TST

First, we will prove the following lemma:

Lemma. Given a positive integer $n$. Prove that the system of linear equations with $n$ variables $(x_1,x_2,\dots ,x_n)$ $$\{\begin{array}{l}{a}_{11}{x}_1+\cdots +a_{1n}{x}_n=b_1,\\ \cdots \\ a_{n1}{x}_1+\cdots +a_{nn}{x}_n=b_n\end{array}(4.1)$$ has exactly one solution if the associated homogeneous system (which means $b_1=\cdots =b_n=0$) has only one solution $x_1=\cdots =x_n=0$.

Proof. Assume that both of $(x_1,\dots ,x_n)$ and $(y_1,\dots ,y_n)$ are the solutions of this system, we have $$\{\begin{array}{rl}& a_{11}(x_1-y_1)+\cdots +a_{1n}(x_n-y_n)=0,\\ & \cdots \\ & a_{n1}(x_1-y_1)+\cdots +a_{nn}(x_n-y_n)=0\end{array}$$ hence by the given condition, we obtain that $x_1-y_1=x_2-y_2=\cdots =x_n-y_n=0$, which means the system has at most one solution.

We will prove this system always has a solution by induction for $n$. It is obvious for $n=1$. Assume that the lemma is proved for $n-1$. It is clear that if $a_{ij}=0$ for all pairs $(i,j)$ then the associated homogeneous system has infinite solutions, hence there must exist $a_{ij}\ne 0$. Without loss of generality, assume that $a_{nn}\ne 0$. The system can be rewritten as follows $$\{\begin{array}{rl}& \sum _{i=1}^{n-1}\left(a_{1i}-a_{ni}\frac{a_{1n}}{a_{nn}}\right)x_i=b_1-b_n\frac{a_{1n}}{a_{nn}},\\ & \sum _{i=1}^{n-1}\left(a_{2i}-a_{ni}\frac{a_{2n}}{a_{nn}}\right)x_i=b_2-b_n\frac{a_{2n}}{a_{nn}},\\ & \cdots \\ & \sum _{i=1}^{n-1}\left(a_{n-1,i}-a_{ni}\frac{a_{n-1,n}}{a_{nn}}\right)x_i=b_{n-1}-b_n\frac{a_{n-1,n}}{a_{nn}},\\ & \sum _{i=1}^n{a}_{ni}{x}_i=b_n\end{array}$$ Clearly, if the system $$\{\begin{array}{rl}& \sum _{i=1}^{n-1}\left(a_{1i}-a_{ni}\frac{a_{1n}}{a_{nn}}\right)x_i=0,\\ & \sum _{i=1}^{n-1}\left(a_{2i}-a_{ni}\frac{a_{2n}}{a_{nn}}\right)x_i=0,\\ & \cdots \\ & \sum _{i=1}^{n-1}\left(a_{n-1,i}-a_{ni}\frac{a_{n-1,n}}{a_{nn}}\right)x_i=0\end{array}$$ has a solution $(y_1,y_2,\dots ,y_{n-1})\ne (0,0,\dots ,0)$ then the homogeneous system with $n$ variables $(x_1,x_2,\dots ,x_n)$ $$\{\begin{array}{l}\sum _{i=1}^{n-1}\left(a_{1i}-a_{ni}\frac{a_{1n}}{a_{nn}}\right)x_i=0,\\ \sum _{i=1}^{n-1}\left(a_{2i}-a_{ni}\frac{a_{2n}}{a_{nn}}\right)x_i=0,\\ ⋮\\ \sum _{i=1}^{n-1}\left(a_{n-1,i}-a_{ni}\frac{a_{n-1,n}}{a_{nn}}\right)x_i=0,\\ \sum _{i=1}^n{a}_{ni}{x}_i=0\end{array}$$ has a root $(y_1,y_2,\dots ,y_{n-1},0)$, which is a contradiction. Thus, applying the assumption for $n-1$, the system $$\{\begin{array}{l}\sum _{i=1}^{n-1}\left(a_{1i}-a_{ni}\frac{a_{1n}}{a_{nn}}\right)x_i=b_1-b_n\frac{a_{1n}}{a_{nn}},\\ \sum _{i=1}^{n-1}\left(a_{2i}-a_{ni}\frac{a_{2n}}{a_{nn}}\right)x_i=b_2-b_n\frac{a_{2n}}{a_{nn}},\\ ⋮\\ \sum _{i=1}^{n-1}\left(a_{n-1,i}-a_{ni}\frac{a_{n-1,n}}{a_{nn}}\right)x_i=b_{n-1}-b_n\frac{a_{n-1,n}}{a_{nn}}\end{array}$$ has exactly one solution $(z_1,\dots ,z_{n-1})$ and note that $$x_n=\frac{b_n-{\sum }_{i=1}^{n-1}{a}_{ni}{b}_i}{a_{nn}},$$ which implies that the lemma is also true for $n$. $\square$

