# Temperature dependence of magnetic susceptibility of nuclear matter: lowest
order constrained variational calculations

M. Bigdeli1,3 111E-mail: m<EMAIL_ADDRESS>G.H. Bordbar 2,3222E-mail:
<EMAIL_ADDRESS>and Z. Rezaei 2 1Department of Physics, Zanjan
University, P.O. Box 45195-313, Zanjan, Iran333Permanent address
2Department of Physics, Shiraz University, Shiraz 71454, Iran444Permanent
address
3Research Institute for Astronomy and Astrophysics of Maragha,
P.O. Box 55134-441, Maragha, Iran

###### Abstract

In this paper we study the magnetic susceptibility and other thermodynamic
properties of the polarized nuclear matter at finite temperature using the
lowest order constrained variational (LOCV) method employing the $AV_{18}$
potential. Our results show a monotonic behavior for the magnetic
susceptibility which indicates that the spontaneous transition to the
ferromagnetic phase does not occur for this system.

###### pacs:

21.65.-f, 26.60.-c, 64.70.-p

##  I INTRODUCTION

The magnetic susceptibility is one of the most important magnetic properties
of the dense matter, its behavior specifies whether the spontaneous phase
transition to a ferromagnetic state occurs. This transition in nuclear matter
could have important consequences for the physical origin of the magnetic
field of the pulsars which are believed to be rapidly rotating neutron stars
with strong surface magnetic fields in the range of $10^{12}-10^{13}$ Gauss
shap ; paci ; gold ; navarro . Considering different stages of the neutron
star formation at different temperatures, the study of the magnetic properties
of the polarized nuclear matter at finite temperature is of special interest
in the description of protoneutron stars. A protoneutron star (newborn neutron
star) is born within a short time after the supernovae collapse. In this stage
the interior temperature of the neutron star matter is of the order 20-50 MeV
burro . The magnetic susceptibility of nucleonic matter is also an useful
quantity to estimate the mean free path of the neutrino in the dense nucleonic
matter which is relevant information for the understanding of the mechanism
underlying the supernova explosion and the cooling process of the neutron
stars iwamoto .

There exists several possibilities of the generation of the magnetic field in
a neutron star. From the nuclear physics point of view, such a possibility has
been studied by several authors using different theoretical approaches [7-30],
but the results are still contradictory. In most calculations, the neutron
star matter is approximated by pure neutron matter at zero temperature. The
properties of the polarized neutron matter both at finite and zero temperature
have been studied by several authors apv ; dapv ; bprrv . Some calculations
show that the neutron matter becomes ferromagnetic for some densities brown ;
rice ; pandh ; marcos ; apv . Some others, using the modern two-body and
three-body realistic interactions, show no indication of the ferromagnetic
transition at any density for the neutron matter and the asymmetrical nuclear
matter kutsb ; fanto ; vida ; bprrv . The results of another calculation show
both behaviors; the D1P force exhibits a ferromagnetic transition whereas no
sign of such transition is found for D1 at any density and temperature dapv .
The influence of the finite temperature on the antiferromagnetic (AFM) spin
ordering in the symmetric nuclear matter with the effective Gogny interaction
within the framework of a Fermi liquid formalism has been studied by Isayev
isay1 ; isay2 . We note that in the symmetric nuclear matter corresponding to
AFM spin ordering, we have
$\Delta\rho_{\uparrow\downarrow}=(\rho_{n\uparrow}+\rho_{p\downarrow})-(\rho_{n\downarrow}+\rho_{p\uparrow})\neq
0$ and $\Delta\rho_{\uparrow\uparrow}=\rho_{\uparrow}-\rho_{\downarrow}=0$. In
this article, we use the lowest order constrained variational (LOCV) formalism
to investigate the possibility of the transition to a ferromagnetic phase for
the polarized hot symmetrical nuclear matter.

The LOCV method has been developed to study the bulk properties of the quantal
fluids OBI1 ; OBI2 ; OBI3 . This technique has been used for studying the
ground state properties of finite nuclei and treatment of isobars BHIM ; MI1 ;
MI2 . Modarres has extended the LOCV method to the finite temperature
calculations and has applied it to the neutron matter, nuclear matter and
asymmetrical nuclear matter in order to calculate the different thermodynamic
properties of these systems Mod93 ; Mod95 ; Mod97 ; MM98 . Few years ago, we
calculated the properties of nuclear matter at zero and finite temperature
using the LOCV method with the new nucleon-nucleon potentials BM97 ; BM98 ;
MB98 . The LOCV method has several advantages with respect to the other many-
body formalism. These are as follows: (i) Since the method is fully self-
consistent, it does not introduce any free parameters into the calculations.
(ii) It considers the constraint in the form of a normalization constraint
Feen  to keep the higher-order terms as small as possible OBI3 ; MI1 ; MI2 ;
MM98 ; BM97  and it also assumes a particular form for the long-range behavior
of the correlation function in order to perform an exact functional
minimization of the two-body energy with respect to the short-range behavior
of the correlation function. (iii) The functional minimization procedure
represents an enormous computational simplification over the unconstrained
methods (i.e. to parameterize the short-range behavior of the correlation
functions) which attempt to go beyond the lowest order.

Recently, we have computed the properties of the polarized neutron matter
bordbig , polarized symmetrical bordbig2  and asymmetrical nuclear matters
bordbig3  and also polarized neutron star matter bordbig3  at zero temperature
using the microscopic calculations employing the LOCV method with the
realistic nucleon-nucleon potentials. We have concluded that the spontaneous
phase transition to a ferromagnetic state in these matters does not occur. We
have also calculated the thermodynamic properties of the polarized neutron
matter at finite temperature bordbig4  such as the total energy, magnetic
susceptibility, entropy and pressure using the LOCV method employing the
$AV_{18}$ potential wiring . Our calculations do not show any transition to a
ferromagnetic phase for a hot neutron matter.

In the present work, we intend to apply the LOCV calculation for the polarized
symmetrical nuclear matter at finite temperature using the $AV_{18}$
potential.

##  II Finite temperature calculations for polarized nuclear matter with the
LOCV method

We consider a system of $A$ interacting nucleons with $A^{(+)}$ spin-up and
$A^{(-)}$ spin-down nucleons. For this system, the total number density
($\rho$) and spin asymmetry parameter ($\delta$) are defined as

$\displaystyle\rho$ $\displaystyle=$ $\displaystyle\rho^{(+)}+\rho^{(-)},$
$\displaystyle\delta$ $\displaystyle=$
$\displaystyle\frac{\rho^{(+)}-\rho^{(-)}}{\rho}.$ (1)

$\delta$ shows the spin ordering of the matter which can have a value in the
range of $\delta=0.0$ (unpolarized matter) to $\delta=1.0$ (fully polarized
matter). To obtain the macroscopic properties of this system, we should
calculate the total free energy per nucleon, $F$,

$\displaystyle F=E-{\cal T}S^{(+)}-{\cal T}S^{(-)}.$ (2)

$E$ is total energy per nucleon and $S^{(i)}$ is the entropy per nucleon
corresponding to spin projection $i$,

$\displaystyle S^{(i)}(\rho,T)$ $\displaystyle=$
$\displaystyle-\frac{1}{A}\sum_{k}\\{[1-n^{(i)}(k,{\cal
T},\rho^{(i)})]\textrm{ln}[1-n^{(i)}(k,{\cal T},\rho^{(i)})]$ (3)
$\displaystyle+n^{(i)}(k,{\cal T},\rho^{(i)})\textrm{ln}n^{(i)}(k,{\cal
T},\rho^{(i)})\\}.$

where $n^{(i)}(k,{\cal T},\rho^{(i)})$ is the Fermi-Dirac distribution
function,

$\displaystyle n^{(i)}(k,{\cal
T},\rho^{(i)})=\frac{1}{e^{\beta[\epsilon^{(i)}(k,{\cal
T},\rho^{(i)})-\mu^{(i)}({\cal T},\rho^{(i)})]}+1}\cdot$ (4)

In the above equation $\beta=\frac{1}{k_{B}{\cal T}}$ , $\mu^{(i)}$ being the
chemical potential which is determined at any adopted value of the temperature
$\cal T$, number density $\rho^{(i)}$ and spin polarization $\delta$, by
applying the following constraint,

$\displaystyle\sum_{k}n^{(i)}(k,{\cal T},\rho^{(i)})=A^{(i)},$ (5)

and $\epsilon^{(i)}$ is the single particle energy of a nucleon. In our
formalism, the single particle energy of a nucleon with momentum $k$ and spin
projection $i$ is approximately written in terms of the effective mass as
follows apv ; dapv ; isay2

$\displaystyle\epsilon^{(i)}(k,{\cal
T},\rho^{(i)})=\frac{\hbar^{2}{k^{2}}}{2{m^{*}}^{(i)}(\rho,{\cal
T})}+U^{(i)}({\cal T},\rho^{(i)}).$ (6)

In fact, we use a quadratic approximation for single particle potential
incorporated in the single particle energy as a momentum independent effective
mass. $U^{(i)}({\cal T},\rho^{(i)})$ is the momentum independent single
particle potential. We introduce the effective masses, $m^{{*}{(i)}}$, as
variational parameters bordbig4 ; fp . We minimize the free energy with
respect to the variations in the effective masses and then we obtain the
chemical potentials and the effective masses of the spin-up and spin-down
nucleons at the minimum point of the free energy. This minimization is done
numerically.

As it is also mentioned in the pervious section, for calculating the total
energy of the polarized symmetrical nuclear matter, we use the LOCV method. We
adopt a trial many-body wave function of the form

$\displaystyle\psi=\cal{F}\phi,$ (7)

where $\phi$ is the uncorrelated ground state wave function (simply the Slater
determinant of plane waves) of $A$ independent nucleons and ${\cal F}={\cal
F}(1\cdots A)$ is an appropriate A-body correlation operator which can be
replaced by a Jastrow form i.e.,

$\displaystyle{\cal F}={\cal S}\prod_{i>j}f(ij),$ (8)

in which ${\cal S}$ is a symmetrizing operator. Now, we consider the cluster
expansion of the energy functional up to the two-body term clark ,

$\displaystyle
E([f])=\frac{1}{A}\frac{\langle\psi|H\psi\rangle}{\langle\psi|\psi\rangle}=E_{1}+E_{2}\cdot$
(9)

For the hot nuclear matter, the one-body term $E_{1}$ is

$\displaystyle E_{1}=E_{1}^{(+)}+E_{1}^{(-)},$ (10)

where

$\displaystyle E_{1}^{(i)}=\sum_{k}\frac{\hbar^{2}{k^{2}}}{2m}n^{(i)}(k,{\cal
T},\rho^{(i)}).$ (11)

The two-body energy $E_{2}$ is

$\displaystyle E_{2}$ $\displaystyle=$
$\displaystyle\frac{1}{2A}\sum_{ij}\langle ij\left|\nu(12)\right|ij-
ji\rangle,$ (12)

where

$\displaystyle\nu(12)=-\frac{\hbar^{2}}{2m}[f(12),[\nabla_{12}^{2},f(12)]]+f(12)V(12)f(12).$
(13)

In above equation, $f(12)$ and $V(12)$ are the two-body correlation and
potential. In our calculations, we use the $AV_{18}$ two-body potential which
has the following form wiring ,

$V(12)=\sum^{18}_{p=1}V^{(p)}(r_{12})O^{(p)}_{12},$ (14)

where

$\displaystyle O_{12}^{(p=1-18)}$ $\displaystyle=$ $\displaystyle 1,\
{\bf\sigma_{1}}\cdot{\bf\sigma_{2}},\ {\bf\tau_{1}}\cdot{\bf\tau_{2}},\
({\bf\sigma_{1}}\cdot{\bf\sigma_{2}})\ ({\bf\tau_{1}}\cdot{\bf\tau_{2}}),\
S_{12},\ S_{12}({\bf\tau_{1}}\cdot{\bf\tau_{2}}),$ (15) $\displaystyle{\bf
L}\cdot{\bf S},\ {\bf L}\cdot{\bf S}({\bf\tau_{1}}\cdot{\bf\tau_{2}}),\ {\bf
L}^{2},\ {\bf L}^{2}({\bf\sigma_{1}}\cdot{\bf\sigma_{2}}),\ {\bf
L}^{2}({\bf\tau_{1}}\cdot{\bf\tau_{2}}),$ $\displaystyle{\bf
L}^{2}({\bf\sigma_{1}}\cdot{\bf\sigma_{2}})({\bf\tau_{1}}\cdot{\bf\tau_{2}}),\
({\bf L}\cdot{\bf S})^{2},\ ({\bf L}\cdot{\bf
S})^{2}({\bf\tau_{1}}\cdot{\bf\tau_{2}}),$ $\displaystyle{\bf T_{12}},\
({\bf\sigma_{1}}\cdot{\bf\sigma_{2}}){\bf T_{12}},\ S_{12}{\bf T_{12}},\
(\bf\tau_{z1}+\bf\tau_{z2}).$

