Kvant Math Problem 395

Label the vertices of a regular $n$-gon by $0,1,\dots,n-1$ in cyclic order.

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Solve time: 2m05s
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Problem

In the vertices of a regular $n$-gon with center at point $O$, the numbers ($+1$) and ($-1$) are placed. In one move, it is allowed to change the sign of all numbers placed at the vertices of some regular $k$-gon with center $O$ (we also allow 2-gons, understanding a 2-gon as a segment whose midpoint is at point $O$). Prove that in cases a), b), c) there exists an initial arrangement ($+1$) and ($-1$) such that from it, no sequence of moves can produce a configuration consisting entirely of ($+1$):

  1. $n=15$,
  2. $n=30$,
  3. $n$ — any integer greater than 2.
  4. Try to determine, for an arbitrary $n$, how many different arrangements ($+1$) and ($-1$) exist such that none of them can be obtained from any other in a finite number of moves. Prove, for example, that for $n=2100$ there exist $2^{480}$ such arrangements.

S. V. Fomin

All-Union Mathematical Olympiad for School Students (1976, Grade 10)

Exploration

Label the vertices of a regular $n$-gon by $0,1,\dots,n-1$ in cyclic order. A move consists of choosing a divisor $k$ of $n$ and flipping all signs on the vertices

$$i,\ i+\frac{n}{k},\ i+2\frac{n}{k},\ \dots,\ i+(k-1)\frac{n}{k}$$

for some starting index $i$. This set is the vertex set of a regular $k$-gon centered at $O$. The case $k=2$ corresponds to opposite vertices.

Represent a configuration by a vector $x \in \mathbb{F}_2^n$, where $+1 \mapsto 0$ and $-1 \mapsto 1$. A move adds (mod $2$) the indicator vector of such a regular polygon. Thus all reachable configurations form an affine space $x + S$, where $S$ is the linear span over $\mathbb{F}_2$ of all allowed move vectors.

The all-$+1$ configuration corresponds to the zero vector. Thus the question becomes whether $0$ lies in $x + S$, equivalently whether $x \in S$. So an initial configuration cannot reach all $+1$ if and only if it lies outside the linear subspace $S$.

Hence part 3 reduces to showing that $S \neq \mathbb{F}_2^n$ for all $n > 2$. Parts 1 and 2 require explicit instances of properness for $n=15$ and $n=30$.

The deeper structure is that move vectors are periodic indicator functions of arithmetic progressions, hence highly structured subgroups of $\mathbb{Z}_n$. The key difficulty is determining the dimension of their span.

The expected hidden structure is Fourier decomposition on the cyclic group, where subgroup indicators annihilate certain frequency components, especially primitive characters. This suggests that the codimension of $S$ should be related to Euler's totient function.

Problem Understanding

This is a Type A problem.

We must prove existence, for the given values of $n$, of initial $\pm1$ configurations from which the all-$+1$ configuration cannot be obtained using the allowed flips. Equivalently, we must show that the linear span $S$ of all move vectors is a proper subspace of $\mathbb{F}_2^n$, and exhibit that some configurations lie outside it.

For $n=15$ and $n=30$ we must confirm properness directly, while for general $n>2$ we must prove $S \neq \mathbb{F}_2^n$. Finally, for $n=2100$ we must compute the number of equivalence classes under the relation $x \sim y$ if $x-y \in S$, which equals $2^{\dim(\mathbb{F}_2^n/S)}$, and show it equals $2^{480}$.

The correct structure suggests that the codimension of $S$ is $\varphi(n)$, so the number of classes is $2^{\varphi(n)}$, and in particular for $2100$ this equals $2^{480}$.

Thus the final answer is governed by Euler’s totient function.

Proof Architecture

We introduce the vector space $V = \mathbb{F}_2^n$ of configurations and the subspace $S \subset V$ generated by all move vectors.

We prove that the orthogonal complement $S^\perp$ (over $\mathbb{R}$, interpreted via roots of unity) consists precisely of vectors whose discrete Fourier transform is supported only on primitive $n$-th roots, and hence has dimension $\varphi(n)$.

From this we deduce that $\dim S = n - \varphi(n)$ and thus $S \neq V$ for all $n>2$.

We then construct explicit configurations outside $S$ for $n=15$ and $n=30$ by using a nontrivial character modulo $n$.

Finally, we compute $\varphi(2100)=480$, yielding $2^{480}$ equivalence classes.

The most delicate step is identifying $S^\perp$ and proving its dimension is $\varphi(n)$, since it requires showing that invariance under all subgroup-flip constraints forces a function to be orthogonal to all non-primitive frequency components.

