^{3r}$ for prime $p\geq 5$ and
$r\geq 1$. It was verified in Lemma 3.1 that $f_{3}$ fixes $5^{3}$ and
$7^{3}$.

Let $n=p^{r}=6m\pm 1$ with $m\geq 2$. We can say $f_{3}(n^{3})=n^{3}$
inductively by

$m^{3}+(5m)^{3}+(5m\pm 2)^{3}=(2m\mp 1)^{3}+(3m\pm 2)^{3}+(6m\pm 1)^{3}.$

∎

Now, we prove Theorem 1.1 for $k=3$. By the above results, we may show only
$f_{3}(p^{r})=p^{r}$ for prime $p\geq 5$ and $r\in\mathbb{N}$.

In 1923 Richmond proved that every positive rational number, say $R$, can be
written as a sum of positive cubes of rational numbers [10]. In especial, he
showed

$\displaystyle R\bigl{(}6\lambda^{2}(\lambda^{3}+3R)^{2}\bigr{)}^{3}$
$\displaystyle=\bigl{(}-(\lambda^{3}+3R)(\lambda^{6}-30\lambda^{3}R+9R^{2})\bigr{)}^{3}$
$\displaystyle+\bigl{(}\lambda^{9}+45\lambda^{6}R-81\lambda^{3}R^{2}+27R^{3}\bigr{)}^{3}$
$\displaystyle+\bigl{(}36\lambda^{3}(3R-\lambda^{3})R\bigr{)}^{3}$

and each term is positive when

$\frac{1-\sqrt{4/3}\sin 20^{\circ}}{1+\sqrt{4/3}\sin 20^{\circ}}\times
3R=.43376\dots\times 3R<\lambda^{3}<3R.$

We can check the representation by the Mathematica code:

    Factor[( -(lambda^3 + 3 R) (lambda^6 - 30 lambda^3 R + 9 R^2) )^3
    + ( lambda^9 + 45 lambda^6 R - 81 lambda^3 R^2 + 27 R^3 )^3
    + ( 36 lambda^3 (3R-lambda^3) R )^3]

Let $R$ be a prime $p>23$. Then the length of interval

$(\sqrt[3]{.43376\dots\times 3p},\sqrt[3]{3p})$

is greater than $1$. So the interval contains at least one integer $\lambda$.

But, since there exist no integer $t$ such that

$.43376\dots\times 3p=1.30128\dots\times p<(pt)^{3}<3p,$

we can choose $\lambda$ relatively prime to $p$ in the interval. As for other
primes $\leq 23$ which are not proved to be fixed, we can choose a suitable
$\lambda$ as follows:

$\displaystyle p=13$ $\displaystyle:\lambda=3\in(2.56707\dots,3.39121\dots)$
$\displaystyle p=19$ $\displaystyle:\lambda=3\in(2.91323\dots,3.8485\dots)$

Then, $\gcd(p,6\lambda^{2}(\lambda^{3}+3p)^{2})=1$ and

$f_{3}\bigl{(}p\bigl{(}6\lambda^{2}(\lambda^{3}+3p)^{2}\bigr{)}^{3}\bigr{)}=f_{3}(p)\,\bigl{(}6\lambda^{2}(\lambda^{3}+3p)^{2}\bigr{)}^{3}$

by Lemma 3.3. Thus, $f_{3}(p)=p$.

For example, if we let $p=29$ and choose $\lambda=4$ from the interval
$(3.3542,4.43105)$, then

$29\cdot 2188896^{3}=6646265^{3}+1906183^{3}+1536768^{3}$

and thus $f_{3}(29)=29$.

If $R=p^{2}\geq 25$, then the interval

$(\sqrt[3]{.43376\dots\times 3p^{2}},\sqrt[3]{3p^{2}})$

contains an integer $\lambda$ relatively prime to $p^{2}$. So,
$f_{3}(p^{2})=p^{2}$ by the similar way.

If $R=p^{3}$, then $f_{3}(p^{3})=p^{3}$ by Lemma 3.3. If $R=p^{r}$ with $p\geq
5$ and $r\geq 4$, then the length of the inteval

$(\sqrt[3]{.43376\dots\times 3p^{r}},\sqrt[3]{3p^{r}})$

is greater than $2$. So we can find $\lambda$ relatively prime to $p^{r}$.
Therefore, we can conclude that $f_{3}(p^{r})=p^{r}$.