In above equation,
$S_{12}=[3({\bf\sigma_{1}}\cdot\hat{r})({\bf\sigma_{2}}\cdot\hat{r})-{\bf\sigma_{1}}\cdot{\bf\sigma_{2}}]$
is the tensor operator and ${\bf
T_{12}}=[3({\bf\tau_{1}}\cdot\hat{r})({\bf\tau_{2}}\cdot\hat{r})-{\bf\tau_{1}}\cdot{\bf\tau_{2}}]$
is the isotensor operator. The above 18 components of the $AV_{18}$ two-body
potential are denoted by the labels
$c,\sigma,\tau,\sigma\tau,t,t\tau,ls,ls\tau,l2,l2\sigma,l2\tau,l2\sigma\tau,ls2,ls2\tau,T,\sigma
T,tT$ and $\tau z$, respectively wiring . In the LOCV formalism, the two-body
correlation $f(12)$ is considered as the following form OBI3 ,

$\displaystyle f(12)$ $\displaystyle=$
$\displaystyle\sum^{3}_{k=1}f^{(k)}(r_{12})P^{(k)}_{12},$ (16)

where

$\displaystyle P_{12}^{(k=1-3)}$ $\displaystyle=$
$\displaystyle\left(\frac{1}{4}-\frac{1}{4}O^{(2)}_{12}\right),\
\left(\frac{1}{2}+\frac{1}{6}O^{(2)}_{12}+\frac{1}{6}O^{(5)}_{12}\right),$
(17)
$\displaystyle\left(\frac{1}{4}+\frac{1}{12}O^{(2)}_{12}-\frac{1}{6}O^{(5)}_{12}\right).$

The operators $O^{(2)}_{12}$ and $O^{(5)}_{12}$ are given in Eq. (15). Using
the above two-body correlation and potential, after doing some algebra we find
the following equation for the two-body energy,

$\displaystyle E_{2}$ $\displaystyle=$
$\displaystyle\frac{2}{\pi^{4}\rho}\left(\frac{h^{2}}{2m}\right)\sum_{JLTSS_{z}T_{z}}\frac{(2J+1)(2T+1)}{2(2S+1)}[1-(-1)^{L+S+T}]\left|\left\langle\frac{1}{2}\sigma_{z1}\frac{1}{2}\sigma_{z2}\mid
SS_{z}\right\rangle\right|^{2}$ (18) $\displaystyle\times\int
dr\left\\{\left[{f_{\alpha}^{(1)^{{}^{\prime}}}}^{2}{a_{\alpha}^{(1)}}^{2}(k_{f}r)+\frac{2m}{h^{2}}\left(\\{V_{c}-3V_{\sigma}+(V_{\tau}-3V_{\sigma\tau})(4T-3)\right.\right.\right.$
$\displaystyle\left.\left.\left.\ \ \ \ \ \ \ \ \
+(V_{T}-3V_{\sigma\tau})(4T)\\}{a_{\alpha}^{(1)}}^{2}(k_{f}r)\
+[V_{l2}-3V_{l2\sigma}\right.\right.\right.$ $\displaystyle\left.\left.\left.\
\ \ \ \ \ \ \ \
+(V_{l2\tau}-3V_{l2\sigma\tau})(4T-3)]{c_{\alpha}^{(1)}}^{2}(k_{f}r)\right)(f_{\alpha}^{(1)})^{2}\right]\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \ \ \
+\sum_{k=2,3}\left[{f_{\alpha}^{(k)^{{}^{\prime}}}}^{2}{a_{\alpha}^{(k)}}^{2}+\frac{2m}{h^{2}}\left(\left\\{V_{c}+V_{\sigma}+(-6k+14)V_{t}+-(k-1)V_{ls}\right.\right.\right.\right.$
$\displaystyle\left.\left.\left.\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+[V_{\tau}+V_{\sigma\tau}+(-6k+14)V_{tz}-(k-1)V_{ls\tau}](4T-3)\right.\right.\right.\right.$
$\displaystyle\left.\left.\left.\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+[V_{T}+V_{\sigma\tau}+(-6k+14)V_{tT}][T(6T_{z}^{2}-4)]+2V_{\tau
z}T_{z}\right\\}{a_{\alpha}^{(k)}}^{2}(k_{f}r)\right.\right.\right.$
$\displaystyle\left.\left.\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+[V_{l2}+V_{l2\sigma}+(V_{l2\tau}+V_{l2\sigma\tau})(4T-3)]{c_{\alpha}^{(k)}}^{2}(k_{f}r)\right.\right.\right.$
$\displaystyle\left.\left.\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+[(V_{ls2}+V_{ls2\tau})(4T-3)]{d_{\alpha}^{(k)}}^{2}(k_{f}r)\right){f_{\alpha}^{(k)}}^{2}\right]\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+\frac{2m}{h^{2}}[[(V_{ls\tau}-2(V_{l2\sigma\tau}+V_{l2\tau})-3V_{ls2\tau})(4T-3)]\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+V_{ls}-2(V_{l2}+V_{l2\sigma})-3V_{ls2}]b_{\alpha}^{2}(k_{f}r)f_{\alpha}^{(2)}f_{\alpha}^{(3)}\right.$
$\displaystyle\left.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+\frac{1}{r^{2}}(f_{\alpha}^{(2)}-f_{\alpha}^{(3)})^{2}b_{\alpha}^{2}(k_{f}r)\right\\},$

where $\alpha=\\{J,L,S,S_{z}\\}$ and the coefficient ${a_{\alpha}^{(1)}}^{2}$,
etc., are as follows,

$\displaystyle{a_{\alpha}^{(1)}}^{2}(x)=x^{2}I_{L,S_{z}}(x),$ (19)
$\displaystyle{a_{\alpha}^{(2)}}^{2}(x)=x^{2}[\beta I_{J-1,S_{z}}(x)+\gamma
I_{J+1,S_{z}}(x)],$ (20) $\displaystyle{a_{\alpha}^{(3)}}^{2}(x)=x^{2}[\gamma
I_{J-1,S_{z}}(x)+\beta I_{J+1,S_{z}}(x)],$ (21) $\displaystyle
b_{\alpha}^{(2)}(x)=x^{2}[\beta_{23}I_{J-1,S_{z}}(x)-\beta_{23}I_{J+1,S_{z}}(x)],$
(22) $\displaystyle{c_{\alpha}^{(1)}}^{2}(x)=x^{2}\nu_{1}I_{L,S_{z}}(x),$ (23)
$\displaystyle{c_{\alpha}^{(2)}}^{2}(x)=x^{2}[\eta_{2}I_{J-1,S_{z}}(x)+\nu_{2}I_{J+1,S_{z}}(x)],$
(24)
$\displaystyle{c_{\alpha}^{(3)}}^{2}(x)=x^{2}[\eta_{3}I_{J-1,S_{z}}(x)+\nu_{3}I_{J+1,S_{z}}(x)],$
(25)
$\displaystyle{d_{\alpha}^{(2)}}^{2}(x)=x^{2}[\xi_{2}I_{J-1,S_{z}}(x)+\lambda_{2}I_{J+1,S_{z}}(x)],$
(26)
$\displaystyle{d_{\alpha}^{(3)}}^{2}(x)=x^{2}[\xi_{3}I_{J-1,S_{z}}(x)+\lambda_{3}I_{J+1,S_{z}}(x)].$
(27)

In above equations, we have

$\displaystyle\beta=\frac{J+1}{2J+1},\ \ \gamma=\frac{J}{2J+1},\ \
\beta_{23}=\frac{2J(J+1)}{2J+1},$ (28) $\displaystyle\nu_{1}=L(L+1),\ \
\nu_{2}=\frac{J^{2}(J+1)}{2J+1},\ \ \nu_{3}=\frac{J^{3}+2J^{2}+3J+2}{2J+1},$
(29) $\displaystyle\eta_{2}=\frac{J(J^{2}+2J+1)}{2J+1},\ \
\eta_{3}=\frac{J(J^{2}+J+2)}{2J+1},$ (30)
$\displaystyle\xi_{2}=\frac{J^{3}+2J^{2}+2J+1}{2J+1},\ \
\xi_{3}=\frac{J(J^{2}+J+4)}{2J+1},$ (31)
$\displaystyle\lambda_{2}=\frac{J(J^{2}+J+1)}{2J+1},\ \
\lambda_{3}=\frac{J^{3}+2J^{2}+5J+4}{2J+1},$ (32)

and

$\displaystyle I_{J,S_{z}}(r,\rho,T)=\frac{1}{2\pi^{6}\rho^{2}}\int
k_{1}^{2}dk_{1}k_{2}^{2}dk_{2}n_{i}(k_{1},T,\rho_{i})n_{j}(k_{2},T,\rho_{j})J_{J}^{2}(|k_{2}-k_{1}|r),$
(33)

where $J_{J}(x)$ is the Bessel’s function .

Now, we minimize the two-body energy Eq. (18) with respect to the variations
in the correlation functions ${f_{\alpha}}^{(k)}$, but subject to the
normalization constraint OBI3 ; BM98 ,

$\displaystyle\frac{1}{A}\sum_{ij}\langle
ij\left|h_{S_{z}}^{2}-f^{2}(12)\right|ij\rangle_{a}=0\cdot$ (34)

In the case of polarized symmetrical nuclear matter, the Pauli function
$h_{S_{z}}(r)$ is as follows

$\displaystyle
h_{S_{z}}(r)=\left\\{\begin{array}[]{ll}\left[1-\frac{1}{2}\left(\frac{\gamma^{(i)}(r)}{\rho}\right)^{2}\right]^{-1/2}&;\
\hbox{$S_{z}=\pm 1$}\\\ 1&;\ \hbox{$S_{z}=0$}\end{array}\right.$ (37)

where

$\displaystyle\gamma^{(i)}(r)=\frac{1}{\pi^{2}}\int n^{(i)}(k,{\cal
T},\rho^{(i)})J_{0}(kr)k^{2}dk.$ (38)

From the minimization of the two-body cluster energy we get a set of coupled
and uncoupled Euler-Lagrange differential equations. The Euler-Lagrange
equations for uncoupled states are

$\displaystyle
g_{\alpha}^{(1)^{\prime\prime}}-\\{\frac{a_{\alpha}^{(1)^{\prime\prime}}}{a_{\alpha}^{(1)}}+\frac{m}{\hbar^{2}}[V_{c}-3V_{\sigma}+(V_{\tau}-3V_{\sigma\tau})(4T-3)$
$\displaystyle+(V_{T}-3V_{\sigma T})[T(6T_{z}^{2}-4)]+2V_{\tau
z}T_{z}+\lambda]$
$\displaystyle+\frac{m}{\hbar^{2}}(V_{l2}-3V_{{l2}\sigma}+(V_{{l2}\tau}-3V_{{l2}\sigma\tau})(4T-3))\frac{c_{\alpha}^{(1)^{2}}}{a_{\alpha}^{(1)^{2}}}\\}g_{\alpha}^{(1)}=0,$
(39)

while for the coupled states, these equations are written as follows,

$\displaystyle
g_{\alpha}^{(2)^{\prime\prime}}-\\{\frac{a_{\alpha}^{(2)^{\prime\prime}}}{a_{\alpha}^{(2)}}+\frac{m}{\hbar^{2}}[V_{c}+V_{\sigma}+2V_{t}-V_{{ls}}+(V_{\tau}+V_{\sigma\tau}+2V_{t\tau}-V_{{ls}\tau})(4T-3)$
$\displaystyle+(V_{T}+V_{\sigma T}+2V_{tT})[T(6T_{z}^{2}-4)]+2V_{\tau
z}T_{z}+\lambda]$
$\displaystyle+\frac{m}{\hbar^{2}}[V_{l2}+V_{{l2}\sigma}+(V_{{l2}\tau}+V_{{l2}\sigma\tau})(4T-3)]\frac{c_{\alpha}^{(2)^{2}}}{a_{\alpha}^{(2)^{2}}}$
$\displaystyle+\frac{m}{\hbar^{2}}[V_{{ls}2}+V_{{ls}2\tau}(4T-3)]\frac{d_{\alpha}^{(2)^{2}}}{a_{\alpha}^{(2)^{2}}}+\frac{b_{\alpha}^{2}}{r^{2}a_{\alpha}^{(2)^{2}}}\\}g_{\alpha}^{(2)}$
$\displaystyle+\\{\frac{1}{r^{2}}-\frac{m}{2\hbar^{2}}[V_{ls}-2V_{l2}-2V_{{l2}\sigma}-3V_{{ls}2}$
$\displaystyle+(V_{{ls}\tau}-2V_{{l2}\tau}-2V_{{l2}\sigma\tau}-3V_{{ls}2\tau})(4T-3)]\\}\frac{b_{\alpha}^{2}}{a_{\alpha}^{(2)}a_{\alpha}^{(3)}}g_{\alpha}^{(3)}=0,$
(40) $\displaystyle
g_{\alpha}^{(3)^{\prime\prime}}-\\{\frac{a_{\alpha}^{(3)^{\prime\prime}}}{a_{\alpha}^{(3)}}+\frac{m}{\hbar^{2}}[V_{c}+V_{\sigma}-4V_{t}-2V_{ls}+(V_{\tau}+V_{\sigma\tau}-4V_{t\tau}-2V_{{ls}\tau})(4T-3)$
$\displaystyle+(V_{T}+V_{\sigma T}-4V_{tT})[T(6T_{z}^{2}-4)]+2V_{\tau
z}T_{z}+\lambda]$
$\displaystyle+\frac{m}{\hbar^{2}}[V_{l2}+V_{{l2}\sigma}+(V_{{l2}\tau}+V_{{l2}\sigma\tau})(4T-3)]\frac{c_{\alpha}^{(3)^{2}}}{a_{\alpha}^{(3)^{2}}}$
$\displaystyle+\frac{m}{\hbar^{2}}[V_{{ls}2}+V_{{ls}2\tau}(4T-3)]\frac{d_{\alpha}^{(3)^{2}}}{a_{\alpha}^{(3)^{2}}}+\frac{b_{\alpha}^{2}}{r^{2}a_{\alpha}^{(2)^{2}}}\\}g_{\alpha}^{(3)}$
$\displaystyle+\\{\frac{1}{r^{2}}-\frac{m}{2\hbar^{2}}[V_{ls}-2V_{l2}-2V_{{l2}\sigma}-3V_{{ls}2}$
$\displaystyle+(V_{{ls}\tau}-2V_{{l2}\tau}-2V_{{l2}\sigma\tau}-3V_{{ls}2\tau})(4T-3)]\\}\frac{b_{\alpha}^{2}}{a_{\alpha}^{(2)}a_{\alpha}^{(3)}}g_{\alpha}^{(2)}=0,$
(41)

