[This is an attempt to rescue the deleted question https://math.stackexchange.com/questions/4230138/sum-of-a-list-of-squares]


Is there a set of $4042$ distinct positive integers that can be partitioned into two sets of $2021$ elements each with equal sums and equal sums of squares, that is, with
$$a_1+a_2+\cdots+a_{2021}=b_1+b_2+\cdots+b_{2021}$$ and
$$a_1^2+a_2^2+\cdots+a_{2021}^2=b_1^2+b_2^2+\cdots+b_{2021}^2? $$

I can do it with three numbers on a side: $1^k+5^k+6^k=2^k+3^k+7^k$ for $k=1,2$,
and four numbers on a side: $1^k+4^k+6^k+7^k=2^k+3^k+5^k+8^k$ for $k=1,2$,
but $2021$ numbers on a side is a bit much for trial and error and/or hand calculation.
Is there an efficient, systematic way to produce these things?

  • $\begingroup$ I have accepted my own answer to this question, but I welcome other answers. $\endgroup$ Apr 2, 2022 at 7:53

1 Answer 1


If $c_1+c_2+\cdots+c_n=d_1+d_2+\cdots+d_n$, then for all $t$ we get $c_1+\cdots+c_n+(d_1+t)+\cdots+(d_n+t)=(c_1+t)+\cdots+(c_n+t)+d_1+\cdots+d_n$ and $c_1^2+\cdots+c_n^2+(d_1+t)^2+\cdots+(d_n+t)^2=(c_1+t)^2+\cdots+(c_n+t)^2+d_1^2+\cdots+d_n^2$
as one sees by just multiplying everything out.

The number of terms on each side is $2n$, which is even, but we want it to be $2021$, which is odd. We can arrange for this to happen, with $n=2022$, by having (for example) $c_2=c_1+t$ but no other equal terms on the two sides of the equations, so there will be exactly one cancellation. If we take $c_1=1$, $c_2=2$, $t=1$, all the other $c_i$ to be $2\bmod4$ (so we don't have any other instances of $c_i=c_j+1$), all the $d_i$ to be $0\bmod4$, except fiddle with $c_n$ and $d_n$ to make $\sum c_i=\sum d_i$, then we win.

  • $\begingroup$ Very noble of you to save the question. +1! $\endgroup$ Feb 9 at 21:42

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