##  4\. $4$-additive uniqueness

Let $f_{4}$ be a multiplicative function satisfying

$f_{4}(a^{3}+b^{3}+c^{3}+d^{4})=f_{4}(a^{3})+f_{4}(b^{3})+f_{4}(c^{3})+f_{4}(d^{3}).$

Trivially, $f_{4}(1)=1$ and $f_{4}(4)=4$.

We show that $f_{4}(n)=n$ through the some steps.

  1. (1)

$f_{4}(2^{r})=2^{r}$ for all positive integers $r$

  2. (2)

$f_{4}(n^{3})=n^{3}$ for all positive integers $n$

  3. (3)

$f_{4}(p^{r})=p^{r}$ for all primes $p$

######  Lemma 4.1.

$f_{4}(2^{r})=2^{r}$.

###### Proof.

If we set $z_{n}=f_{4}(n)$, then the following Mathematica code yields

$f_{4}(n)=n\text{ for }n=1,2,2^{2},2^{3},3,3^{2},3^{3},\text{ and }5.$

    Solve[z1 == 1 && z4 == 4 && z11 == z8 + z1 + z1 + z1
     && z2 z9 == z8 + z8 + z1 + z1 && z25 == z8 + z8 + z8 + z1
     && z2 z3 z5 == z27 + z1 + z1 + z1 && z37 == z27 + z8 + z1 + z1
     && z4 z11 == z27 + z8 + z8 + z1 && z8 z7 == z27 + z27 + z1 + z1
     && z9 z7 == z27 + z27 + z8 + z1 && z2 z5 z7 == z27 + z27 + z8 + z8
     && z2 z37 == z64 + z8 + z1 + z1 && z8 z11 == z64 + z8 + z8 + z8
     && z4 z25 == z64 + z27 + z8 + z1,
     {z1, z2, z3, z4, z5, z7, z8, z9, z11, z25, z27, z37, z64}]

We use induction. We obtain $f_{4}(2^{3m+1})=2^{3m+1}$ is from

$\displaystyle f_{4}(3\cdot 5\cdot 2^{3m+1})$ $\displaystyle=15\cdot
f_{4}(2^{3m+1})$ $\displaystyle=f_{4}(30\cdot
2^{3m})=f\bigl{(}(1^{3}+1^{3}+1^{3}+3^{3})\,2^{3m}\bigr{)}$
$\displaystyle=30\cdot f_{4}(2^{3m})$

Trivially, $f_{4}(2^{3m+2})=2^{3m+2}$ by

$\displaystyle f_{4}(2^{3m+2})$
$\displaystyle=f_{4}(2^{3m}+2^{3m}+2^{3m}+2^{3m}).$

Next, $f_{4}(2^{3(m+1)})=2^{3(m+1)}$ from

$\displaystyle f_{4}(9\cdot 2^{3m+1})$
$\displaystyle=f_{4}(9)\,f_{4}(2^{3m+1})=9\cdot 2^{3m+1}$
$\displaystyle=f_{4}(9\cdot 2\cdot
2^{3m})=f\bigl{(}(1^{3}+1^{3}+2^{3}+2^{3})\,2^{3m}\bigr{)}$
$\displaystyle=f_{4}(2^{3m})+f_{4}(2^{3m})+f_{4}(2^{3(m+1)})+f_{4}(2^{3(m+1)})$
$\displaystyle=2\cdot 2^{3m}+2\cdot f_{4}(2^{3(m+1)}).$

∎

######  Lemma 4.2.

$f_{4}(n^{3})=n^{3}$.

###### Proof.

We have $f_{4}(2^{r})=2^{r}$ and $f_{4}(3^{3})=3^{3}$. For odd integer
$n=2s+1$ with $s\geq 2$ we use induction and the following equality:

$\displaystyle(2s+1)^{3}+(2s+1)^{3}+(s+2)^{3}+(s-1)^{3}$
$\displaystyle=(s+1)^{3}+s^{3}+\bigl{(}2(s+1)\bigr{)}^{3}+(2s)^{3}.$

∎

By the above Lemma, if $n$ can be written as a sum of four positive cubes,
then $f_{4}(n)=n$.

