I've learned that $[0,2^{n+1}-1]$ can be divided into two sets $A,B$, each with $2^n$ elements, such that \begin{equation} A^k:=\sum_{a\in A}a^k,\;\; B^k:=\sum_{b\in B}b^k,\;\; \text{and}\;\;A^k=B^k \end{equation} for all $k=0,1,\cdots,n$. In fact, one set has all elements with even number of $1$'s in their binary expression and the other set with odd number of $1$'s. For example $n=2$, we have $A=\{0,3,5,6\}$, $B=\{1,2,4,7\}$ and (define $0^0=1$) \begin{equation} 4=4,\quad 0+3+5+6=1+2+4+7,\quad 0^2+3^2+5^2+6^2=1^2+2^2+4^2+7^2. \end{equation} Using induction, one can prove the above properties like this: for $n+1$, for $k=0,\cdots,n+1$, \begin{align} A'^k&=\sum_{a\in A}(a0)^k+\sum_{b\in B}(b1)^k=2^k\left(\sum_{a\in A}a^k+\sum_{b\in B}b^k\right)+\sum_{i=0}^{k-1}\sum_{b\in B}\binom{k}{i}(2b)^i\\ &=2^k\left(\sum_{a\in A}a^k+\sum_{b\in B}b^k\right)+\sum_{i=0}^{k-1}\sum_{b\in B}\binom{k}{i}(2a)^i=\sum_{a\in A}(a1)^k+\sum_{b\in B}(b0)^k=B'^k. \end{align} It is also simple to show their sums of the next power are not equal.
My question is, how to prove or disprove this is the only partition such that their sums of $k-$th power, for $k=0,\cdots,n$, are equal?