where

$\displaystyle g_{\alpha}^{(i)}(r)=f_{\alpha}^{(i)}(r)a_{\alpha}^{(i)}(r).$
(42)

The primes in the above equations mean differentiation with respect to $r$.
The Lagrange multiplier $\lambda$ is introduced by the normalization
constraint, Eq. (34). Now, we can calculate the correlation functions by
numerically solving these differential equations and then using these
correlation functions, the two-body energy is obtained. Finally, we can
compute the energy and then the free energy of the system.

##  III Results and discussion

We have presented the effective masses of the spin-up and spin-down nucleons
as functions of the spin polarization ($\delta$) at $\rho=0.5fm^{-3}$ and
${\cal T}=20$ MeV in Fig. 1. It is seen that the difference between the
effective masses of spin-up and spin-down nucleons increases by increasing the
polarization. We also see that the effective mass of spin-up nucleons
increases by increasing the polarization whereas the effective mass of spin-
down nucleons decreases by increasing the polarization. These behaviors have
been also seen for the effective mass of the neutron in the case of spin
polarized hot neutron matter apv ; dapv ; bprrv ; bordbig4 . A similar
qualitative behavior of the nucleon effective mass as a function of the
isospin asymmetric parameter
$\beta=\frac{\rho_{n}-\rho_{p}}{\rho_{n}+\rho_{p}}$ has been already found in
non-polarized isospin asymmetric nuclear matter, as it has been already
discussed in Refs. bomb1 ; li .

The free energy per nucleon of the polarized hot nuclear matter versus the
total number density ($\rho$) for different values of the spin polarization
($\delta$) at ${\cal T}=10$ and $20$ MeV is shown in Fig. 2. It can be seen
that for each value of the temperature, the free energy increases by
increasing both density and polarization. From Fig. 2, it is seen that the
free energy of the polarized hot nuclear matter decreases by increasing the
temperature. We have seen that above a certain values of the temperature and
spin polarization, the free energy does not show any bound states for the
polarized hot nuclear matter. From Fig. 2 we also see that for all given
temperatures there is no crossing of the free energy curves for different
polarizations and the difference between the free energy of the nuclear matter
at different polarizations increases by increasing the density. This indicates
that the spontaneous transition to the ferromagnetic phase does not occur in
the hot nuclear matter. We have also compared the free energy per nucleon of
the unpolarized case of the nuclear matter at different temperatures in Fig.
3. It is seen that the free energy of unpolarized nuclear matter decreases by
increasing the temperature. We can see that above a certain temperature, the
free energy does not show the bound state (the minimum point of the free
energy) for the unpolarized nuclear matter.

For the polarized hot nuclear matter, the magnetic susceptibility, $\chi$,
which characterizes the response of the system to the magnetic field can be
calculated by using the following relation,

$\displaystyle\chi=\frac{\mu^{2}\rho}{\left(\frac{\partial^{2}F}{\partial\delta^{2}}\right)_{\delta=0}},$
(43)

where $\mu$ is the magnetic moment of the nucleons. Fig. 4 shows the ratio
${{\chi_{F}}/{\chi}}$ as a function of the temperature at $\rho=0.16fm^{-3}$,
where $\chi_{F}$ is the magnetic susceptibility for a noninteracting Fermi
gas. As it can be seen from this figure, this ratio is inversely proportional
to absolute temperature without any anomalous change in its behavior. This
indicates that hot nuclear matter is paramagnetic. The ratio
${{\chi_{F}}/{\chi}}$ has been also shown versus the total number density at
temperature ${\cal T}=20$ MeV in Fig. 5. A magnetic instability would require
${{\chi_{F}}/{\chi}<0}$. It is seen that the value of ${{\chi_{F}}/{\chi}}$ is
always positive and monotonically increasing up to highest density and does
not show any spontaneous phase transition to the ferromagnetic phase for the
hot nuclear matter.

The difference between the entropy per nucleon of the fully polarized and
unpolarized cases of the nuclear matter is plotted as a function of the total
number density at ${\cal T}=20$ MeV in Fig. 6. It is seen that for all given
values of the density, this difference is negative. This shows that the fully
polarized case of hot nuclear matter is more ordered than the unpolarized
case. We also see that the magnitude of this deference decreases by increasing
the density. The entropy per nucleon of the polarized hot nuclear matter
versus the spin polarization for fixed density $\rho=0.5fm^{-3}$ and
temperature ${\cal T}=20$ has been presented in Fig. 7. It is shown that the
entropy decreases by increasing the polarization. It is also shown that the
highest value of the entropy occurs for the unpolarized case of the hot
nuclear matter. For the polarized hot nuclear matter, the following condition
for the effective mass prevents the anomalous behavior of the entropy versus
the spin polarization apv ,

$\displaystyle\frac{m^{*}(\rho,\delta=1.0)}{m^{*}(\rho,\delta=0.0)}<2^{2/3},$
(44)

where $m^{*}(\rho,\delta=1.0)$ and $m^{*}(\rho,\delta=0.0)$ are the effective
masses of the fully polarized and unpolarized nuclear matter, respectively.
This condition was first derived in Ref. apv  for the particular case of the
Skyrme interaction where the effective mass is independent of the momentum and
temperature and therefore the single particle potential is purely parabolic.
In our approach, the effective mass depends on both density and temperature
but is independent of the momentum. In other words, a similar rigorous
condition can not be obtained straightforwardly. However, within this
approximation, one can use the condition of Eq. (44). From our result for the
effective mass at ${\cal T}=20MeV$ for $\rho=0.5fm^{-3}$ (Fig. 1), we have
found that this ratio is $1.24$. We see that this value is smaller than the
above limiting value which indicates that the entropy of polarized case of the
hot nuclear matter is always smaller than the entropy of unpolarized case.
This so-considered ”natural” behavior was also found in the case of Gogny dapv
and in the BHF analysis of Ref. bprrv . In contrast, for Skyrme forces the
entropy per particle of the polarized phase is seen to be higher than the non-
polarized one above a certain density dapv .

Finally, we have plotted the pressure of the polarized hot nuclear matter as a
function of the total number density ($\rho$) for different polarizations at
${\cal T}=10$ and $20$ MeV in Fig. 8. For all values of temperature and
polarization, it is seen that the pressure increases by increasing the
density. For this system, we see that at each temperature the equation of
state becomes stiffer as the polarization increases. For each polarization, it
is found that the pressure of the polarized hot nuclear matter increases by
increasing the temperature.

##  IV Summary and Conclusions

The lowest order constrained variational (LOCV) method has been used for
calculating the susceptibility of the polarized hot nuclear matter and some of
the thermodynamic properties of this system such as the effective mass, free
energy, entropy and the equation of state. In our calculations, we have
employed the $AV_{18}$ potential. Our results show that the spontaneous
transition to the ferromagnetic phase does not occur for the hot nuclear
matter. We have seen that the spin polarization substantially affects the
thermodynamic properties of the hot nuclear matter.

###### Acknowledgements.

This work has been supported by Research Institute for Astronomy and
Astrophysics of Maragha. We wish to thank Shiraz University and Zanjan
University Research Councils.

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Figure 1: The effective mass of spin-up (full curve) and spin-down (dashed
curve) nucleons versus the spin polarization ($\delta$) for density
$\rho=0.5fm^{-3}$ at ${\cal T}=20$ MeV.

Figure 2: The free energy per nucleon of the polarized hot nuclear matter as a
function of the total number density ($\rho$) for different values of the spin
polarization ($\delta$) at ${\cal T}=10$ (a) and ${\cal T}=20$ MeV (b).

Figure 3: The free energy per nucleon of the nuclear matter versus the total
number density ($\rho$) for unpolarized case at ${\cal T}=0$, $10$ and $20$
MeV .

Figure 4: The magnetic susceptibility of the hot nuclear matter versus the
temperature at $\rho=0.16fm^{-3}$ .

Figure 5: The magnetic susceptibility of the hot nuclear matter versus the
total number density ($\rho$) at ${\cal T}=20$ MeV.

Figure 6: As Fig. 4 but for the entropy difference of the fully polarized and
the unpolarized cases.

Figure 7: The entropy per nucleon as a function of the spin polarization
($\delta$) for density $\rho=0.5fm^{-3}$ at ${\cal T}=20$ MeV .

Figure 8: The equation of state of the polarized hot nuclear matter for
different values of the spin polarization ($\delta$) at ${\cal T}=10$ (a) and
${\cal T}=20$ MeV (b).


# Escape Room combined with European Board Game Concepts for self-adjusted
Challenge Levels: An educational Eurogame Escape Room in Physics

Sascha Albert Bräuninger , Damian Alexander Motz , Matthias Lüpke , Hermann
Seifert  0009-0008-1809-4101 0000-0003-2013-9117 0000-0001-7855-8301 Institute
for General Radiology and Medical Physics, University of Veterinary Medicine
Hannover Foundation, Bischofsholer Damm 15, 30173 Hanover, Lower Saxony,
Germany<EMAIL_ADDRESS>
###### Abstract

We present an educational escape room in physics (ERP) covering the main
disciplines of physics as taught in the education of neighboring sciences, in
this case veterinary medicine, extended by a European board game system of
victory points (VP). The puzzles in physics are mandatory to master the ERP
requiring to be solved in a sequential order. In our study, we show the growth
of knowledge by knowledge-transfer cards separated into disciplines of physics
as demonstrated by critical exemplary questions of test groups. Tactical
secondary puzzles, the so called Eurogame puzzles (EPs), are able to be solved
in parallel in any arbitrary and independent sequence, allowing a self-
adjusted challenge autolevelled by the students by choice. The motivation is
improved by a scoring list to solve, at least, a few optional EPs as
additional game-like adventure and challenge collecting VP. A second part of
the study demonstrates the attraction of the introduced European board game
concept as an important tool for gamificiation and open-by-choice EPs not
being necessarily related to physics directly. EPs are able to smooth the
contrast of playing games and learning, providing an additional bridge to
overcome the fear of physics. A strong advantage is the flexibility of the
European board concept, choosing from a broad range of optional EPs to
fascinate and motivate students taking into account individual preferences
improving escape rooms by decision-making skills. The concept is able be
applied to every escape room independent of the taught subject or topic.

  *

Keywords: Escape Room of physics, Gamification, European Boardgames,
Eurogames, Victory Points

##  1 Introduction

###  1.1 General Remarks and Abbreviations

In everyday life, escape rooms (ERs) and board games are tools of cooperative
or competitive ways of analogue entertainment entering the field of education.
See Table 1 for abbreviations as used in this report. The following part
introduces briefly the concept of ERs in higher-level education, its
advantages and challenges. The second part discusses basics of European board
game concepts and the victory point system as used in European board games.
This part is important because it is less discussed in the scientific
literature, playing a dominant role in the eEERP which combines a sequential
ER concept and Eurogame-like ideas to generate an efficient modern concept of
gamification.

