I am trying to solve a problem related to harmonic functions, but I do not see how to begin. The problem is
Let $Ω \subset \mathbb{R}^{n}$ be an open and connected set. Let $u : \Omega \to \mathbb{R}$ be a harmonic function. Suppose that for a pair of points $(x_0, y_0) \in \Omega \times \Omega$, $u$ satisfies $u(x_0) + u(y_0) = M$. Then there are infinitely many pairs $(x, y) \in \Omega \times \Omega$ such that $u(x) + u(y) = M$.
I think that it has something to do with mean value properties. Any tip to approach this problem?.
Suppose on the contrary, that there are only finitely many such pairs, then we can find $r>0$ such that the ball $B_r(x_0,y_0)\subset \Omega\times \Omega$ contains no other such point, i.e. $\forall(x,y)\in B_r(x_0,y_0)\setminus \{(x_0,y_0)\}: u(x)+u(y)\neq M$.
Now we define the harmonic function $\tilde{u}:B_r(x_0,y_0)\to \mathbb{R}$ by $\tilde{u}(x,y)=u(x)+u(y)$. If $M=\max_{(x,y)\in B_r(x_0,y_0)} \tilde{u}(x,y)$, then we can conclude by the maximum principle that $\tilde{u}$ is constant on $B_r(x_0,y_0)$ and this contradicts the assumption that $(x_0,y_0)$ is the only point in the ball that attains the value $M$. Similarly, we can use the minimum principle to conclude that $M\neq \min_{(x,y)\in B_r(x_0,y_0)} \tilde{u}(x,y)$.
Therefore we have shown that there exists points $(x_1,y_1), (x_2,y_2)\in B_r(x_0,y_0)$ such that \begin{align} \tilde{u}(x_1,y_1)<M<\tilde{u}(x_2,y_2). \end{align} Now we can choose a contiuous curve $c:[0,1]\to B_r(x_0,y_0)$ with $c(0)=(x_1,y_1)$ and $c(1)=(x_2,y_2)$ such that $c(t)\neq (x_0,y_0)$ for all $t\in[0,1]$. If we denote the composition by $f(t)=\tilde{u}(c(t))$, then by construction we have $f(0)<M<f(1)$ and by the intermediate value theorem we can conclude that there has to be a point $t'$ such that $\tilde{u}(c(t'))=f(t')=M$.
Since $c(t')\neq (x_0,y_0)$ we have found another point in $B_r(x_0,y_0)$ that attains the value $M$, which contradicts our assumption that there are only finitely many such points in $\Omega\times\Omega. $