Exist a theorem known as British Flag Theorem. It say that in a rectangle $ABCD$ we have $PA^2 - PB^2 + PC^2 - PD^2 = 0$, for any point $P$ in the plane.
I was thinking in a type of converse of this theorem. Given a polygon with vertices $A_1,A_2,\ldots,A_n$, consider the function
$$f(P) = \sum_{i=1}^{n} (-1)^{i} {PA_i}^2, $$
where $P$ is a point in the plane. So I conjectured that
If a polygon with vertices $A_1,A_2,\ldots,A_n$ is such that $f(P) = 0$, for all point $P$ in the plane, then this polygon is a rectangle.
I can to prove easily that if $n = 4$, then this is true. Somebody know this problem? This conjecture is true? If this question is easy, but requires a expert argument, can to give me a hint of how to solve this?
I think you can use linear algebra to get a necessary and sufficient condition for your $f$ to be zero.
Write points in the plane as vectors. Then the hypothesis is that $A_1, \ldots, A_n$ are such that $$\sum_{i=1}^n (-1)^i ||P-A_i||^2 = 0$$ for all $P$. You can expand this into $$\sum_{i=1}^n (-1)^i\langle P-A_i, P-A_i \rangle = 0$$ or $$(\sum_{i=1}^n (-1)^i)||P||^2 + \sum_{i=1}^n (-1)^i ||A_i||^2 -2\langle P , \sum_{i=1}^n (-1)^i A_i \rangle = 0.$$ If this is true for $P$, it must also be true for $kP$ for any scalar $k$. So for a fixed $P$ you have a polynomial of degree $2$ in $k$, which must vanish identically. It follows that a necessary and sufficient condition to have $f(P)$ identically zero is
You can find many polygons satisfying these. For example, let all the $A_i$ have equal length and let $n$ be even. Then you just need $A_1-A_2 + \cdots - A_{n-1} + A_n =0$.