There's a folklore problem:

Let $x_1, \cdots, x_{23} \in \mathbb{Z}$ be the weights of $23$ soccer players. Now Master Yoda want's to form two soccer teams with $11$ players each. Turns out for any $1 \leq i \leq 23$, one can partition $\{1, \cdots, n \} - \{ i \}$ into two disjont sets $A, B$ with $|A| = |B| = 11$ such that $\displaystyle \sum_{k \in A} x_k = \sum_{k \in B} x_k$. Prove that all numbers must be equal.

The solution is well known and is not very hard for $\mathbb{Z}$.

I'm wondering replacing $\mathbb{Z}$ by which commutative ring with unit $R$ makes the problem false.

If $R = \mathbb{Q}$, then it's also same as $\mathbb{Z}$ (and the answer is affirmative), just multiply everything by the LCM of the numerators to reduce it to the case $R = \mathbb{Z}$.

If $R = \mathbb{R}$, then also the problem is true, but you need a lemma by Dirichlet (which is proven by PHP) to reduce it to the case $R = \mathbb{Z}$.

If $R = \mathbb{C}$, then also the problem is true. Because if $\displaystyle \sum_{k \in A} z_k = \sum_{k \in B} z_k \Rightarrow \sum_{k \in A} \text{Re}(z_k) = \sum_{k \in B} \text{Re}(z_k)$, and by the previous one $R = \mathbb{R}$ applied to the real components, you get $\text{Re}(z_i) = \text{Re}(z_j)$ for all $i, j$. Similarly you prove the imaginary components are same, so all numbers are same.

If $R = \mathbb{Q}[x], \mathbb{C}[x], \mathbb{R}[x], M_{m,n}(\mathbb{Q}), M_{m,n}(\mathbb{Z}), M_{m,n}(\mathbb{C}), M_{m,n}(\mathbb{R}) $, even then the problem is true since you can look at the problem "component wise" and reduce it to the above cases.

However I have no idea whether the problem is true when $R = \mathbb{Z}_p$ for some prime $p$ or in some other rings (ring of rational functions etc).

Is it true for all rings, or are there some rings for which this problem doesn't hold ?

If it's false for some rings, are there any characterizations for such rings ?

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    The statement is false if the characteristic of the ring is $p\in\{3,5,7,11\}$. Choose $p$ players of weight $1$ and $23-p$ players of weight $0$. If a player of weight $0$ misses the game, team A drafts all players of weight $1$. If a player of weight $1$ misses the game, the remaining players of weight $1$ are distributed equally among teams $A$ and $B$. – Philipp Lampe Aug 10 at 8:43
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    This problem can be more generally asked in the context of abelian groups rather than rings. That is to say, $\mathbb{Z}$ can be replaced with any abelian group. In this more general setup, it will be true over any $\mathbb{Q}$-vector space because you can project to $\mathbb{Q}$ multiple times. Moreover, you can embed any torsion free abelian group in a vector space over $\mathbb{Q}$, so it will also be true in that case. – Uriya First Aug 10 at 10:02
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    @alxchen Typically it is inappropriate to post the same question simultaneously in math.se AND mathoverflow. If a question sits in math.se for a long time with no luck, then you can consider crossposting (and crosslinking) and explaining what happened, but posting simultaneously is usually frowned upon. Most often, it belongs in one and not the other. This one I can imagine living in either community, however... – rschwieb Aug 10 at 18:15
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    "Folklore"? 34th Putnam competition, 1973, problem B-1: Let $a_1,a_2,\dots,a_{2n+1}$ be a set of integers such that, if any one of them is removed, the remaining ones can be divided into two sets of $n$ integers with equal sums. Provve $a_1=a_2=\cdots=a_{2n+1}$. – Gerry Myerson Aug 11 at 22:51

Observation. If the result is true for some abelian groups $G_1$ and $G_2$ then it's also true for any extension $0 \to G_1 \to G \to G_2 \to 0$. Moreover if the result holds for $G$ then the same holds for any subgroup $H$ of $G$.