Table 1: List of abbreviations as used in this publication. To support readability, we have restricted ourselves to use just one or two abbreviations per sentences. Abbreviation | Synonym
---|---
EP | Eurogame puzzle, Europuzzle
ER | Escape room
ERP | Escape room of/in physics
EER | Eurogame escape room
EERP | Eurogame escape room of/in physics
PP | Puzzle of physics or physics puzzle
VP |  Victory points (exceptionally, plural)
prefix e | educational (e.g. eER means educational ER)
suffix s |  plural (e.g. EPs - Eurogame puzzles)
dc | direct current
ac | alternating current

###  1.2 Educational Escape Rooms

In general, an ER is

> a live-action team-based game where players discover clues, solve puzzles,
> and accomplish tasks in one or more rooms in order to accomplish a specific
> goal (usually escaping from the room) in a limited amount of time. [1]

Historically, ERs have been built for pure entertainment. Nowadays, the
attention has been extended to ERs for the purpose of education as well [1, 2,
3], called eERs. Game-based eERs have advantages compared to conventional
lectures and practical courses promoting

  * teamwork and collaboration, social interaction and communication

  * higher level enjoyment

  * engagement

  * increased learning gain

  * critical thinking, problem-solving and creativity

being the main advantages [4]. As we will point out in our eEERP, this list is
able to be extended by decision-making and setting priorities by gauging
puzzles.

###  1.3 Eurogames

Eurogames are also called European-style games, European board games or
German-style board games [5, 6]. Eurogames are a specific class of board games
or table-top games showing characteristics:

  * •

Player conflict is indirect,

  * •

No player elimination,

  * •

Symmetric start conditions,

  * •

Score point/victory point system,

  * •

Low-level of luck and randomness.

The latter points are important, supporting an intellectual multiplayer
challenge realised by a contest of collecting score points. Table 2 shows
selected Eurogames and details, namely the publisher, game mechanics and the
associated number of score points typically received in the game per match.
Settlers of Catan is a famous example of a Eurogame although the start
conditions are slightly asymmetric. One should keep in mind that, the
definition of Eurogames is not that rigid and sharp from an academic point of
view as expected. Usually, Eurogames are thematic games, mostly not related to
real events of history (in contrast to historical board games). Sometimes
elements may be mixed, e.g., Imperial uses a map and fractions of the First
World War. Therefore, Eurogames are not as abstract as chess or checkers.
Surprisingly, the aforementioned characteristics (see bullet points above) do
not contain a real board as a requirement and the description European-style
board game should be understood as a concept. Carcassonne is a pure Eurogame
although it contains tiles working by placing tiles on the table instead of
having a conventional board. In Eurogames, an asymmetry develops mostly over
time, e.g. in Puerto Rico and Caylus by buying buildings. Interestingly,
playing the cardgame Skat with score points fulfills a few of the
aforementioned points, especially, one can argue the high asymmetry (2 versus
1) is averaged out because a match contains several hands. However, the amount
of luck is a matter of discussion which depends on the number of played hands
and the missing theme (French-suited deck) ends up with the result that Skat
is just a Eurogame-like cardgame being extremely on the border to be like. One
could try to extend this statement to Rummy or Canasta, but various subgames
in Skat (Null, $\clubsuit,\spadesuit,\heartsuit,\diamondsuit$, Grand), are
definitively missing deepness and higher randomness in Rummy and Canasta
excluding this games. This shows the high complexity of the topic and fluent
transition. The most important and clear criterion for exclusion is player
elimination which is prohibited in Eurogames. VP of various matches are not
necessarily comparable because of the interaction of players and match-
dependent dynamics. Additionally, start conditions may be match-dependent.
111Issues of Eurogames are rarely discussed in the scientific literature. The
main part of this subsection roots in decades of experience playing Eurogames
and visits to the Internationale Spieletage SPIEL/Essen Game Fair (world’s
greatest board game fair) of the main author.

Table 2: Selected Eurogames using a victory point system to identify the best player of the match. Typical VP are mentioned, being strongly dependent on the number of players and dynamics of the match. Eurogame | Publisher | Mechanics | typical VP
---|---|---|---
Settlers of Catan | Kosmos | Dice rolling | max. 10
Carcassonne | Hans im Glück | Area majority, tile placing | 70-125
Puerto Rico | Alea | Action drafting | 25-65
Caylus | Ystari Games | Worker placement | 80-125
Imperial | PD Games | Investment, area majority | 75-180

##  2 Eurogame Escape Room in Physics

###  2.1 Goal, Room Concept, Capacity and Restrictions

Coming back to the above definition of an ER, the specific goal of our ERP was
not to escape from the room. We used an ER concept of two rooms. The ER
started in the first room (so-called mirror room, starting room) and the
intermediate goal was to enter the second locked room which was named
Schrödinger’s working room, see S. A. Bräuninger et al. [7, 8]. The entrance
required an opening of eight boxes by solving the first eight PPs. Therefore,
our ER was an inverted ER taking into account fire protection issues as
stipulated by the university. The main goal was to finish all PPs. The
secondary (not mandatory) goal was to collect VP by solving the puzzles using
a victory point system as in Eurogames. The ER was team-based solved by 2-4
students. However, ideally the aimed occupancy was 3 students. The processing
time was 2.5 hours. The students are hereinafter called players as well. The
instructor is hereinafter called e-game master (educational-game master). The
e-game master was available by phone combined with diluted visits to check the
work flow. The starting room was not locked to allow the players to use the
restroom and mobile phones were allowed to be taken into the eERP. However,
digital detoxification (digidetox) was strongly urged. The ERP could be freely
chosen as a compulsory elective subject. Approximately 85% of the students
were female (studying veterinary medicine at University at the Veterinary
Medicine Hannover Foundation, year 2023). The eERP is a spectral one, which
means it covers a variety of main disciplines, in this case in physics
(mechanics, optics, electromagnetism: dc/ac, atomic physics, solid states,
thermodynamics, X-ray).

###  2.2 Rules, Educational Descriptions and Documents

The ER containes four kinds of documents to support the understanding,
providing a transfer of knowledge of physics, atmosphere and instructions to
master the puzzles.

  * •

Story: The story document contains a story-like introduction to explore the
loss of Schrödinger’s cat. This document is a cosmetic one for thematic
embellishment. It provides an atmosphere without transferring important
knowledge to the reader to master the ER.

  * •

Rulebook: The rulebook explains the eurogame system of VP containing a list of
the values of points for PPs and EPs. This list can be found in the Appendix
in table 5 and will be discussed in the following subsection as well. The
rulebook discusses the general principle of the ER (opening boxes and
treasures by codes, solving the PPs in a sequential order).

  * •

Knowledge transfers: Every PP contains a document of 1-2 pages (DIN A4)
presenting important knowledge of physics strongly associated with the puzzle.
Every asked question in the questionnaire targeted in our presented study
could be answered successfully by the transfer documents. The knowledge cards
are unassigned, containing one digit for the next code of a box/treasure.
Therefore, the players had to combine the PPs and the knowledge transfer
documents.

  * •

Puzzle instructions: Short supplements of 1 page (DIN A5) contained tips,
safety remarks and experimental issues. Puzzle instructions were only added to
PPs.

The visit to the eEERP started with an audio explanation of a few minutes
summarising the game rules. This highlighted important key rules.

###  2.3 Victory Point Concept

The accumulation of VP is one motor of motivation. The effective ’match’ of
the game is counted over one semester to compare the collected VP of groups of
players. This is called a run. The students visited the ER collected points in
accordance with the Eurogame concept to compare their VP per run. Every group
which successfully completed the ER accumulated points for two kinds of
puzzles: PPs and EPs. Both kinds of puzzles are explained in the following
subsections.
Coming back to the above definition of an ER, the primary specific goal of our
eEER was to complete all PPs successfully to master the ER. The secondary goal
was to collect VP by solving all PPs and EPs. Table 5 (see Appendix) shows the
victory points $P_{i}$ of PPs and points $p_{i}$ of EPs as used in our ERP in
2023/2024. The points for PPs have two purposes. First of all, without solving
any EP the players would end up with 0 VP (after mastering), this being a
frustration-like number of points. To avoid this, PPs give points, therefore
providing improved motivation. The second reason is to offer support if the
students have problems solving individual puzzles in physics. Here, we halved
the points to sell ideas. Values of points scaled with the difficulty of the
puzzle. Therefore, the maximum of VP was

$p=\sum_{\mathrm{PPs}}P_{i}+\sum_{\mathrm{EPs}}p_{i}=\sum_{i=1}^{10}P_{i}+\sum_{i=1}^{16}p_{i}=40+126=166.$
(1)

Realistically, it was impossible to solve all puzzles because the EPs were
quantitatively overestimated, offering a broad proposal to motivate the
players by selecting their individual preferences according to skills,
strengths and passion. Thus, the players should focus on the mandatory PPs,
solving the optional EPs diluted, i.e., only a few in parallel.

###  2.4 Puzzles of Physics

Figure 1: a) Miniature Stirling engine which runs on several heat sources,
e.g., cup of hot water, b) train experiment to calculate the velocity of a
moving model train on railroads by stopping a grey clock. Figure 2: a) dog
which consists of tetris-like blocks containing magnets inside to stabilise
the construction of this puzzle b) separated blocks emphasizing the added
numbers and symbols to achieve equations after puzzling the dog.

The puzzle structure is sequential according to A. Veldkamp et al. [2]. PPs
are mandatory to finish the ERP successfully. Table 3 shows the list of PPs
and transferred knowledge. In total, 10 PPs were developed. Exemplarily, there
are three PPs presented in Figure 1 and 2 which are discussed in the
following:

  * Stirling engine: Stirling machine which must be put on a cup of hot water; see Figure 1 a). Parts of the equipment are hidden. The machine is a final puzzle without generating a new code. The photo shows the Kontax KS90 Silver Stirling Engine from the United Kingdom (high quality product). The main knowledge of transfer is that the efficiency $\eta$ of a heat engine is

$\eta=\left|\frac{\Delta W}{\Delta Q}\right|=1-\frac{T_{2}}{T_{1}}<1$ (2)

where the second equality concerns a Carnot engine. The latter one is
mentioned for didactic simplicity being the famous example of literature. Heat
$\Delta Q$ can not be transformed completely in mechanical work $\Delta W$.

  * Train experiment: A miniaturised train was used as puzzle in kinematics, see Figure 1 b). It drives in a circle and the task was to calculate the velocity and a transformation of units to get the next code for the final treasure. The experiment contains a miniaturized train with rails (scale 1:120) produced by Berliner Bahnen TT Hobby with a trail of 12 mm. For details, see Escape room in physics: Train experiment [9] which contains a description to extend the puzzle by an error calculation. The result of the PP is

$v=\frac{2\pi r}{t}\approx\frac{180.6\,\mathrm{cm}}{16\,\mathrm{s}}\approx
11.3\,\mathrm{cm/s}.$ (3)

where $r$ is the radius of the circle of around $r=d/2$ with $d\approx 57.5$
cm and $t\approx 16$ s the time per circle of the train. The first two digits
”1” and ”1” are the first two digits for the code of the next box. The third
digit is given by the correct dedicated knowledge transfer card. The players
must match all knowledge cards to the corresponding PPs because the knowledge
cards are matched. An important issue is to identify a circular motion as
accelerated movement in which the direction changes (change of direction of
$\vec{v}$). The absolute value is constant: $|\vec{v}|=v$.

  * Magnetic dog puzzle: Magnetic tetris-like stones/blocks in 3D are used to visualise ferromagnetic forces; CUBIMAG Pro produced by HCM Kinzel from Germany, see Figure 2 a). This puzzle has a high gamification and difficulty level. The blocks are stabilised by small magnets (grey pieces) inside the blocks. We added numbers and mathematical symbols (”+”,”-”,”=”), see Figure 2 b). After assembling, it yields two digits (two equations). The individual blocks can be turned because of the symmetry recalling symmetry operations in molecules and solid-state physics. This must be considered. Additionally, small experiments are supplemented, a demonstrative add-on without any VP, see Appendix B. In our case, the quantity of interest is the Curie temperature $T_{C}$ of the material and the magnetic susceptibility is

$\chi=\frac{C}{T-T_{C}}$ (4)

in which the magnet shows spontaneous magnetization below $T_{C}$ and room
temperature should be understood as $\approx 293$ K $=T<T_{C}$ to adapt the
Kelvin-scale as well. $C$ is a material-dependent magnetic constant.

