If we split a set \(S\) of integers into two subsets \(A\) and \(B\) whose intersection is empty and whose union is the whole of \(S\), and such that
the sum of the elements of \(A\) is equal to the sum of the elements of \(B\)
and the sum of the squares of the elements of \(A\) is equal to the sum of the squares of the elements of \(B\),
then we say that we have found a balanced partition of \(S\) into two subsets.
Find a balanced partition of the set \(\{1, 2, 3, 4, 5, 6, 7, 8\}\) into two subsets \(A\) and \(B\), each of size 4.
Given that \(a_1, a_2, \ldots, a_m\) and \(b_1, b_2, \ldots, b_m\) are sequences with
\[\sum_{k=1}^m a_k = \sum_{k=1}^m b_k \quad \text{and} \quad \sum_{k=1}^m a_k^2 = \sum_{k=1}^m b_k^2,\]
show that
\[\sum_{k=1}^m a_k^3 + \sum_{k=1}^m (c + b_k)^3 = \sum_{k=1}^m b_k^3 + \sum_{k=1}^m (c + a_k)^3\]
for any real number \(c\).
Find, with justification, a balanced partition of the set \(\{1, 2, 3, \ldots, 16\}\) into two subsets \(A\) and \(B\), each of size 8, which also has the property that
the sum of the cubes of the elements of \(A\) is equal to the sum of the cubes of the elements of \(B\).
You are given that the sets \(A = \{1, 3, 4, 5, 9, 11\}\) and \(B = \{2, 6, 7, 8, 10\}\) form a balanced partition of the set \(\{1, 2, 3, \ldots, 11\}\).
Let \(S = \{n^2, (n+1)^2, (n+2)^2, \ldots, (n+11)^2\}\), where \(n\) is any positive integer. Find, with justification, two subsets \(C\) and \(D\) of \(S\) whose intersection is empty and whose union is the whole of \(S\), and such that
the sum of the elements of \(C\) is equal to the sum of the elements of \(D\).
Note that
\begin{align*}
&& \sum_{k=1}^m a_k^2 + \sum_{k=1}^m (c + b_k)^2 &= \sum_{k=1}^m a_k^2 + mc^2 + 2c \sum_{k=1}^m b_k +\sum_{k=1}^m b_k^2 \\
&&&= \sum_{k=1}^m (a_k+c)^2 + \sum_{k=1}^m b_k^2
\end{align*}
Therefore if we take our balanced subsets of \(\{1,2,3,4,5,6,7,8\}\) and take \(A \cup (B+8)\) and \(B \cup (A+8)\) they will be balanced (including the cubes) so: \(A = \{1,4,6,7,10,11,13,16\}, B = \{2,3,5,8,9,12,14,15\}\)