Now, we are ready to prove $f_{4}(n)=n$ for arbitrary positive integer $n$. We
need a deep result about sums of four positive cubes. It is a well-known
conjecture that all sufficiently large positive natural numbers may be written
as the sum of four positive cubes. But, it is known that

$\frac{\\#\mathcal{E}_{4}(N)}{N}\to 0\quad\text{ as }N\to\infty,$

where $\mathcal{E}_{4}(N)$ is the set of numbers $\leq N$ whose elements
cannot be represented as sums of four positive cubes [5, 13, 2, 3].

Consider $f_{4}(p^{r})$ for $p$ prime. Denote the set of positive integers
$\leq N$ by $[N]$ and the subset of $[N]$ whose elements are not divisible by
$p$ by $[N]_{p}$. Then, every element of $[N]_{p}\setminus\mathcal{E}_{4}(N)$
can be written as a sum of four positive cubes.

Note that

$\displaystyle\frac{\\#\bigl{(}p^{r}([N]_{p}\cap\Sigma^{3}_{4})\bigr{)}}{\\#[p^{r}N]}$
$\displaystyle=\frac{\\#\bigl{(}p^{r}([N]_{p}\setminus\mathcal{E}_{4}(N))\bigr{)}}{\\#[p^{r}N]}$
$\displaystyle=\frac{\\#\bigl{(}[N]_{p}\setminus\mathcal{E}_{4}(N)\bigr{)}}{p^{r}N}\to\frac{1}{p^{r}}\bigl{(}1-\frac{1}{p}\bigr{)}>0$

as $N\to\infty$, where $\Sigma^{3}_{4}$ denotes the set of sums of $4$
positive cubes. Thus, there exists a positive integer $M$ such that $p\nmid
M$, $M\in\Sigma^{3}_{4}$ and $p^{r}M\in\Sigma^{3}_{4}$.

Then, $f_{4}(p^{r})=p^{r}$ since

$\displaystyle p^{r}M=f_{4}(p^{r}M)=f_{4}(p^{r})\,f_{4}(M)=f_{4}(p^{r})M.$

We are done.

##  5\. $k$-additive uniqueness

Let $k\geq 5$. Suppose a multiplicative function $f_{k}$ satisfies

$f_{k}(a_{1}^{3}+a_{2}^{3}+\dots+a_{k}^{3})=f_{k}(a_{1}^{3})+f_{k}(a_{2}^{3})+\dots+f_{k}(a_{k}^{3})$

for all positive integers $a_{1},a_{2},\dots,a_{k}$.

######  Lemma 5.1.

Either $f_{k}(n^{3})=n^{3}$ for all $n\in\mathbb{N}$.

###### Proof.

Note that

$\displaystyle f_{k}\bigl{(}1^{3}+3^{3}\cdot 4^{3}+(k-3)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}9^{3}+2^{3}\cdot 5^{3}+(k-3)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+5^{3}+5^{3}+(k-3)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}2^{3}+3^{3}+2^{3}\cdot 3^{3}+(k-3)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+2^{3}+2^{3}\cdot 5^{3}+(k-3)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}4^{3}+2^{3}\cdot 3^{3}+9^{3}+(k-3)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+1^{3}+1^{3}+2^{3}\cdot
3^{3}+(k-4)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}3^{3}+4^{3}+4^{3}+4^{3}+(k-4)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+1^{3}+2^{3}+8^{3}+(k-4)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}3^{3}+3^{3}+5^{3}+7^{3}+(k-4)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+4^{3}+5^{3}+8^{3}+(k-4)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}2^{3}+2^{3}+7^{3}+7^{3}+(k-4)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+1^{3}+3^{3}+9^{3}+(k-4)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}2^{3}+4^{3}+7^{3}+7^{3}+(k-4)1^{3}\bigr{)},$
$\displaystyle f_{k}\bigl{(}1^{3}+1^{3}+3^{3}+4^{3}+4^{3}+(k-5)1^{3}\bigr{)}$
$\displaystyle=f_{k}\bigl{(}2^{3}+2^{3}+2^{3}+2^{3}+5^{3}+(k-5)1^{3}\bigr{)}.$

