You have a set S of 2N+1 real numbers, with the following property: if you remove any one element, you can partition the remaining 2N elements into two sets, A and B, each of size N, such that the sum of the N numbers in set A equals the sum of the N numbers in set B. Prove that all the numbers in the original set are equal.

Spoiler Alert - Hint Ahead

Start by trying to solve the riddle for the special case where the set contains only natural numbers instead of general reals.

Spoiler Alert - Solution Ahead!

We will solve the riddle for the case of natural numbers, integers, rational numbers, and finally real numbers. Note that everywhere below I use the regular set notation to denote multisets.

Solving for the Natural Numbers

Theorem 1: If the numbers in S are all natural numbers, then they are all equal.

Lemma 1: If \(S \subset \mathbb{Z}\), then all the numbers in S must have the same parity, i.e. either all numbers are even, or all numbers are odd.

Proof of Lemma 1: Indeed, let’s consider two numbers \(x, y \in S\). Since you can split the remaining numbers into 2 sets of equal sums that must mean that \((\sum_{a \in S}a) - x\) is even and \((\sum_{a \in S}a) - y\) is even. Therefore x and y have the same parity. \(\blacksquare\)

Lemma 2: S is a set satisfying the above property i.f.f. the set \(\{x - t \mid x \in S, t \in \mathbb{Z}\}\) is satisfying the above property.

Lemma 3: If S contains only even numbers, then S satisfies the above property, i.f.f. the set \(\{\frac{x}{2} \mid x \in S \}\) satisfies the above property.

The proofs of Lemmas 2 and 3 are trivial.

Proof of Theorem 1: By Lemma 1, all elements of S are either even or odd. If all elements are odd, by Lemma 2, we can subtract 1 from all elements of S, making all elements even. If all the elements are even, by Lemma 3, we can divide all elements of by 2. Consider the sequence of sets you get by repeated application of this process, diving by 2 if all elements are even, and removing 1 if all elements are odd, until one of the elements of the set, x, becomes 0. Call the set at that point S’. Let y be another element of S’. If y is not 0, y must be an even number from Lemma 1. Keep diving all elements by 2 (from Lemma 3) at some point the image of y will be an odd number, leading to a contradiction. Therefore for all \(y \in S', y = x\). Therefore all elements of S’ are equal. Therefore all elements of S are equal. \(\blacksquare\)

Solving for the Integers

Theorem 2: If the numbers in S are all integers, then they are all equal.

Proof of Theorem 2: Let \(m = \text{min}(S)\). The set \(S' = \{x - m \mid x \in S\} \subset \mathbb{N}\) satisfies the conditions of Theorem 1. Therefore all elements of S’ are equal. Therefore all elements of S are equal. \(\blacksquare\)

Solving for the Rationals

Theorem 3: If the numbers in S are all rational numbers, then they are all equal.

Proof of Theorem 3: Let \(S = \{\frac{a_i}{b_i} \mid 0 \leq i \leq N\}\). Let \(g = \prod_{0 \le i \le N}b_i\). Let \(S' = \{x \times g \mid x \in S\}\). Then \(S' \subset \mathbb{Z}\) and S’ satisfies the condition of Theorem 2. So all the elements of S’ are equal. So all the elements of S are equal. \(\blacksquare\)

Solving for the Reals

Theorem 4: If the numbers in S are general real numbers, then they are all equal.

Proof of Theorem 4: Look at field extension of \(\mathbb{Q}\) in \(\mathbb{R}\) generated by the elements of S. You can look at its elements as a finite dimensional vector field over \(\mathbb{Q}\). The claim is true i.f.f. it is true separately in each dimension. Each of the dimensions separately satisfies the conditions for Theorem 3, therefore the values in each dimension are all equal, therefore the values in all dimensions are equal, therefore the numbers in S are all equal. \(\blacksquare\)