Proposition. Let $G$ be an abelian group, and let $P$ be the set of primes $p$ such that $G$ contains an element of order $p$. Then the result is true for $G$ if and only if it is true for $\mathbf{Z}/p\mathbf{Z}$ for all $p \in P$.

I'm grateful to darij grinberg for simplifying my initial proof of this proposition.

Proof. The direct implication follows from the fact that $G$ contains $\mathbf{Z}/p\mathbf{Z}$ for every $p \in P$.

Conversely, let $x_1,\ldots,x_{23} \in G$ satisfying the assumption of the problem. Let $H$ be the subgroup of $G$ generated by the $x_i$. The structure theorem of finitely generated abelian groups tells us that $H$ is a direct sum of copies of $\mathbf{Z}$ and $\mathbf{Z}/p^k\mathbf{Z}$ with $p \in P$. Since $\mathbf{Z}/p^k\mathbf{Z}$ is an iterated extension of $\mathbf{Z}/p\mathbf{Z}$, the initial observation shows that the result is true for all $\mathbf{Z}/p^k\mathbf{Z}$ with $p$ in $P$, and thus for $H$, so that all $x_i$ are equal. QED

So it remains to study the case of $\mathbf{Z}/p\mathbf{Z}$ where $p$ is prime.

Following darij grinberg's comments at this link: let $M$ be a matrix of the form $M = (\pm \delta_{i \neq j})_{1 \leq i,j \leq 23}$ where $\delta$ is the Kronecker symbol and the number of $+$ signs is 11 in each row. The upper-left $22 \times 22$ minor is odd (in particular nonzero), therefore the rank of $M$ is $22$ and its kernel is generated by the vector $(1,\ldots,1)$. This solves the original problem over $\mathbf{Z}$, and over $\mathbf{Z}/2\mathbf{Z}$.

Now over $\mathbf{Z}/p\mathbf{Z}$, the result is false if and only there exists a matrix $M$ of this form with rank $<22$, which amounts to say that the before last elementary divisor of $M$ is divisible by $p$. There are only finitely many such matrices, so the result is true if $p$ is large enough. In fact, by Hadamard's inequality, the abolute value of the $22 \times 22$ minor is bounded by $21^{11}$, so the result is true for every $p>21^{11}$.

Using Magma (I can share the code if you're interested) I generated $\approx 5 \cdot 10^7$ random matrices of this form and found that the result is false for at least 1471 values of $p$, in particular all $2<p<3529$. The largest value I got is $p=36285031$.

For a given value of $p$, it seems challenging to decide whether the result holds over $\mathbf{Z}/p\mathbf{Z}$, since one apparently needs to check all ${{22}\choose{11}}^{23}$ matrices of this form. It would be interesting to devise a better method.

  • I don't see why if the result is true for $G_1$ and $G_2$, then the result is also true for any extension of the groups. I can see those elements in the extension have the same image in $G_2$, but why does this imply that they are equal? – awllower Aug 11 at 20:40
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    @awllower Say the $x_i$ all have image $z \in G_2$. Choose any element $y \in G$ projecting to $z$, then all the $x_i-y$ belong to $G_1$. Since the result holds in $G_1$, all the $x_i-y$ are equal and we are done. – François Brunault Aug 11 at 21:08
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    You can assume right away that $G $ is finitely generated, since you can always replace $G $ by the subgroup spanned by $x_1, x_2, \ldots, x_{23}$. Then, the $p $-primary part of $G $ is obtained from a bunch of copies of $\mathbf{Z}/p $ by extension; you don't need any fancy results about Prüfer groups. – darij grinberg Aug 11 at 21:24
  • @darij grinberg Good point, this simplifies much of my argument. – François Brunault Aug 11 at 21:31

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