Then, the following Mathematica code yields

$\displaystyle t_{n}=f_{k}(n^{3})=1\quad\text{ for }n=2,3,4,5,7,8,9$ or
$\displaystyle t_{n}=f_{k}(n^{3})=n^{3}\quad\text{ for }n=2,3,4,5,7,8,9.$

    ΨSolve[1 + t3 t4 == t9 + t2 t5 && 1 + t5 + t5 == t2 + t3 + t2 t3
    Ψ && 1 + t2 + t2 t5 == t4 + t2 t3 + t9 && 3 + t2 t3 == t3 + 3 t4
    Ψ && 2 + t2 + t8 == 2 t3 + t5 + t7 && 1 + t4 + t5 + t8 == 2 t2 + 2 t7
    Ψ && 2 + t3 + t9 == t2 + t4 + 2 t7 && 2 + t3 + 2 t4 == 4 t2 + t5,
    Ψ{t2, t3, t4, t5, t7, t8, t9}]

Now, we can say that either $f_{k}(n^{3})=1$ or $f_{k}(n^{3})=n^{3}$ for
$n\leq 10$.

We use induction on two cases. Assume that $n\geq 11$ and $f_{k}(m)=1$ for
$m\leq n$. Then, $f_{k}(n^{3})=1$ inductively by

$\displaystyle
f_{k}\bigl{(}n^{3}+(n-3)^{3}+(n-3)^{3}+\overbrace{2^{3}+\cdots+2^{3}}^{(k-3)\text{
summands}}\,\bigr{)}$
$\displaystyle=f_{k}\bigl{(}(n-1)^{3}+(n-1)^{3}+(n-4)^{3}+\underbrace{2^{3}+\cdots+2^{3}}_{(k-5)\text{
summands}}+1^{3}+3^{3}\bigr{)}.$

Note that $6^{3}=3^{3}+4^{3}+5^{3}$ and $7^{3}=1^{3}+1^{3}+5^{3}+6^{3}$. If
$N^{3}=a_{1}^{3}+a_{2}^{3}+\dots+a_{r}^{3}$ with $a_{i}\geq 1$, then

$\displaystyle(6N)^{3}$
$\displaystyle=(6a_{1})^{3}+(6a_{2})^{3}+\cdots+(6a_{r-1})^{3}+6^{3}a_{r}^{3}$
$\displaystyle=(6a_{1})^{3}+(6a_{2})^{3}+\cdots+(6a_{r-1})^{3}+(3a_{r})^{3}+(4a_{r})^{3}+(5a_{r})^{3}$
$\displaystyle(7N)^{3}$
$\displaystyle=(7a_{1})^{3}+(7a_{2})^{3}+\cdots+(7a_{r-1})^{3}+7^{3}a_{r}^{3}$
$\displaystyle=(7a_{1})^{3}+(7a_{2})^{3}+\cdots+(7a_{r-1})^{3}+a_{r}^{3}+a_{r}^{3}+(5a_{r})^{3}+(6a_{r})^{3}.$

Thus, multiplying $6^{3}$ or $7^{3}$ repeatedly, we can find a large cubic
number $N^{3}$ which is a sum of $k$ positive cubes
$a_{1}^{3},a_{2}^{3},\dots,a_{k}^{3}$ starting with any cubic number.

However, then, we obtain a contradiction:

$1=f_{k}(N^{3})=f_{k}(a_{1}^{3})+f_{k}(a_{2}^{3})+\dots+f_{k}(a_{k}^{3})=1+1+\dots+1=k.$

So we can exclude the case $f(n^{3})=1$.

By the equality used in the previous case, we can obtain $f_{k}(n^{3})=n^{3}$
for all $n$ inductively. ∎

Now, we prove Theorem 1.1 for $k\geq 5$. We may consider $f_{k}(p^{r})$. If
$p^{r}$ can be written as a sum of $k$ positive cubes, then $f(p^{r})=p^{r}$
trivially. But, if $p^{r}$ is not the sum of $k$ positive cubes, we need some
tricks.