Back to our problem, let $a_1,a_2,\dots ,a_{2019}$ be the remaining numbers on 2019 faces and denote $a_{2020}=4,a_{2021}=26,a_{2022}=2022$. Next, we write $b_{i,j}=1$ if the face containing $a_i$ has a common edge with the face containing $a_j$, otherwise we write $b_{i,j}=0$. Denote $$b_{ii}=-\sum _{j=1,j\ne i}^n{b}_{ij}.$$ By the given conditions, we have the following system $$\{\begin{array}{l}\sum _{j=1}^{2019}{b}_{1,j}{a}_j=-4b_{1,2020}-26b_{1,2021}-2022b_{1,2022},\\ \\ \sum _{j=1}^{2019}{b}_{2,j}{a}_j=-4b_{2,2020}-26b_{2,2021}-2022b_{2,2022},\\ \\ ⋮\\ \\ \sum _{j=1}^{2019}{b}_{2019,j}{a}_j=-4b_{2019,2020}-26b_{2019,2021}-2022b_{2019,2022}.\end{array}$$ Applying the lemma, it is clear that we only need to prove the system $$\{\begin{array}{l}\sum _{j=1}^{2019}{b}_{1,j}{a}_j=0,\\ \\ \sum _{j=1}^{2019}{b}_{2,j}{a}_j=0,\\ \\ ⋮\\ \\ \sum _{j=1}^{2019}{b}_{2019,j}{a}_j=0,\end{array}$$ has exactly one solution $a_1=a_2=\cdots =a_{2019}=0$. Assume that this system has another solution, which means there exists $j$ such that $a_j\ne 0$. Without loss of generality, assume that $$a_1=\max \{a_i:1\le i\le 2019\}>0.$$ We observe that $$M\left(\sum _{i=2}^{2019}{b}_{1,i}\right)\ge \sum _{i=2}^{2019}{b}_{1,i}{a}_i=a_1\left(\sum _{i=2}^{2022}{b}_{1,i}\right)\ge M\left(\sum _{i=2}^{2019}{b}_{1,j}\right),$$ which means all the equalities must attain, or $a_i=M$ if $b_{1,i}=1$ and $$b_{1,2020}=b_{1,2021}=b_{1,2022}=0.$$ Similarly, we obtain that for every $a_i=M$ then all the faces that have a common edge with $a_i$ contain $M$ and the face that contains $a_i$ has no common edge with the faces that contain $a_{2022},a_{2021}$ and $a_{2020}$. Denote $A=\{i\ge 2019:a_i=M\},B=\{1,2,\dots ,2022\}\setminus A$. Clearly, there exists $i\in A,j\in B$ such that the face containing $a_i$ has a common edge with the face containing $a_j$, which is a contradiction. Hence, the assumption is wrong, which means there exists $a_1$ such that $a_1<0$. Consider the solution $$(c_1,c_2,\dots ,c_{2019})=(-a_1,-a_2,\dots ,-a_{2019}),$$ we have a solution with the biggest number positive, which is a contradiction. Thus, $a_i=0$ for all $i$ from 1 to 2019 is the only solution of that homogeneous system. Applying the lemma, the system has exactly one solution. $\square$