Table 3: Overview of the ten PPs. Additionally, corresponding topics in the field of physics are presented and important laws or quantitative relations are transferred to the reader by knowledge cards. Puzzle/Quest in Physics | Topic | Knowledge Transfer
---|---|---
Nobel prize (start-up) | History of physics | Important physicists, Nobel prize modalities
Apple puzzle | Mechanics and forces | Newton’s law of motion
Fluorescence | Atomic physics | Excitation of atoms by UV-light
Electric conductivity | Electricity (dc) | Ohm’s law
Total internal reflection | Geometric optics | Reflection and refraction
50 Hertz sound experiment | Electricity (ac) |  Frequency $\nu=\frac{1}{T}$ as inverted period time $T$
X-ray | Radio diagnostics | Absorption coefficients
Dog puzzle | Solid state physics | Force of ferromagnetism, symmetry
Train experiment [9] | Kinematics | Mechanical circular motion
Cup Stirling machine | Thermodynamics | Heat engine, inverted fridge

###  2.5 Eurogame puzzles

EPs are optional and arbitrary puzzles. It is not mandatory to solve EPs, but
rather they are an opportunity to collect additional VP. The puzzle structure
is open according to A. Veldkamp et al. [2]. EPs are independent of each other
and independent of the PPs. A few EPs are located in the second room,
therefore these puzzles are available later on. An advantage of EPs is the
possibility to include commercial and arbitrary puzzles in every ER, in the
present case in an eERP, thus offering an optional supplement. EPs do not
necessarily have anything to do with the topic of education, in this case
physics. To avoid confusion, a EP is not a Eurogame. EPs are arbitrary
commercial or self-made puzzles that earn points after solving, comparable to
Eurogame elements, e.g. by building a building (Caylus, Carcassonne) or by
shipping resources (Puerto Rico). Once again, EPs are not mandatory and fully
optional puzzles. Exceptionally, the team of players is motivated to earn more
VP, thus accepting the challenge. A mark/grade was not given. It was only
possible to fail or pass the ERP ’examination’. Table 4 shows the list of the
16 EPs in the eEERP. Exemplarily, three EPs are presented:

  * 24 dexterity cubes: One group of EPs are 24 cubes requiring mechanical dexterity; these were bought from the company Furado from China, see Figure 3 a). The aim is self-explained by logic and the first part of the challenge to find it out. The second part is mechanical. For instance, it contains four holes inside and this recess should be filled with four metal balls. Each of the 24 EPs has individual challenges promoting surgical dexterity which is needed in veterinary medicine.

  * 6 Wood puzzles: Figure 3 b) shows the presented six wood puzzles from Casa Vivente. Two sets of the puzzle were bought. One set is already assembled and the second one should be assembled. The 6 EPs require a combination of dexterity and spatial sense/imagination.

  * Magnetic snake puzzle: The players may build a snake after finishing the dog puzzle as shown in Figure 2, using this equipment twice.

Figure 3: a) Twenty-four dexterity cubes distributed in various boxes in the mirror room matching the modern environment (eight are shown) b) six wood puzzles located in the rustic room called Schrödinger’s workroom. Table 4: EPs supplemented by a short description. Number | EBG puzzle | Description
---|---|---
1 | Backgammon | UV-letters on the board yielding a name of a physicist
2 | Biochemistry | Order chemical reaction fragments based on tyrosine and Iodine
3 | Domino | UV-letters on domino stones yielding a name of a physicist
4 | Einstein lock | Open the complicated lock puzzle
5 | Bottle safe | Open the safe made of wood and cord
6 | 24 dexterity cubes | For instance, put metal balls in a hole
7 | 6 Wood puzzle | Build up cubes consisting of small pieces of wood
8 | Card game | Order cards to yield the name of a physicist
9 | 10 knots | Knot knots following instructions
10 | Magnet blocks | Build an animal (here a snake) with building blocks
11 | Periodic table | Abbreviations of elements yielding a name of a physicist
12 | Wood bottle | Build a bottle made of wooden puzzle pieces
13 | Chess | Best moves in Scandinavian defence yields a name by letters
14 | 3x3 square | Assemble squares to a large 3x3 square
15 | 5 Text challenges | Logical exercises in envelopes as written text
16 | Hidden card | A card is very well hidden in a drawer

##  3 Details of the study

###  3.1 Group of interest

The students being educated in physics can be divided into three groups, as
follows:

  * •

Group A: Students who require a high-end education in physics (main study
being physics, biophysics, quantum engineering, …)

  * •

Group B: Students who require an intermediate level of education in physics
(main study being in chemistry, machine engineering, mechanical engineering,
…)

  * •

Group C: Students who should have a basic knowledge of physics (main study
being in medicine, veterinary medicine, biology, …)

The education in physics of these different groups is characterised by
specific dedicated and methodological challenges. Empirically, the levels of
motivation and interest are low in group C. The target group of our study was
group C and the eER was visited by students of veterinary medicine in their
first semester aged between 18-22 years. The students had already attended 2/3
of the lectures in physics covering the main disciplines of physics. The
majority of the students passed their experimental examination in three
laboratory courses: current of liquids, light microscopy/polarimetry and
electricity (dc). By choosing the eEERP, the education is organized in three
parts, see Figure 4.

Figure 4: Education in physics at the University of Veterinary Medicine
Hannover, Foundation includes visiting lectures, participating in the
laboratory course supplemented by the optional eEERP. Arrows visualise
dominant transfer of knowledge and skills.

###  3.2 Educational objectives

Group C should be taught a basic knowledge of physics and the goals of the
eERs are

  * •

Stabilisation/Repetition and application of knowledge in physics as learned in
the lecture

  * •

Slight extension of the knowledge by exemplification, demonstration and
individual cases associated with veterinary medicine

  * •

Training of experimental skills and hands-on-mentality

The latter point is important because it is a matter of interest in veterinary
medicine when examining patients. Especially, the corona pandemic led to a
strong request for the training of practical skills after the intensive
online-teaching. It should be mentioned that because of the broad range of
disciplines in physics, a processing time of 2.5 hours and the required
intensity of our eEERP should not be used for Group A, at least taking into
account the here presented knowledge-transfer documents.

###  3.3 Questions and procedure

The study contains two questionnaires. The first one contains 12 questions on
physics from various disciplines which can be found in the Supplementary. This
questionnaire was answered right at the beginning of the ERP by all players
individually, being separated from each other to avoid interaction (called
pretest). After the successful completion of the eEERP, they received a blank
questionnaire with the identical questions again (called posttest). One
example of a question in physics was:

> What is the single-phase alternating current frequency in the USA?
>
>   * $\square$
>
> 50 Hz
>
>   * $\square$
>
> 55 Hz
>
>   * $\square$
>
> 60 Hz
>
>   * $\square$
>
> I do not know
>
>

We added the answer ’$\square$ I do not know’ to avoid random choices. Honesty
was asked for when filling out the questionnaire. The questionnaires were
anonymous. The posttest was supplemented by an additional questionnaire asking
questions concerning the VP system, Eurogame and room concept called EER-test.
This was the second part of the study. All tests were compiled in such a way
to ensure that all answers were given by the same player.

##  4 Results

The evaluation contains two parts. The first part is summary of a set of
various straight-forward analyses of unmodified raw data according to
statistical methods for statistical significance. The second part presents a
demonstrative and simple understandable modified analysis using a strategy to
suppress large errors assuming honesty.

###  4.1 Raw data analysis: Test of Significance

Pretests and posttests were statistically evaluated without further
normalisation or modification and, thus, sorted into raw data. In the
following, this evaluation strategy is be presented as an overview of the main
results (for details, see Supplementary). The counting of the number of
correct, wrong as well as neutral answers formed the basis for the evaluation.
Four complementary statistical analysis approaches, in detail

  * •

Approach 1 was based on the total number of students $N_{1}=25$,

  * •

Approach 2 was based on the total number of questions $N_{2}=12$,

  * •

Approach 3 was based on the group-separated total number of questions
$N_{3}=12$,

  * •

Approach 4 was based on the question-separated total number of student groups
$N_{4}=8$,

were performed with the focus on a statistical comparison of the pre- and
posttest averaged number of correct answers. To achieve statistically
confirmed conclusions, statistical tests in the form of

  * •

David tests for normal (Gaussian) distribution [10],

  * •

F-tests [10, 11],

  * •

standard two-tailed t-tests (equal variances) [12],

  * •

Grubbs tests for outliers [10, 11]

were applied to the data. Overall, the evaluations revealed a statistically
(highly) significant increase in the averaged numbers of correct answers in
the posttest and, therefore, after the students had completed the ERP and
proved the success in their physics knowledge transfer by the developed eEERP.
Furthermore, the examination also demonstrated that most of the students
renounced guessing correct answers in the questions.

Figure 5: Overview of normalised increase in quantified knowledge of players
with bemusement before (only players with incorrect answer or ’I do not know’
in the pretest were considered because of measurability). Left scale: 0 means
no player has learned anything, 1 means every player has given the correct
answer after the eEERP.

###  4.2 Increase in Content Knowledge

To visualise the increase in knowledge, a sheet of questions is generated
divided into disciplines of physics as shown in Figure 5. Some players knew
the right answer in the pretest, therefore, these answers are not considered
in the statistic and pretests and posttests were excluded concerning the
corresponding questions/answers. For instance, three players knew (pretest)
that the ac-frequency is 60 Hz (USA) and were excluded. Most of the test
groups decided ’$\square$ I do not know’ and after the eEERP a considerable 16
players (72% or 0.72) chose 60 Hz, these students not knowing the answer
before. The possible error has two extremes: random choice of the answer
yields an error of 25%, meaning just 1/4 of the students chose the correct
answer. This assumes random arbitrary decision and is an effective scattered
background (see grey 25%-line). The second extreme is that the players decided
honestly, choosing ’$\square$ I do not know’ if they had truly no answer. In
this case, the effective error is zero and no background exists caused by
randomness. The truth is something in between with a tendency to the second
extreme. Therefore, the presented error was chosen to be 10% which still
overestimates the reality (request for honesty). One can clearly adumbrate
that the ’balance point’ averaged over all bars is located above the 25%-line
threshold of random choices in the posttest. This is consistent with the above
analysis of significance. The mean values are $\approx 0.44\pm 0.15$ and
$0.47\pm 0.13$ without considering the last question (efficiency of heat
engines); for reasons, see Discussion.

###  4.3 eEER, Room Concept and General Questions

General issues of the eER, the level of acceptance and quality of the Eurogame
concept were investigated by the EER-test. Figure 6 shows three of the rated
questions. A total of 80% rated the existence of a room concept (mirror room,
rustic Schrödinger’s workroom) to be very important to provide an adventure-
like atmosphere. Interestingly, no player rated an element of interest to be
unimportant. This is surprising because the story of an escape room is
cosmetic and wastes time to finish the ERP. However, 40% of the respondents
evaluated it to be just moderately important. A considerable 68% judged the
consideration of all main disciplines of physics to be very important which
emphasises the relevance of a spectral eERP. Figure 7 shows four questions
concerning the eEERP concept to investigate the Eurogame concept by VP,
reflecting the success of the eEER concept, for details see below.

Figure 6: Circle diagram of general questions concerning an ERP.
Interestingly, the choice ’unimportant’ was not elected.

##  5 Discussion

###  5.1 Questions and Tests

In general, an effort was made to find questions which could not be answered
purely from the lecture so as to be able to measure the increase in knowledge,
e.g., 3 students out of 25 could answer the 60-Hz-question above, avoided too
high a zero-rate and wasting data points for statistics. Therefore, the
questions were very specific which was retrospectively a successful strategy.
We asked two questions per discipline in physics. Exceptionally,
thermodynamics (entropy, heat engine, fridge, efficiency) covered four
questions associated with one knowledge transfer card and PP. These four
topics are less discussed in the physics lecture being a measure of knowledge
which clearly exceeded the lecture. The analysis also indicated the dependence
of this success on the individual student group due to group inhomogeneities
in willingness to learn, learning ability and group dynamics. Exemplarily,
Figure 8 shows a comparison of two groups of 3 players (group separated
treatment; for details, see Supplementary). Group 5 showed a higher level of
significance of knowledge increase. The increase is significant for the
question of entropy, heat engine and fridge. The question concerning
efficiency of a heat engine had a low value of correctness. It is worth
mentioning that, the puzzle Stirling engine was the last PP in the eEERP.
Assuming a decrease in concentration among the players, the success of the
increase was counted to be positive. Probably the notation of the last answers
$\eta<1$, $\eta=1$ or $\eta>1$ was difficult to cope with and to remember
after the first confrontation without a deep understanding. Statistically, the
last question can be counted as an outlier. The eEER-test contained two kinds
of questions: general eER questions and EER questions. A considerable 80%
rated a room concept to be very important, probably seen as a break, some sort
of adventure. The EPs and the VP system were rated as good (80%) and it could
motivate players (87%), significantly felt challenged (76%). Therefore, the
concept of VP was considered successful.

Figure 7: Percentage-bar diagrams of eEER questions. Figure 8: Statistical
evaluation based on the total number of questions ($N_{3}=12$), group-
separated and averaged numbers of correct answers per question of pre-
(orange) and posttest (green) in comparison.

###  5.2 Educational Escape Rooms for Physics

C. Lathwesen et al. found just six protocols of eERP [13]. All targeted
secondary education focusing on a specific topic. According to C. Lathwesen et
al.:

> Nevertheless, we were able to identify some gaps in the research and
> development on this topic. First of all, there is still a need for new
> proposals mainly in the subjects of physics and biology [13].