As mentioned in §4, almost all positive integers can be written as sums of
four positive cubes. If $n$ is the sum of four positive cubes, then $n+(k-4)$
is a sum of $k$ positive cubes. So, almost all positive numbers can be written
as sums of $k$ positive cubes. In particular, all positive integers but
finitely many can be written as sums of positive $k$ cubes when $k\geq 9$,
because all sufficiently large integers can be written as sums of nonnegative
seven cubes [11] and $1072$ can be written as sums of $r$ positive cubes with
$2\leq r\leq 8$. Thus, we can say

$\frac{\\#\mathcal{E}_{k}(N)}{N}\to 0\quad\text{ as }\quad N\to\infty,$

where $\mathcal{E}_{k}(N)$ denotes the set of positive integers $\leq N$ which
cannot be written as sums of $k$ positive cubes for $k\geq 5$. Let
$\Sigma^{3}_{k}$ be the set of sums of $k$ positive cubes. Then, since

$\displaystyle\frac{\\#\bigl{(}p^{r}([N]_{p}\cap\Sigma^{3}_{k})\bigr{)}}{\\#[p^{r}N]}$
$\displaystyle=\frac{\\#\bigl{(}p^{r}([N]_{p}\setminus\mathcal{E}_{k}(N))\bigr{)}}{\\#[p^{r}N]}$
$\displaystyle=\frac{\\#\bigl{(}[N]_{p}\setminus\mathcal{E}_{k}(N)\bigr{)}}{p^{r}N}\to\frac{1}{p^{r}}\bigl{(}1-\frac{1}{p}\bigr{)}>0$

as $N\to\infty$, there exists $M$ such that $p\nmid M$, $M\in\Sigma^{3}_{k}$
and $p^{r}M\in\Sigma^{3}_{k}$.

Thus, $f_{k}(p^{r})=p^{r}$ by

$\displaystyle p^{r}M=f_{k}(p^{r}M)=f_{k}(p^{r})\,f_{k}(M)=f_{k}(p^{r})M.$

The proof is completed.

##  6\. Commutative conditions with sums of cubes

As a variant problem, we may consider multiplicative functions $g_{k}$
satisfying

$g_{k}(x_{1}^{3}+x_{2}^{3}+\cdots+x_{k}^{3})=g_{k}(x_{1})^{3}+g_{k}(x_{2})^{3}+\cdots+g_{k}(x_{k})^{3}$

for fixed $k\geq 2$. We can solve this problem in the similar way.

######  Theorem 6.1.

If a multiplicative function $g_{2}$ satisfies

$g_{2}(a^{3}+b^{3})=g_{2}(a)^{3}+g_{2}(b)^{3}$

for $a,b\in\mathbb{N}$, then $g_{2}(n)=n$ unless $3\,|\,n$ and $3^{2}\nmid n$.
Besides, $g_{2}(3)=3,\frac{3(-1+\sqrt{-3})}{2}$, or
$\frac{3(-1-\sqrt{-3})}{2}$.

###### Proof.

We obtain

$g_{2}(n)=n\text{ for }n=1,2,4,5,7,8,9,13,19$

and $g_{2}(3)^{3}=27$ by using the following Mathematica code.

    Solve[g1 == 1 && g2 == g1^3 + g1^3 && g9 == g2^3 + g1^3
      && g4 g7 == g3^3 + g1^3 && g5 g7 == g3^3 + g2^3
      && g5 g13 == g4^3 + g1^3 && g8 g9 == g4^3 + g2^3
      && g7 g13 == g4^3 + g3^3 && g2 g9 g7 == g5^3 + g1^3
      && g7 g19 == g5^3 + g2^3 && g8 g19 == g5^3 + g3^3,
    {g1, g2, g3, g4, g5, g7, g8, g9, g13, g19}]

Note that $g_{2}(27)=27$ from $g_{2}(2)\,g_{2}(27)=g_{2}(3)^{3}+g_{2}(3)^{3}$.
So, $g_{2}(3)^{3}=g_{2}(3^{3})$, but we cannot conclude $g_{2}(3)=3$. We can
show that $g_{2}$ fixes powers of $2$ and $3$ except for $g_{2}(3)$ by the
same method as Lemma 2.3. But, we cannot use

