Here is an old problem I have no solution for it. I don't know where did I find it. However I find it very interesting. What about you?

Let the function $f: \mathbb{R}^2 \to \mathbb{R}$ be such that for each square $ABCD\in \mathbb{R}^2$ we have $$f(A)+f(C)= f(B)+f(D)$$

Show that $f$ is constant function.


I'll show, that the function $f(x,y)=x+y$ gives a counterexample.
First I assume that the square is a rotation of the square with corners $(1,0)$, $(0,1)$, $(-1,0)$ and $(0,-1)$. Say we rotate this square by an angle $\varphi$. Then the corners are $(cos(\varphi),sin(\varphi))$, $(-sin(\varphi), cos(\varphi))$, $(-cos(\varphi),-sin(\varphi))$ and $(sin(\varphi), -cos(\varphi))$. This square satisfies the given functional equation. If you have an arbitrary square in the plain, it is obtained from one of the rotations by linear transformations, i.e. by multiplying with scalars and translation. By linearity of $f$ the functional equation also holds here.

  • $\begingroup$ If you want, you can replace f by any linear map here. $\endgroup$ – tiefi Jul 20 '17 at 7:32

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