Circumcircle and Square Let $P$ be a point on the circumcircle of square $ABCD$. Find all integers
$n > 0$ such that the sum
$$S^n(P) = PA^n + PB^n + PC^n + PD^n$$
is constant with respect to point $P$.
 A: $PA^2+PB^2+PC^2+PD^2$ is the moment of inertia of $\{A,B,C,D\}$ with respect to $P$. By the parallel axis theorem, such sum is constant if $P$ lies on a circle centered at the centroid of $\{A,B,C,D\}$. In a square the centroid and the circumcenter are the same point, hence the statement holds for $n=2$.
Let $X$ be the midpoint of the $AB$ arc in the circumcircle of $ABCD$. It is pretty simple to check that $S^n(X)\neq S^n(A)$ if $n\neq 2$, hence $n=2$ is the only solution.
A: Say $x$ is a side of $ABCD$ and $a=PA$, $b=PB$...
If $n=1$ then the sum is not constant:
Say $P$ is on smaller arc $BC$. Then by Ptolomey theorem for $ABPC$ we have $$ax =bx\sqrt{2}+cx$$ and by Ptolomey theorem for $DBPC$ we have $$dx= bx+cx\sqrt{2}$$ 
so $$a+d = (b+c)(1+\sqrt{2}) \Longrightarrow a+b+c+d = (2+\sqrt{2})(b+c)$$ 
Since obviously $b+c$ is not constant aslo $a+b+c+d$ is not constant. 
If $n=2$ the sum is constant:
Then by theorem of Pythagoras we have $$a^2+c^2 = 2x^2\;\;\;{\rm and}\;\;\;b^2+d^2 = 2x^2\Longrightarrow S^2(P) = 4x^2$$ 