$g_{2}(9\cdot
2^{3m})=g_{2}(2^{3m}+2^{3(m+1)})=g_{2}(2^{m})^{3}+g_{2}(2^{m+1})^{3}$

to show $g_{2}(2^{3(m+1)})=2^{3(m+1)}$. Instead, we use

$\displaystyle g_{2}(9\cdot 2^{3(m+1)})$ $\displaystyle=g_{2}(9\cdot 8\cdot
2^{3m})=g_{2}\bigl{(}(2^{3}+4^{3})\,2^{3m}\bigr{)}$
$\displaystyle=g_{2}(2^{m+1})^{3}+g_{2}(2^{m+2})^{3}.$

Next, $g_{2}$ fixes $p$ and $p^{2}-p+1$ or $\frac{p^{2}-p+1}{3}$ for prime
$p\geq 5$ as in Lemmas 2.4 and 2.5. Here, we cannot say whether $g_{2}$ fixes
$p^{3}$ or not, because the condition yields $g_{2}(p)^{3}=p^{3}$ not for
$g_{2}(p^{3})$.

It can be derived similarly that

$g_{2}(p^{2}+3)=p^{2}+3,\qquad g_{2}(p^{2}+1)^{3}=(p^{2}+1)^{3},\qquad
g_{2}(p^{4}+3)=p^{4}+3$

as in Lemma 2.6. Then, we can conclude that $g_{2}(p^{r})=p^{r}$ for $r=3m+1$
and $r=3m+2$ by calculating inductively

$\displaystyle
g_{2}\\!\left(p^{3m}\left(\frac{p+1}{2}\right)^{3}+p^{3m}\left(\frac{p-1}{2}\right)^{3}\right)=g_{2}(p^{3m+1})\,g_{2}\\!\left(\frac{p^{2}+3}{4}\right),$
$\displaystyle
g_{2}\\!\left(p^{3m}\left(\frac{p^{2}-1}{2}\right)^{3}+p^{3m}\left(\frac{p^{2}+1}{2}\right)^{3}\right)=g_{2}(p^{3m+2})\,g_{2}\\!\left(\frac{p^{4}+3}{4}\right).$

As for $g_{2}(p^{3m})=p^{3m}$, we use induction on the equality

$\displaystyle g_{2}(p^{3m}+p^{3m})$
$\displaystyle=g_{2}(p^{m})^{3}+g_{2}(p^{m})^{3}$
$\displaystyle=g_{2}(p^{3m})\,g_{2}(2)=2g_{2}(p^{3m}).$

∎

######  Theorem 6.2.

Fix $k\geq 3$. If a multiplicative function $g_{k}$ satisfies

$g_{k}(a_{1}^{3}+a_{2}^{3}+\dots+a_{k}^{3})=g_{k}(a_{1})^{3}+g_{k}(a_{2})^{3}+\dots+g_{k}(a_{k})^{3}$

for all $a_{1},a_{2},\dots,a_{k}\in\mathbb{N}$, then $g_{k}$ is the identity
function.

The proof is similar to that of Theorem 1.1. But, we use $g_{k}(n)^{3}=n^{3}$
instead of $g_{k}(n^{3})=n^{3}$.

For $k=3$, we obtain $g_{3}(n)=n$ for $n=1,2,3,4,5,9,11,43$ by the Mathematica
code, if we set $h_{n}=g_{3}(n)$:

    Solve[h1 == 1 && h3 == h1^3 + h1^3 + h1^3
     && h2 h5 == h2^3 + h1^3 + h1^3 && h4 h9 == h3^3 + h2^3 + h1^3
     && h43 == h3^3 + h2^3 + h2^3 && h5 h11 == h3^3 + h3^3 + h1^3
     && h2 h3 h11 == h4^3 + h1^3 + h1^3 && h9 h11 == h4^3 + h3^3 + h2^3
     && h3 h43 == h4^3 + h4^3 + h1^3,
    {h1, h2, h3, h4, h5, h9, h11, h43}]

It can be shown similarly that $g_{3}$ fixes $2^{3m+1}$ and $2^{3m+2}$. But,
we cannot use $g_{3}(5\cdot 2^{3m+1})$ as in the proof of Lemma 3.2 to show
that $2^{3(m+1)}$ is fixed. So, instead, we use

$\displaystyle g_{3}(3\cdot 2^{3(m+1)})$
$\displaystyle=g_{3}\bigl{(}(2^{3}+2^{3}+2^{3})\,2^{3m}\bigr{)}$
$\displaystyle=g_{3}(2^{m+1})^{3}+g_{3}(2^{m+1})^{3}+g_{3}(2^{m+1})^{3}.$

By the same reason, we use

$g_{3}(8\cdot 3^{3(m+1)})=g_{3}(6^{3}\cdot
3^{3m})=g_{3}\bigl{(}(3^{3}+4^{3}+5^{3})\,3^{3m}\bigr{)}$

to show $3^{m+1}$ is fixed. This method can be also applied to Lemma 3.2.

