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For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets such that the sums of the elements in the two subsets are equal or differ by one?

For example : for $n = 10$, we can form the subsets $S_1 = \{100 , 64 , 25 , 4\}$ and $S2 = \{1 , 9 , 16, 36, 49, 81\}$. $S_1$ adds up to $193$ and $S_2$ adds up to $192$.

Can we also identify the elements that we can assign to individual subsets that satisfies this property?

halrankard2
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Vaibhav Tyagi
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    It is not clear what you mean by two subsets having "difference $0$ or $1$". Please put the question into the main post (not just the title), and clarify all details. – halrankard2 Sep 07 '20 at 19:00
  • I apologise for ambiguous structure of the question. I am new to this platform . I made some changes in the formulation. Please see now. – Vaibhav Tyagi Sep 07 '20 at 19:25
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    Might be useful: it's easy to see that if $n$ is "good", then $n+8$ is also "good". Just note that $(n+1)^2-(n+2)^2-(n+3)^2+(n+4)^2-(n+5)^2+(n+6)^2+(n+7)^2-(n+8)^2=0$. – richrow Sep 07 '20 at 19:28
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    @richrow: That looks very useful to me! It means that we just have to check $n=6,7,\ldots,13$ and we are done. – TonyK Sep 07 '20 at 19:33
  • This is exactly something i was looking for. Thank you. – Vaibhav Tyagi Sep 07 '20 at 19:39
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    @richrow Feel free to flesh that out to an answer :-) The good old ${1,4,6,7}$ vs. ${2,3,5,8}$ trick. – Jyrki Lahtonen Sep 07 '20 at 19:49
  • Wonder whether something similar holds for cubes also? Richrow's trick can be generalized to higher powers, Thue-Morse style. Well known, locally see for example here. – Jyrki Lahtonen Sep 07 '20 at 19:57
  • Cubes worked for $n=10$ to $21$ before my method ran out of space, starting with $1^3+8^3+10^3$ for $n=10$ – Empy2 Sep 07 '20 at 20:25

2 Answers2

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The difference of sums is of the from $S:=\sum_{k=1}^n s_kk^2$ where each $s_k=\pm1$. Our task is to find $s_k$ such that the sum is $0$ or $1$.

Observe that $$\tag1a^2-(a+1)^2-(a+2)^2+(a+3)^2=4.$$ Therefore, we can pick four consecutive signs such that they contribute either $+4$ or $-4$. Hence with $8$ consecutive signs, we can achieve zero contribution. Here are the smallest non-negative sums we can achieve for some small $n$ with the eight distinct residues $\bmod 8$: $$ \begin{align}S_0&=0&=0\\ S_1&=1^2&=1\\ S_6 &= 1^2-2^2{+3^2-4^2-5^2+6^2}&=1\\ S_7 &= 1^2+2^2-3^2{+4^2-5^2-6^2+7^2}&=0\\ S_{10}&=-1^2+2^2-3^2-4^2{+5^2-6^2-7^2+8^2}-9^2+10^2&=1\\ S_{11}&=-1^2+2^2-3^2-4^2-5^2+6^2+7^2+8^2-9^2+10^2-11^2&=0\\ S_{12}&=-1^2-2^2-3^2-4^2-5^2-6^2-7^2-8^2+9^2+10^2-11^2+12^2&=0\\ S_{13}&=-1^2-2^2-3^2-4^2-5^2-6^2-7^2+8^2+9^2-10^2+11^2+12^2-13^2&=1 \end{align}$$ We conclude that we can reach sum $=0$ or $=1$ at least when $n$ si one of $0,1,6,7,10,11,12,13$ plus a multiple of $8$. In particular, this covers all $n\ge 6$.

Oscar Lanzi
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Hagen von Eitzen
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Hagen von Eitzen's analysis can be extended to find subsets whose sums differ by exactly $1$, simply by noticing that $2^2 - 1^2 = 3.$

Thus, for example $1^2 + (3^2 + 6^2)$ must differ from
$2^2 + (4^2 + 5^2)$ by exactly $1$.

Having constructed this baseline example, in a manner (again) very similar to Hagen von Eitzen's analysis, you can (for example) construct
$\{1,3,6,7,10,12,13\}$ and $\{2,4,5,8,9,11,14\}.$

user2661923
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