Let the set $A=${$ 1,2,3, \dots ,48n+24$ } , where $ n \in \mathbb {N^*}$ . Prove that there exist a subset $B $ of $A $ with $24n+12$ elements with the property : the sum of the squares of the elements of the set $B $ is equal to the sum of the squares of the elements of the set $A$ \ $B $ .
Problem
Source: Moldova TST 2018
Tags: combinatorics, number theory
HKIS200543
06.04.2018 15:23
Easy but annoying.
We'll construct each of the sets $B$. Let $\Gamma(S)$ be the sum of the squares of the elements of $S$ and define. Then it suffices to construct $B$ such that
\[ \Gamma(B) = \frac{1}{2} \sum_{i=1}^{48n+24} i^2 .\](We'll define the sum of the squares up to $48n+24$ as $f(n)$.
Proceed by induction on $n$. For the base case of $n=0$, note that $B=\{24, 23, 22, 21, 20, 2 \}$ works.
Assume that there exists a set $B_{n-1}$ such that $\Gamma(B_{n-1}) = \frac{1}{2}f(n)$. Then it suffices to find a subset $T$ of $\{48n-23, 48n-22, \dots 48n+24 \}$ such that $\Gamma(T) = \frac{1}{2} \Gamma(\{48n-23, 48n-22, \dots 48n+24 \})$ I claim
\[ T = \{ 48n+k \mid k \in K \} \]works, where $K = \{2, 3, \dots 17, -17, -18, \dots -23 \} $. The proof is a simple calculation.
\begin{align*}
\Gamma(T) &= \sum_{k \in K} (48n+k)^2 \\
&= \sum_{k \in K} (48^2 n^2 + 96 kn + k^2 \\
&= |{K}| 48^2 n^2 + 96 n \sum_{k \in K} k + \sum_{k \in K} k^2 \\
&= 24 \cdot 48^2 n^2 + 96 n \cdot 12 + \left( 17^2 - 1^2 + \sum_{j=1}^{23} j^2 \right) \\
&= 24 \cdot 48^2 n^2 + 96 n \cdot 12 + \left( 288 + \sum_{j=1}^{23} j^2 \right) \\
&= \frac{1}{2}\left( 48 \cdot 48^2 n^2 + 24 \cdot 96n + 24^2 + 2 \sum_{j=1}^{23} j^2 \right) \\
&= \frac{1}{2}\Gamma(\{48n-23, 48n-22, \dots 48n+24 \})
\end{align*}
Thus $B_{n-1} \cup T$ works as $B_n$ and the induction is complete.
reality95
06.04.2018 16:31
Pathetic problem, clearly $(4k + 4)^2 + (4k + 1)^2 - (4k + 2)^2 - (4k + 3)^2 = 4$ so choose $8k + 1, 8k + 4, 8k + 6, 8k + 7 \in B$ with $0 \le k \le \frac{48n + 24}{8} - 1$.
Abbas11235
06.05.2018 14:59
This is easy for TST. $(k+2)^2+(k+3)^2+(k+5)^2+(k+8)^2=(k+1)^2+(k+4)^2+(k+6)^2+(k+7)^2$ for $k\geq 0$.