Dividing the squares $1^2,2^2,\ldots,54^2$ into three equal groups with the same total sum

Question

Is it possible to divide the squares $1^2,2^2,\ldots,54^2$ into three groups, each of which contains $18$ squares, such that the sum of squares within each group is the same for all three groups?

Answer 1

We can partition $9$ consecutive squares $n^2, (n+1)^2, \ldots,(n+8)^2$ in an almost equal fashion: $$\begin{align} (n+0)^2+(n+4)^2+(n+8)^2&=3n^2+24n+80\\ (n+1)^2+(n+5)^2+(n+6)^2&=3n^2+24n+62\\ (n+2)^2+(n+3)^2+(n+7)^2&=3n^2+24n+62\end{align} $$ Each group has three members and only one group has a sum too large by $18$. By rotating the role of the too large partition, we can partition any $27$ consecutive squares into three groups of 9 members each and of equal sum. As we can do this with $1^2,\ldots 27^2$ as well as $28^2,\ldots, 54^2$, the answer to the problem statement is: yes.

Answer 2

I. Squares: A second solution can be found using Hagen von Eitzen's elegant approach. Notice that,

$$(n+a_1)^2+(n+a_2)^2+(n+a_3)^2 = 3n^2+(a_1+a_2+a_3)n+(a_1^2+a_2^2+a_3^2)\\(n+b_1)^2+(n+b_2)^2+(n+b_3)^2 = 3n^2+(b_1+b_2+b_3)n+(b_1^2+b_2^2+b_3^2)$$

Hence one must solve,

$$a_1^k+a_2^k+a_3^k = b_1^k+b_2^k+b_3^k,\quad k = 1,2\tag1$$

This is the Prouhet-Tarry-Escott problem. However, the OP added the interesting constraint that the terms must be taken from a defined set of consecutive integers.

But $(1)$ with terms taken from the nine integers $0,1,2,3\dots8$ has a second solution, namely,

$$(n + 0)^2 + (n + 5)^2 + (n + 7)^2 = 3n^2+24n+74\\ (n + 1)^2 + (n + 3)^2 + (n + 8)^2 = 3n^2+24n+74\\ (n + 2)^2 + (n + 4)^2 + (n + 6)^2 = 3n^2+24n+\color{red}{56}$$

By rotating the role this time of the too small partition, we can partition any $27$ consecutive squares into $3$ groups of $9$ members each of equal sum. Hence,

Solution 1: for powers $k =0,1,2$

$${0, 4, 8, 11, 12, 16, 19, 23, 24} \overset{k}= {1, 5, 6, 9, 13, 17, 20, 21, 25} \overset{k}= {2, 3, 7, 10, 14, 15, 18, 22, 26}$$

Solution 2: for powers $k =0,1,2$

$${0, 5, 7, 10, 12, 17, 20, 22, 24} \overset{k}= {1, 3, 8, 11, 13, 15, 18, 23, 25} \overset{k}= {2, 4, 6, 9, 14, 16, 19, 21, 26}$$

II. Cubes: The impulse is to see if it can be generalized to cubes. To have a system true for more powers, say $k=1,2,3$, by Bastien's Theorem we need more addends, hence,

$$a_1^k+a_2^k+a_3^k+a_4^k = b_1^k+b_2^k+b_3^k+b_4^k,\quad k = 1,2,3\tag2$$

with terms only from the twelve integers $0,1,2,3\dots11$. This has only one solution, so,

$$(n + 0)^3 + (n + 4)^3 + (n + 7)^3 + (n + 11)^3 = 4n^3 + 66n^2 + 558n + 1738\\ (n + 1)^3 + (n + 2)^3 + (n + 9)^3 + (n + 10)^3 = 4n^3 + 66n^2 + 558n + 1738\\ (n + 3)^3 + (n + 5)^3 + (n + 6)^3 + (n + 8)^3 = 4n^3 + 66n^2 + \color{red}{402n + 880}$$

Unfortunately, the difference extends up to the linear term, hence the same approach will not work.