Our contribution follows this request. Interestingly, spectral eERP are extrem
discussed. Most of them covered a narrow topic/discipline. Additionally, l.
Hayden et al. requested:

> Overall, the reviewed articles indicate that more studies are needed on the
> use of escape rooms for educational purposes. It seems that research on this
> topic has reached a new phase, and there is a need for structured research
> and transparency in the research design and methods for data collection and
> analysis [3].

It is unconventional to discuss the topic of transparency and structure in a
paper here because it should be self-evident. However, the above citation
shows it is not. In physics and natural science, research is very structured
and transparent. Therefore, the effort was made to choose a structure as is
usual in physics/chemistry publications to cover the range of topics (Eurogame
strategies and broad range of disciplines in physics). Probably the above
criticism is a result of the interdisciplinary field involving various
subjects and writing styles, preparing an eER just secondarily in parallel
with other research. In this work, various ways of data analysis have been
presented for scrupulous interpretation.

###  5.3 Comparison of eEERP and Eurogames

The common advantages of eERs prevail. The mandatory sequence of PPs requires
prioritisation. This is an improvement, adding decision-making and setting
priorities for the list of promoted skills of eERs. This is an advantage of
the eEERP together with the fun-factor of the EPs. The adapted main point is
the VP system of Eurogames which allows to introduce puzzles without a strong
topical connection to the subject of interest, in this case physics. EPs have
a character of pure entertainment, exceptions such as argumentation that
dexterity cubes are able to train surgical dexterity being excluded. The
player interaction can be understood to be like a cooperative Eurogame. Groups
of players of various teams do not interact, just for clarity. We are strongly
propose adopting the ideas of EERs including EPs and applying the here
presented concept to other eER. The score list of VP is able to be reset, the
decision for every new run being left to the e-game master.

##  6 Conclusions

We presented an eEERP instrumentalised Eurogame concepts of VP as a method of
motivation for knowledge transfer in physics. We combined a strategy of
repetition and application of knowledge gained from the lecture (main part,
3/4) and the introduction of new knowledge (1/4), mainly in the field of
thermodynamics and topics strongly related with the main field of the players,
in this case veterinary medicine, e.g., the name of proteins associated with
the orientation of birds in the magnetic field of earth. An increase in
knowledge was verified together with positive feedback on the Eurogame
concept. Players have been motivated and challenged as shown by the test
groups. By the eEER, the list of advantages promoting skills can be extended
to setting priorities because of a conflict of choice between adventure-like
EPs and PPs. However, PPs are mandatory to master the eEERP. An advantage is
the flexibility of the European board concept of VP, choosing from of a range
of optional EPs to fascinate and captivate students and at the same time
taking into account individual preferences. The Eurogame concept using
arbitrary EPs to collect VP can be applied to every eER and our findings
support its implementation.

## Acknowledgments

Special thanks go to Elisabeth Schaper and Christin Kleinsorgen (University of
Veterinary Medicine Hannover Foundation) for helpful discussions regarding the
project proposal. The project was supported by the program InnovationPlus
(Niedersächsisches Ministerium für Wissenschaft und Kultur).

## Authors declaration

The authors have no conflicts to disclose.

## References

## References

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  * [3] Lene Hayden Taraldsen, Frode Olav Haara, Mari Skjerdal Lysne, Pernille Reitan Jensen, and Eirik S. Jenssen.  A review on use of escape rooms in education – touching the void.  Education Inquiry, 13(2):169–184, 2022.
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##  Appendix A List of victory points

Table 5 shows the victory points $P_{i}$ of PP and points $p_{i}$ of EPs as
used in our EERP in 2023/2024. Statistically, the students earned $\approx
20.1$ VP per player visited the eEERP.

Table 5: Overview of VP of the PPs and EPs. Puzzle |  Victory points of PPs: $P_{i}$ |  Victory points of EPs: $p_{i}$
---|---|---
Nobel prize (start-up) | 2 |
Apple puzzle | 6 |
Fluorescence | 3 |
Electric conductivity | 4 |
Total internal reflection | 3 |
50 Hertz sound experiment | 4 |
X-ray | 4 |
Dog puzzle | 8 |
Train experiment [9] | 4 |
Cup Stirling machine | 2 |
Backgammon |  | 3
Biochemistry |  | 3
Domino |  | 2
Einstein lock |  | 4
Bottle safe |  | 5
24 dexterity cubes |  | 24 total, 1 per cube
6 Wood puzzle |  | 24, 4 per puzzle
Card game |  | 4
10 knots |  | 10, 1 per knot
Magnet blocks |  | 6
Periodic table |  | 3
Wood bottle |  | 4
Chess |  | 7
3x3 square |  | 5
Text challenges |  | 20 total, 4 per text card
Hidden card |  | 2

##  Appendix B Supplements of the magnetic dog

We supplemented small magnetic experiments, being an add-on without any VP.
This is shown in Figure 9 presenting a commercial solution Magnets Experiment
Set from Burnur Science. Various conventional ferromagnets are colourised to
visualize North- and South Pole. The set contains

  * two bar magnets with wheels: magnet cars,

  * an electromagnet with paper clips to test it,

  * two sets of ring magnet levitated by forces,

  * a compass and a horseshoe magnet,

  * iron fragments, magnetic butterflies and a manual.

Figure 9: Small experiments with magnets located on mirror foil to support the
dog puzzle. Keep in mind that, the transparent box in the background contains
an oil-Fe-particle suspension to reveal magnetic field lines. It is not part
of the presented set, belonging to of the equipment used at physics of the
university.


# Exploring Functional Identities: From Division Rings to Matrix Algebras

Daniel Kawai  Daniel Kawai, São Paulo University, Rua Matão, 1010, CEP
05508-090, São Paulo, Brazil
E-mail<EMAIL_ADDRESS>and  Bruno Leonardo Macedo Ferreira  Bruno
Leonardo Macedo Ferreira, Federal University of Technology-Paraná-Brazil,
Federal University of ABC and São Paulo University, Avenida Professora Laura
Pacheco Bastos, 800, 85053-510, Guarapuava, Brazil, Av. dos Estados, 5001 -
Bangú, Santo André - SP, 09280-560 and Rua Matão, 1010, CEP 05508-090, São
Paulo, Brazil
E-mail<EMAIL_ADDRESS>
###### Abstract.

In this paper, we tackle unresolved inquiries by Ferreira et al. [3] in their
recent publication, “Functional Identity on Division Algebras”. We delve into
the intricate behavior of additive functions on matrix algebras over division
rings through rigorous analysis and theorem-proving. Our findings offer
valuable insights into the nature of these functions and their implications
for algebraic structures.

Keywords: Functional identity, division algebras, additive functions, matrix
algebras.

2020 MSC: 16K20; 16R60.

The first author was supported by FAPESP number 2022/14579-0.

##  1\. Introduction

Over the decades, research in functional equations and identities has been
pivotal in understanding the properties of mappings in algebraic structures
and their relevance across various mathematical disciplines.

In a pioneering study in 1987, Vukman [4] studied a result on Cauchy’s
functional equations in division rings. He established conditions under which
any additive mapping on a division ring with certain characteristics must be
identically zero. He showed that if $D$ is a division ring with
characteristics different from $2$ and $f:D\rightarrow D$ is an additive
function satisfying:

(1) $f(x)=-x^{2}f(x^{-1})$

for all $x\neq 0$, then $f(x)=0$ for all $x\in D$. This result, while
fundamental, left additional questions to be explored regarding functional
identities in more general algebraic structures.

Subsequently, Bre$\check{s}$ar [1] expanded this line of inquiry by
investigating functional identities involving two additive mappings, $F$ and
$G$, in a division ring. He presented a characterization theorem, revealing
specific forms that these mappings must adhere to, enriching our understanding
of the structure of these mappings.

Motivated by Bre$\check{s}$ar’s results and intending to extend these
investigations further, this article sets out to resolve the questions raised
by Catalano [2] in 2018. Catalano explored functional identities in division
rings, relating additive mappings to derivations and shedding light on their
underlying properties.

In 2023, Ferreira et al. [3] studied the functional equation:

(2) $x^{-1}f(x)+g(x^{-1})=0$

and proved the following theorem.

######  Theorem 1.1.

Let $A=M_{n}(D)$ be the algebra of $n\times n$ square matrices over a division
ring $D$ with characteristics different from $2$. Let $f,g:A\rightarrow A$ be
additive functions, satisfying identity (2) for every invertible $x$. Then
$f(x)=g(x)=0$, for all $x\in A$.

This work aims to address and resolve the open questions raised in the article
[3], which delves into the investigation of functional identities in algebra.
They posed the following two questions:

######  Question 1.1.

Let $D$ be an algebra such that there exist additive functions
$f,g:D\rightarrow D$ satisfying the following equation:

(3) $x^{-n}f(x)+g(x^{-1})=0$

for every invertible $x$, where $n$ is an integer such that $n>2$. Is it
possible to characterize the functions $f$ and $g$?

######  Question 1.2.

If $D$ is a field of characteristic $2$ and there exist additive functions
$f,g:D\rightarrow D$ satisfying identity (2) for every $x\neq 0$, is $D$ a
perfect field?

Note that Vukman’s article deals with the particular case of equation (3)
where $n=2$ and $f=g$, and the main results of Ferreira et al. article [3]
deal with the particular case of equation (3) where $n=1$.

The central focus of this work is to provide solutions to a modified version
of Vukman’s functional identity, specifically $x^{-1}f(x)+g(x^{-1})=0$, where
$f$ and $g$ are additive mappings on a division algebra. Our approach aims to
characterize the additive mappings that preserve the identity (3), exploring
theoretical aspects.

Furthermore, we extend our investigation to matrix algebras over division
algebras, examining analogous functional equations and providing
characterizations for additive mappings that preserve these identities. We
hope that this study will contribute not only to the understanding of
functional equation theory but also to the advancement of knowledge in various
areas of mathematics.

##  2\. Main Results

In this section, we present our main theorems, which address open questions
left by the recent article published by Ferreira et al. [3]. Before presenting
our primary results, it is pertinent to revisit the well-known identity
attributed to Hua, an important element for the development of this paper.

Consider any two elements $a,b\in\mathcal{A}$, ensuring $ab\neq 0,1$. We
encounter the pivotal Hua’s identity:

$a-\left[a^{-1}+\left(b^{-1}-a\right)^{-1}\right]^{-1}=aba.$

######  Theorem 2.1.

Let $n$ be an odd integer and $m$ a positive integer. Let $D$ be a division
ring with $\mathrm{Char}(D)\neq 2$. Let $A=M_{m}(D)$ be the algebra of
$m\times m$ matrices with entries in $D$. Let $A^{*}$ be the set of invertible
matrices. Let $f,g:A\rightarrow A$ be additive functions such that:

(4) $x^{-n}f(x)+g(x^{-1})=0\quad\text{if $x\in A^{*}$}$

Then $f(x)=g(x)=0$ for all $x\in A$.

###### Proof.

We proceed by delineating several steps.

Step 1: To begin, we show that $f(x)=0$ for all $x\in A^{*}$. Let $x\in
A^{*}$. Utilizing (4) with $x\mapsto x$ and $x\mapsto-x$, where $n$ is odd,
yields:

$x^{-n}f(x)+g(x^{-1})=0,$ $x^{-n}f(x)-g(x^{-1})=0,$

implying $2x^{-n}f(x)=0$, and thus $f(x)=0$.

Step 2: Next, we establish that $g(x)=0$ for all $x\in A^{*}$. For $x\in
A^{*}$, employing (4) with $x\mapsto x^{-1}$ yields:

$0=x^{n}f(x^{-1})+g(x)=g(x),$

confirming $g(x)=0$.

Step 3: Consider $h\in\\{f,g\\}$. We aim to prove that $h(x)=0$ for all $x\in
A$ such that $x^{2}=x$. For $x\in A$ satisfying $x^{2}=x$, observe:

$(1-2x)^{2}=1-4x+4x^{2}=1,$

thus $1-2x\in A^{*}$. Additionally, $1\in A^{*}$, hence:

$0=h(1-2x)=-2h(x),$

leading to $h(x)=0$.

Step 4: We conclude the proof. Let $h\in\\{f,g\\}$. For each $(i,j)$, let
$e_{ij}\in A$ denote the matrix with the $(i,j)$ entry being $1$ and the
others $0$. Then, for each $i$, $e_{ii}^{2}=e_{ii}$, implying $h(e_{ii})=0$.

  * •

For $i$ and $a\in D$, if $a\neq 0$, then $1+(a-1)e_{ii}\in A^{*}$, hence
$h(1+(a-1)e_{ii})=0$, and consequently $h(ae_{ii})=0$. Furthermore, as $h$ is
additive, $h(0e_{ii})=h(0)=0$.

  * •

For each $(i,j)$ with $i\neq j$ and for $a\in D$, $1+ae_{ij}\in A^{*}$, thus
$h(1+ae_{ij})=0$, and hence $h(ae_{ij})=0$.

