Proving $ \sum_{x \in A} x^k = \sum_{x \in B} x^k $ for a partitioning of a specific set.

Asked Nov 12 '21 at 19:27

Active Nov 12 '21 at 19:51

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For each $n \in \mathbb {N}$ prove that we can partition set $\{1,2,...,2^n\}$ into subsets $A$ and $B$ so that for each $0\le k \le n-1$ we have: $$ \sum_{x \in A} x^k = \sum_{x \in B} x^k $$

For this question, I thought of using mathematical induction and binomial expansion but I don't know how to use it here.

I would really appreciate it if someone could help me out.

edited Nov 12 '21 at 19:51

asked Nov 12 '21 at 19:27

ArithEgo

1

How is this even true? When $n=1$ we have ${ 1,2 }$ and I don't see how for any $k$ except $0$ it's possible to partition these. If $k=1$ you can't have $1 = 2$. – A. Thomas Yerger Nov 12 '21 at 19:31
@AlfredYerger i'm sorry i made a mistake in translating the question to english. i edited it. i guess the problem is solved. – ArithEgo Nov 12 '21 at 19:52
It's still not right though, because if $n$ is odd, $n-1$ is even, and the range from $0$ to $n-1$ is odd. So for $k=0$, there are an odd number of $1$'s to partition into equal sets. – A. Thomas Yerger Nov 12 '21 at 20:50
You should be able to do a better job writing your questions. Check out our guide for new askers for several pointers and general ways of improving the questions. Otherwise your questions will attract negative attention. – Jyrki Lahtonen Nov 12 '21 at 21:04
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@AlfredYerger This is a relatively well known problem due to Prouhet, Tarry and Escott. A solution is related to the Thue-Morse sequence. – Jyrki Lahtonen Nov 12 '21 at 21:08

Proving $ \sum_{x \in A} x^k = \sum_{x \in B} x^k $ for a partitioning of a specific set.

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