<s>[INST] prove that the sum of the elements in two subsets is the same 
Eight consecutive positive integers are partitioned into two subsets such that the sum of the squares in each subset is the same. Prove that the sum of the elements in each subset is also the same, assuming that the smallest element is at least 56.

The method below is mostly a brute force approach.

Is there a better method than brute force?

For a set $S, $ let $f(S)$ denote the sum of the squares of its elements and let $g(S)$ denote the sum of the elements of S.
Let the integers be $x + 1,\cdots, x + 8$. The sum of the squares in each subset must be $\frac{1}2 \sum_{i=1}^8 (x+i)^2 = \frac{1}2 (8x^2 + 72x + 204) = 4x^2 + 36x+102.$
First we need to show that the partitions must in fact have the same number of elements. If one partition has more than $4$ elements, then its sum of squares will be at least $(x+1)^2 + (x+2)^2 + (x+3)^2 + (x+4)^2 + (x+5)^2 = 5x^2 + 30x + 55,$ which is too large (for $x\ge 11,$ since we would have $x^2 - 6x - 47 > 0$). So no subset can have more than 4 elements, implying than both subsets must have exactly 4 elements.
One of the subsets, say $S_1$, must contain $x+8$, and so the sum of squares of the remaining three elements in this case must be $3x^2 + 20 x +38.$ Let $S_2$ be the other subset. Suppose this subset also contains $x+7$. Then the remaining two elements have a sum of squares equal to $2x^2 + 6x -11.$ The minimum possible sum of squares of two elements is $(x+1)^2 + (x+2)^2 = 2x^2 + 6x + 5$, so we cannot have $x+7$ in $S_1$. If $x+6$ is in $S_1$, then  $f(S_1\backslash \{x+8, x+6\}) = 2x^2 +8x+2$. The only two possible remaining elements are $x+1$ and $x+2$, as $(x+1)^2 + (x+3)^2 > 2x^2 + 8x+2$, and is the second smallest sum of squares of elements. But then $f(S_1) < 2X^2 + 8x+2$, a contradiction.
So suppose $x+5\in S_1$. $f(S_1\backslash \{x+8,x+5\} ) = 2x^2 + 10x + 13,$ and in this case one can verify that $S_1 = \{x+8, x+5, x+2, x+3\}$, since all other choices of two remaining elements give either a smaller or larger sum of squares than $2x^2 + 10x+13$. In this case, $g(S_1) = 4x+18$, so $g(S_1) = g(S_2)$.
We know $x+7$ and $x+6$ are in $S_2 $. So we need to find the remaining two elements of $S_2$. If $x+5$ is in $S_2$, then the square of the remaining element in $S_2$ must be $ 4x^2 + 36x + 102 - ((x+7)^2 + (x+6)^2 + (x+5)^2) = x^2 - 8,$ which is clearly impossible.
So the only possibility is $S_1 = \{x+8, x+5, x+2, x+3\}$.
 [/INST] Here is a counterexample, if the two partition sets don't have to have the same number of elements, using $\{2,3,4,5,6,7,8,9\}$:
$2^2+3^2+4^2+7^2+8^2=142=5^2+6^2+9^2$.
But $2+3+4+7+8=24$ and $5+6+9=20$.
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