Now, we can say that $g_{3}(n)^{3}=n^{3}$ for all $n\in\mathbb{N}$ by the same
method of Lemma 3.3. But, we are not yet sure $g_{3}(n^{3})=n^{3}$. So, we are
not sure that $\bigl{(}6\lambda^{2}(\lambda^{3}+3p^{r})^{2}\bigr{)}^{3}$ is
fixed by $g_{3}$ in the LHS of Richmond’s representation

$g_{3}\bigl{(}p^{r}\bigl{(}6\lambda^{2}(\lambda^{3}+3p^{r})^{2}\bigr{)}^{3}\bigr{)}=g_{3}(p^{r})\,g_{3}\bigl{(}\bigl{(}6\lambda^{2}(\lambda^{3}+3p^{r})^{2}\bigr{)}^{3}\bigr{)}.$

However, $g_{3}$ fixes $(6N)^{3}$, since

$\displaystyle g_{3}\bigl{(}(6N)^{3}\bigr{)}$
$\displaystyle=g_{3}\bigl{(}(3N)^{3}+(4N)^{3}+(5N)^{3}\bigr{)}$
$\displaystyle=g_{3}(3N)^{3}+g_{3}(4N)^{3}+g_{3}(5N)^{3}$
$\displaystyle=(3N)^{3}+(4N)^{3}+(5N)^{3}=(6N)^{3}.$

So, $g_{3}(p^{r})=p^{r}$ for all prime $p$ and $r\geq 1$.

For $k=4$, we obtain $g_{4}(n)=n$ for $n=1,2,3,4,5,7,9,11,37$, if we set
$s_{n}=g_{4}(n)$:

    Solve[s1 == 1 && s4 == s1^3 + s1^3 + s1^3 + s1^3
     && s11 == s2^3 + s1^3 + s1^3 + s1^3
     && s2 s9 == s2^3 + s2^3 + s1^3 + s1^3
     && s2 s3 s5 == s3^3 + s1^3 + s1^3 + s1^3
     && s37 == s3^3 + s2^3 + s1^3 + s1^3
     && s4 s11 == s3^3 + s2^3 + s2^3 + s1^3
     && s9 s7 == s3^3 + s3^3 + s2^3 + s1^3
     && s2 s5 s7 == s3^3 + s3^3 + s2^3 + s2^3
     && s2 s37 == s4^3 + s2^3 + s1^3 + s1^3,
    {s1, s2, s3, s4, s5, s7, s9, s11, s37}]

Then, we can show that $g_{4}$ fixes $2^{3m+1}$ and $2^{3m+2}$ by the similar
ways as in Lemma 4.1. Also, $g_{4}(2^{3(m+1)})=2^{3(m+1)}$ can be derived by

$\displaystyle g_{4}(7\cdot 2^{3(m+1)})$ $\displaystyle=g_{4}(7\cdot
2^{3}\cdot 2^{3m})=g_{4}\bigl{(}(1^{3}+1^{3}+3^{3}+3^{3})\,2^{3m}\bigr{)}$
$\displaystyle=g_{4}(2^{m})^{3}+g_{4}(2^{m})^{3}+g_{4}(3\cdot
2^{m})^{3}+g_{4}(3\cdot 2^{m})^{3}.$

Thus, we can conclude that $g_{4}(p^{r})=p^{r}$ by the similar way as in §4.

The proof for $k\geq 5$ is almost identical to §5. If we set
$t_{n}=g_{k}(n)^{3}$, $t_{n}=1$ or $t_{n}=n^{3}$ for $n\leq 10$ as in Lemma
5.1. Thus, we can obtain $g_{k}(n)^{3}=1$ or $g_{k}(n)^{3}=n^{3}$ for all $n$.
But, the former case is impossible by $g_{k}(k)=k$.

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