Let $x\in A$, then $x$ takes the form $x=\sum_{i,j}a_{ij}e_{ij}$ where
$a_{ij}\in D$, yielding:

$h(x)=h\left(\sum_{i,j}a_{ij}e_{ij}\right)=\sum_{i,j}h\left(a_{ij}e_{ij}\right)=0$

Therefore, $h(x)=0$ for all $x\in A$. ∎

And what about the scenario when $n$ is even? We have obtained fragmented
results.

######  Theorem 2.2.

Let $n$ be an even integer such that $n>2$. Let $D$ be a division ring with
$\mathrm{Char}(D)>n$. Let $f,g:D\rightarrow D$ be additive functions such
that:

(5) $x^{-n}f(x)+g(x^{-1})=0\quad\text{if $x\neq 0$}$

Then $f(x)=g(x)=0$ for all $x\in D$.

###### Proof.

Once again, we proceed with the proof of Theorem 4.1 by delineating several
steps.

Step 1: Let $x\neq 0$. Then, utilizing (5) with $x\mapsto x$ and $x\mapsto
x^{-1}$, we have:

(6) $x^{-n}f(x)+g(x^{-1})=0$

and also:

$x^{n}f(x^{-1})+g(x)=0$

implying:

(7) $x^{-n}g(x)+f(x^{-1})=0$

Consequently, from (6) and (7), we obtain:

$x^{-n}(f+g)(x)+(f+g)(x^{-1})=0$ $x^{-n}(f-g)(x)-(f-g)(x^{-1})=0$

Step 2: Let $h\in\\{f+g,f-g\\}$. Taking $e=-1$ if $h=f+g$ and $e=1$ if
$h=f-g$, we have:

(8) $h(x)=ex^{n}h(x^{-1})\quad\text{if $x\neq 0$}$

Let $a\neq 0$ and $a\neq 1$, then by Hua’s identity, we have:

(9) $a^{2}=a-(a^{-1}+(1-a)^{-1})^{-1}$

thus:

$\begin{array}[]{rcl}h(a^{2})&=&h(a-(a^{-1}+(1-a)^{-1})^{-1})\\\
&=&h(a)-h((a^{-1}+(1-a)^{-1})^{-1})\\\
&=&h(a)-e(a^{-1}+(1-a)^{-1})^{-n}h(a^{-1}+(1-a)^{-1})\\\
&=&h(a)-ea^{n}(1-a)^{n}h(a^{-1}+(1-a)^{-1})\\\
&=&h(a)-ea^{n}(1-a)^{n}(h(a^{-1})+h((1-a)^{-1}))\\\
&=&h(a)-ea^{n}(1-a)^{n}h(a^{-1})-ea^{n}(1-a)^{n}h((1-a)^{-1})\\\
&=&h(a)-e^{2}a^{n}(1-a)^{n}a^{-n}h(a)-e^{2}a^{n}(1-a)^{n}(1-a)^{-n}h(1-a)\\\
&=&h(a)-(1-a)^{n}h(a)-a^{n}h(1-a)\\\ &=&h(a)-(1-a)^{n}h(a)-a^{n}(h(1)-h(a))\\\
&=&h(a)-(1-a)^{n}h(a)-a^{n}h(1)+a^{n}h(a)\\\
&=&(1+a^{n}-(1-a)^{n})h(a)-a^{n}h(1),\end{array}$

thus:

$h(a^{2})=(1+a^{n}-(1-a)^{n})h(a)-a^{n}h(1).$

The above equation also holds for $a=0$ and $a=1$ as well. As $n$ is even and
$n\geq 4$, we may express this as:

$\left(h(a^{2})-nah(a)\right)+\sum_{k=3}^{n-1}(-1)^{k-1}\binom{n}{k-1}a^{k-1}h(a)+\left(-na^{n-1}h(a)+a^{n}h(1)\right)=0$

Now, let:

$\begin{array}[]{rcl}t_{0}(x)&=&0\\\ t_{1}(x)&=&0\\\
t_{2}(x)&=&h(x^{2})-nxh(x)\\\
t_{k}(x)&=&(-1)^{k-1}\binom{n}{k-1}x^{k-1}h(x)\quad\text{se $3\leq k\leq
n-1$}\\\ t_{n}(x)&=&-nx^{n-1}h(x)+x^{n}h(1)\end{array}$

Each $t_{k}(x)$ is homogeneous with degree $k$ in $x$. Therefore, for all
$a\in D$, we have:

(10) $t_{0}(a)+t_{1}(a)+\cdots+t_{n}(a)=0$

Let $a\in D$, then for all $m\in{0,\dots,n}$, by applying (10) with $ma$
instead of $a$, we derive:

$m^{0}t_{0}(a)+m^{1}t_{1}(a)+\cdots+m^{n}t_{n}(a)=0,$

hence, if we set:

$V=\begin{pmatrix}0^{0}&0^{1}&\cdots&0^{n}\\\ 1^{0}&1^{1}&\cdots&1^{n}\\\
&&\vdots&\\\ n^{0}&n^{1}&\cdots&n^{n}\end{pmatrix},\quad\quad
X=\begin{pmatrix}t_{0}(a)\\\ t_{1}(a)\\\ \vdots\\\ t_{n}(a)\end{pmatrix},$

then the matrix identity $VX=0$ holds. Thus, if $U$ is the adjoint111The
adjoint matrix of $A$ is the transpose of the cofactor matrix of $A$. of $V$,
we have $UVX=0$, and since $UV=\det(V)I$, we get $\det(V)X=0$. By
Vandermonde’s determinant identity, we have:

$\det(V)=\prod_{0\leq i<j\leq n}(j-i),$

but also $\mathrm{Char}(D)>n$, so $X=0$, particularly $t_{2}(a)=t_{n}(a)=0$,
thereby yielding:

(11) $h(a^{2})=nah(a)$ (12) $na^{n-1}h(a)=a^{n}h(1)$

Utilizing (12) with $1$ in place of $a$, we obtain $(n-1)h(1)=0$, yet
$\mathrm{Char}(D)>n$, implying $h(1)=0$. Hence, for all $a\in D$ with $a\neq
0$, by (12), we have $na^{n-1}h(a)=a^{n}h(1)=0$, yet $\mathrm{Char}(D)>n$,
hence $h(a)=0$. Moreover, as $h$ is additive, we have $h(0)=0$ as well.

Step 3: Consequently, we conclude that $h(x)=0$ for all $x\in D$. Therefore,
for all $x\in D$, we have $f(x)+g(x)=f(x)-g(x)=0$, implying $2f(x)=0$, hence
$f(x)=0$ and also $g(x)=0$, completing the proof. ∎

######  Remark 2.1.

It is essential to consider the conditions on the characteristic of the ring
$D$. This necessity becomes evident in the following example.

######  Example 2.1.

Let $n$ be any integer, and $p$ be a prime such that $p-1$ divides $n-2$.
Consider $D=\mathbb{Z}_{p}$, the ring of integers modulo $p$, and the
functions:

$f(x)=x\quad\text{and}\quad g(x)=-x.$

Then $D$ is a division ring and $f,g:D\rightarrow D$ are additive functions
satisfying equation (3) for every $x\neq 0$, for the Fermat’s little theorem
implies $x^{p-1}=1$ for every $x\neq 0$. However, $f(1)\neq 0$.

Example 2.1 illustrates the necessity of the condition $\mathrm{Char}(D)\neq
2$ in Theorem 2.1, as for any integer $n$, the prime number $p=2$ satisfies
$p-1\mid n-2$. Furthermore, it underscores the significance of retaining the
condition $\mathrm{Char}(D)>n$ in Theorem 2.2. A division ring always has a
prime or infinite characteristic. Moreover, for even integers $n$ such that
$n>2$, there may exist several prime numbers $p$ satisfying $p-1\mid n-2$. For
instance, if $n=4$ or $n=6$, then every prime number $p$ such that $p\leq n$
satisfies $p-1\mid n-2$.

Now, we endeavor to provide a negative response to Question 1.2, illustrating
that within the non-perfect field $D=\mathbb{Z}_{2}(t)$ of rational functions
with coefficients in $\mathbb{Z}_{2}$, numerous solutions $(f,g)$ to the
functional equation (2) exist, where $f$ and $g$ are additive.

######  Theorem 2.3.

Let $D=\mathbb{Z}_{2}(t)$ denote the field of rational functions with
coefficients in $\mathbb{Z}_{2}$. For any $A,B\in D$, there exists a unique
pair $(f,g)$ consisting of additive functions $f,g:D\rightarrow D$ satisfying
$f(1)=A$, $f(t)=B$, and equation (2) for every $x\in D$ such that $x\neq 0$.

###### Proof.

1) Uniqueness: Let $f,g:D\rightarrow D$ be additive functions satisfying
$f(1)=A$, $f(t)=B$, and equation (2) for all $x\neq 0$. Then, for every
$a,b\in D$ such that $ab\neq 0,1$, using the Hua identity and the fact that
$\mathrm{Char}(D)=2$, we have:

$aba=a+(a^{-1}+(b^{-1}+a)^{-1})^{-1}.$

This yields:

$\displaystyle f(aba)$ $\displaystyle=f(a+(a^{-1}+(b^{-1}+a)^{-1})^{-1})$
$\displaystyle=f(a)+f((a^{-1}+(b^{-1}+a)^{-1})^{-1})$
$\displaystyle\overset{\eqref{moraes}}{=}f(a)+(a^{-1}+(b^{-1}+a)^{-1})^{-1}g(a^{-1}+(b^{-1}+a)^{-1})$
$\displaystyle=f(a)+(aba+a)g(a^{-1}+(b^{-1}+a)^{-1})$
$\displaystyle=f(a)+(aba+a)(g(a^{-1})+g((b^{-1}+a)^{-1}))$
$\displaystyle=f(a)+(aba+a)g(a^{-1})+(aba+a)g((b^{-1}+a)^{-1})$
$\displaystyle\overset{\eqref{moraes}}{=}f(a)+(aba+a)a^{-1}f(a)+(aba+a)(b^{-1}+a)^{-1}f(b^{-1}+a)$
$\displaystyle=f(a)+(ab+1)f(a)+ab(a+b^{-1})(b^{-1}+a)^{-1}f(b^{-1}+a)$
$\displaystyle=abf(a)+abf(b^{-1}+a)$ $\displaystyle=abf(a)+ab(f(b^{-1})+f(a))$
$\displaystyle=abf(a)+abf(b^{-1})+abf(a)$
$\displaystyle=abf(a)+abb^{-1}g(b)+abf(a)$ $\displaystyle=abf(a)+ag(b)+abf(a)$
$\displaystyle=ag(b),$

which simplifies to:

(13) $f(aba)=ag(b).$

By $f(0)=g(0)=0$ and (2), equation (13) also holds for $a,b\in D$ such that
$ab=0$ or $ab=1$. Thus, (13) holds for all $a,b\in D$. By (13) with $a=1$, for
all $b\in D$, we have $f(b)=g(b)$. So $f=g$, and for all $a\in D$, we have:

(14) $f(aba)=af(b).$

For every $n\in\mathbb{Z}$, by (14) with $a=t$ and $b=t^{n}$, we have:

(15) $f(t^{n+2})=tf(t^{n}).$

Therefore, by using induction on $n$ and (15), for all $n\in\mathbb{Z}$, we
have:

$f(t^{2n})=t^{n}A,\quad f(t^{2n+1})=t^{n}B.$

Moreover, if $P(t)$ is a polynomial with coefficients in $\mathbb{Z}_{2}$ and
$p\in\mathbb{Z}$, then it’s evident that:

(16) $f(P(t^{2})t^{2p})=P(t)t^{p}A,\quad f(P(t^{2})t^{2p+1})=P(t)t^{p}B.$

Any element $x\in D$ can be represented as:

$x=\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t},$

where $P(t),Q(t),R(t),S(t)$ are polynomials with coefficients in
$\mathbb{Z}_{2}$. Hence, by representing:

$P(t)=\sum_{p=0}^{\hat{p}}P_{p}t^{p},\quad
Q(t)=\sum_{q=0}^{\hat{q}}Q_{q}t^{q},$

where $P_{p},Q_{q}\in\mathbb{Z}_{2}$, we have:

$\displaystyle f(x)$
$\displaystyle=f\left(\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t}\right)$
$\displaystyle=\sum_{p=0}^{\hat{p}}P_{p}f\left(\frac{t^{2p}}{R(t^{2})+S(t^{2})t}\right)+\sum_{q=0}^{\hat{q}}Q_{q}f\left(\frac{t^{2q+1}}{R(t^{2})+S(t^{2})t}\right)$
$\displaystyle\overset{\text{\eqref{moraes}}}{=}\sum_{p=0}^{\hat{p}}\frac{P_{p}t^{2p}}{R(t^{2})+S(t^{2})t}f\left(\frac{R(t^{2})+S(t^{2})t}{t^{2p}}\right)+\sum_{q=0}^{\hat{q}}\frac{Q_{q}t^{2q+1}}{R(t^{2})+S(t^{2})t}f\left(\frac{R(t^{2})+S(t^{2})t}{t^{2q+1}}\right)$
$\displaystyle=\sum_{p=0}^{\hat{p}}\frac{P_{p}t^{2p}f(R(t^{2})t^{-2p}+S(t^{2})t^{-2p+1})}{R(t^{2})+S(t^{2})t}+\sum_{q=0}^{\hat{q}}\frac{Q_{q}t^{2q+1}f(R(t^{2})t^{-2q-1}+S(t^{2})t^{-2q})}{R(t^{2})+S(t^{2})t}$
$\displaystyle\overset{\text{\eqref{18}}}{=}\sum_{p=0}^{\hat{p}}\frac{P_{p}t^{p}(R(t)A+S(t)B)}{R(t^{2})+S(t^{2})t}+\sum_{q=0}^{\hat{q}}\frac{Q_{q}t^{q}(R(t)B+S(t)tA)}{R(t^{2})+S(t^{2})t}$
$\displaystyle=\frac{P(t)(R(t)A+S(t)B)}{R(t^{2})+S(t^{2})t}+\frac{Q(t)(R(t)B+S(t)tA)}{R(t^{2})+S(t^{2})t}$
$\displaystyle=\frac{P(t)R(t)+Q(t)S(t)t}{R(t^{2})+S(t^{2})t}A+\frac{P(t)S(t)+Q(t)R(t)}{R(t^{2})+S(t^{2})t}B,$

thus:

(17)
$f\left(\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t}\right)=\frac{P(t)R(t)+Q(t)S(t)t}{R(t^{2})+S(t^{2})t}A+\frac{P(t)S(t)+Q(t)R(t)}{R(t^{2})+S(t^{2})t}B.$

2) Existence: It remains to show that for any $A,B\in D$, there exists a
function $f:D\rightarrow D$ such that (17) holds for any polynomials
$P(t),Q(t),R(t),S(t)$ with coefficients in $\mathbb{Z}_{2}$, and to prove that
indeed $f$ is additive, $f(1)=A$, $f(t)=B$, and if $g=f$, then $f$ and $g$
satisfy (2) for all $x\neq 0$.

This task can be divided into steps. Notice that for every polynomial $P(t)$,
we have:

(18) $P(t^{2})=P(t)^{2},$

because the coefficients are in $\mathbb{Z}_{2}$.

Step 1: We will prove the existence of a function $f:D\rightarrow D$
satisfying (17) for any polynomials $P(t),Q(t),R(t),S(t)$ with coefficients in
$\mathbb{Z}_{2}$. In other words, we will show that the function $f$ as
described in (17) is well-defined.

Let $P(t),\hat{P}(t),Q(t),\hat{Q}(r),R(t),\hat{R}(t),S(t),\hat{S}(t)$ be
polynomials such that:

$\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t}=\frac{\hat{P}(t^{2})+\hat{Q}(t^{2})t}{\hat{R}(t^{2})+\hat{S}(t^{2})t}.$

Then:

$(P(t^{2})+Q(t^{2})t)(\hat{R}(t^{2})+\hat{S}(t^{2})t)=(\hat{P}(t^{2})+\hat{Q}(t^{2})t)(R(t^{2})+S(t^{2})t).$

Hence:

$\displaystyle(P(t^{2})\hat{R}(t^{2})+Q(t^{2})\hat{S}(t^{2})t^{2})+(P(t^{2})\hat{S}(t^{2})+Q(t^{2})\hat{R}(t^{2}))t$
$\displaystyle=(\hat{P}(t^{2})R(t^{2})+\hat{Q}(t^{2})S(t^{2})t^{2})+(\hat{P}(t^{2})S(t^{2})+\hat{Q}(t^{2})R(t^{2}))t.$

Both sides of the above equality are polynomials over $t$, so we can compare
their coefficients and obtain the following equalities:

(19) $P(t)\hat{R}(t)+Q(t)\hat{S}(t)t=\hat{P}(t)R(t)+\hat{Q}(t)S(t)t$ (20)
$P(t)\hat{S}(t)+Q(t)\hat{R}(t)=\hat{P}(t)S(t)+\hat{Q}(t)R(t)$

Multiplying (19) by $R(t)\hat{R}(t)$ and by $S(t)\hat{S}(t)t$, and multiplying
(20) by $R(t)\hat{S}(t)t$ and by $S(t)\hat{R}(t)t$, we obtain respectively:

$\displaystyle
P(t)R(t)\hat{R}(t)^{2}+Q(t)R(t)\hat{R}(t)\hat{S}(t)t=\hat{P}(t)\hat{R}(t)R(t)^{2}+\hat{Q}(t)\hat{R}(t)R(t)S(t)t$
$\displaystyle
P(t)S(t)\hat{R}(t)\hat{S}(t)t+Q(t)S(t)\hat{S}(t)^{2}t^{2}=\hat{P}(t)\hat{S}(t)R(t)S(t)t+\hat{Q}(t)\hat{S}(t)S(t)^{2}t^{2}$
$\displaystyle
P(t)R(t)\hat{S}(t)^{2}t+Q(t)R(t)\hat{R}(t)\hat{S}(t)t=\hat{P}(t)\hat{S}(t)R(t)S(t)t+\hat{Q}(t)\hat{S}(t)R(t)^{2}t$
$\displaystyle
P(t)S(t)\hat{R}(t)\hat{S}(t)t+Q(t)S(t)\hat{R}(t)^{2}t=\hat{P}(t)\hat{R}(t)S(t)^{2}t+\hat{Q}(t)\hat{R}(t)R(t)S(t)t$

By summing up the above four equations, and using $\mathrm{Char}(D)=2$ and
(18), we obtain:

$(P(t)R(t)+Q(t)S(t)t)(\hat{R}(t^{2})+\hat{S}(t^{2})t)=(\hat{P}(t)\hat{R}(t)+\hat{Q}(t)\hat{S}(t)t)(R(t^{2})+S(t^{2})t)$

so that:

(21)
$\frac{P(t)R(t)+Q(t)S(t)t}{R(t^{2})+S(t^{2})t}=\frac{\hat{P}(t)\hat{R}(t)+\hat{Q}(t)\hat{S}(t)t}{\hat{R}(t^{2})+\hat{S}(t^{2})t}$

Now, multiplying (19) by $S(t)\hat{R}(t)$ and by $R(t)\hat{S}(t)$, and
multiplying (20) by $S(t)\hat{S}(t)t$ and by $R(t)\hat{R}(t)$, we obtain
respectively:

$\displaystyle
P(t)S(t)\hat{R}(t)^{2}+Q(t)S(t)\hat{R}(t)\hat{S}(t)t=\hat{P}(t)\hat{R}(t)R(t)S(t)+\hat{Q}(t)\hat{R}(t)S(t)^{2}t$
$\displaystyle
P(t)R(t)\hat{R}(t)\hat{S}(t)+Q(t)R(t)\hat{S}(t)^{2}t=\hat{P}(t)\hat{S}(t)R(t)^{2}+\hat{Q}(t)\hat{S}(t)R(t)S(t)t$
$\displaystyle
P(t)S(t)\hat{S}(t)^{2}t+Q(t)S(t)\hat{R}(t)\hat{S}(t)t=\hat{P}(t)\hat{S}(t)S(t)^{2}t+\hat{Q}(t)\hat{S}(t)R(t)S(t)t$
$\displaystyle
P(t)R(t)\hat{R}(t)\hat{S}(t)+Q(t)R(t)\hat{R}(t)^{2}=\hat{P}(t)\hat{R}(t)R(t)S(t)+\hat{Q}(t)\hat{R}(t)R(t)^{2}$

By adding the above four equations, and using $\mathrm{Char}(D)=2$ and (18),
we have:

$(P(t)S(t)+Q(t)R(t))(\hat{R}(t^{2})+\hat{S}(t^{2})t)=(\hat{P}(t)\hat{S}(t)+\hat{Q}(t)\hat{R}(t))(R(t^{2})+S(t^{2})t)$

so that:

(22)
$\frac{P(t)S(t)+Q(t)R(t)}{R(t^{2})+S(t^{2})t}=\frac{\hat{P}(t)\hat{S}(t)+\hat{Q}(t)\hat{R}(t)}{\hat{R}(t^{2})+\hat{S}(t^{2})t}$

Multiplying (21) by $A$, multiplying (22) by $B$, and summing, we obtain:

$\frac{P(t)R(t)+Q(t)S(t)t}{R(t^{2})+S(t^{2})t}A+\frac{P(t)S(t)+Q(t)R(t)}{R(t^{2})+S(t^{2})t}B$
$=\frac{\hat{P}(t)\hat{R}(t)+\hat{Q}(t)\hat{S}(t)t}{\hat{R}(t^{2})+\hat{S}(t^{2})t}A+\frac{\hat{P}(t)\hat{S}(t)+\hat{Q}(t)\hat{R}(t)}{\hat{R}(t^{2})+\hat{S}(t^{2})t}B$

Step 2: Let’s prove that $f$ is additive. Take $x$ and $y$ from $D$. By
bringing the fractions to a common denominator, we can express them as
polynomials $P(t)$, $\hat{P}(t)$, $Q(t)$, $\hat{Q}(t)$, $R(t)$, and $S(t)$ as
follows:

$x=\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t},\quad\quad
y=\frac{\hat{P}(t^{2})+\hat{Q}(t^{2})t}{R(t^{2})+S(t^{2})t},$

resulting in:

$x+y=\frac{(P(t^{2})+\hat{P}(t^{2}))+(Q(t^{2})+\hat{Q}(t^{2}))t}{R(t^{2})+S(t^{2})t}.$

Therefore,

$\displaystyle f(x+y)$
$\displaystyle\overset{\eqref{17}}{=}\frac{(P(t)+\hat{P}(t))R(t)+(Q(t)+\hat{Q}(t))S(t)t)}{R(t^{2})+S(t^{2})t}A$
$\displaystyle+\frac{(P(t)+\hat{P}(t))S(t)+(Q(t)+\hat{Q}(t))R(t)}{R(t^{2})+S(t^{2})t}B$
$\displaystyle=\frac{P(t)R(t)+Q(t)S(t)t}{R(t^{2})+S(t^{2})t}A+\frac{P(t)S(t)+Q(t)R(t)}{R(t^{2})+S(t^{2})t}B$
$\displaystyle+\frac{\hat{P}(t)R(t)+\hat{Q}(t)S(t)t}{R(t^{2})+S(t^{2})t}A+\frac{\hat{P}(t)S(t)+\hat{Q}(t)R(t)}{R(t^{2})+S(t^{2})t}B$
$\displaystyle\overset{\eqref{17}}{=}f(x)+f(y).$

Step 3: Let’s establish that $f(1)=A$ and $f(t)=B$. Indeed, we have:

$1=\frac{1+0t}{1+0t},$

thus,

$f(1)\overset{\eqref{17}}{=}\frac{1\cdot 1+0\cdot 0t}{1+0t}A+\frac{1\cdot
0+0\cdot 1}{1+0t}B=A.$

Moreover:

$t=\frac{0+1t}{1+0t},$

therefore,

$f(t)\overset{\eqref{17}}{=}\frac{0\cdot 1+1\cdot 0t}{1+0t}A+\frac{0\cdot
0+1\cdot 1}{1+0t}B=B.$

Step 4: Let’s show that if $g=f$, then $(f,g)$ satisfy (2) for all $x\neq 0$.
Let $x\neq 0$, then we have polynomials $P(t)$, $Q(t)$, $R(t)$, and $S(t)$
such that:

$x=\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t},$

hence,

$x^{-1}=\frac{R(t^{2})+S(t^{2})t}{P(t^{2})+Q(t^{2})t},$

thus,

$\displaystyle xf(x^{-1})$
$\displaystyle\overset{\eqref{17}}{=}\frac{P(t^{2})+Q(t^{2})t}{R(t^{2})+S(t^{2})t}\cdot\left(\frac{R(t)P(t)+S(t)Q(t)t}{P(t^{2})+Q(t^{2})t}A+\frac{R(t)Q(t)+S(t)P(t)}{P(t^{2})+Q(t^{2})t}B\right)$
$\displaystyle=\frac{P(t)R(t)+Q(t)S(t)t}{R(t^{2})+S(t^{2})t}A+\frac{P(t)S(t)+Q(t)R(t)}{R(t^{2})+S(t^{2})t}B$
$\displaystyle\overset{\eqref{17}}{=}f(x),$

so, using $\mathrm{Char}(D)=2$, we find:

$x^{-1}f(x)+f(x^{-1})=0.$

Hence, if $g=f$, then $(f,g)$ satisfy (2) for all $x\neq 0$.

Conclusion: Thus, given any $A,B\in D$, we have proven both uniqueness and
existence of the additive functions $f,g:D\rightarrow D$ satisfying (2) for
every $x\neq 0$. Therefore, the theorem is established. ∎

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