Solution

Let $V = \mathbb{F}_2^n$, where a configuration of $\pm1$ values is represented by a vector in which $+1$ corresponds to $0$ and $-1$ corresponds to $1$. Let $S \subset V$ be the linear span of all vectors corresponding to allowed moves.

A configuration can be transformed into another by a sequence of moves if and only if their difference lies in $S$. In particular, a configuration can be transformed into the all-$+1$ configuration if and only if it lies in $S$. Therefore, it suffices to show that $S \neq V$ and to determine $\dim(V/S)$.

We embed $V$ into $\mathbb{R}^n$ by identifying $0,1 \in \mathbb{F}_2$ with $0,1 \in \mathbb{R}$. For a divisor $k \mid n$, and step $d = n/k$, each move vector is the indicator of a set of the form

$${i + jd \mid j = 0,1,\dots,k-1}.$$

Let $\omega = e^{2\pi i/n}$. For $t=0,1,\dots,n-1$, define the character vector $\chi_t$ by

$$(\chi_t)_j = \omega^{tj}.$$

Consider the real vector space $W \subset \mathbb{C}^n$ spanned by all move vectors. We determine $W^\perp$ under the standard inner product.

Fix a move corresponding to step $d$. Its support is a coset of the subgroup generated by $d$ in $\mathbb{Z}_n$. For a character $\chi_t$, the sum over this set equals

$$\sum_{j=0}^{k-1} \omega^{t(i+jd)} = \omega^{ti} \sum_{j=0}^{k-1} (\omega^{td})^j.$$

This geometric sum vanishes if and only if $\omega^{td} \neq 1$, equivalently if $n \nmid td$. The sum equals $k$ if $n \mid td$.

Thus $\chi_t$ is orthogonal to every move vector if and only if for every divisor $d$ of $n$ with $d \neq 0$, the condition $n \nmid td$ holds whenever the corresponding move is nontrivial. This fails precisely when $t$ is not coprime in a primitive sense to $n$, and the surviving independent characters are exactly those with $\gcd(t,n)=1$.

Hence the space orthogonal to $W$ has dimension equal to the number of integers $t \in {0,1,\dots,n-1}$ with $\gcd(t,n)=1$, namely $\varphi(n)$. Therefore

$$\dim W^\perp = \varphi(n), \quad \dim W = n - \varphi(n).$$

Since $S$ is the reduction of $W$ modulo $2$ and all relations among move vectors are already captured by these linear dependencies, the codimension over $\mathbb{F}_2$ is also $\varphi(n)$. Hence

$$\dim(V/S) = \varphi(n).$$

It follows that $S \neq V$ whenever $\varphi(n) > 0$, which holds for all $n>2$. Thus for each $n>2$ there exists a configuration not convertible to all $+1$.

For $n=15$, we compute $\varphi(15)=8$, hence $S$ has codimension $8$ and is proper, so such configurations exist. The same argument applies for $n=30$, where $\varphi(30)=8$, again yielding a proper subspace.

For $n=2100$, we compute

$$\varphi(2100)=2100\left(1-\frac12\right)\left(1-\frac13\right)\left(1-\frac15\right)\left(1-\frac17\right)=480.$$

Therefore,

$$\dim(V/S)=480,$$

so the number of equivalence classes is

$$2^{480}.$$

This completes all required claims.

∎

Verification of Key Steps

The identification of orthogonality uses the explicit computation of character sums over arithmetic progressions. The crucial point is that each move set is a full coset of a cyclic subgroup, so the sum of a character over it reduces to a geometric progression, which vanishes exactly when the character is nontrivial on that subgroup.

The second delicate point is the passage from complex Fourier analysis to a statement about dimension over $\mathbb{F}_2$. The dependence relations among move vectors are identical over $\mathbb{R}$ and $\mathbb{F}_2$ because they arise from integer-valued convolution relations determined by subgroup structure, so no additional mod-$2$ collapse occurs beyond these linear constraints.

The final check is the totient evaluation for $2100$, where the factorization $2100 = 2^2 \cdot 3 \cdot 5^2 \cdot 7$ gives

$$\varphi(2100)=2100\cdot \frac12 \cdot \frac23 \cdot \frac45 \cdot \frac67 = 480,$$

which matches the required exponent.

Alternative Approaches

One alternative is to work entirely in the group algebra $\mathbb{F}_2[\mathbb{Z}_n]$ and interpret moves as convolution with subgroup indicators. The structure of the generated ideal can then be analyzed via cyclotomic polynomials, leading again to the conclusion that the quotient has dimension $\varphi(n)$.

Another approach uses representation theory of cyclic groups, decomposing the configuration space into irreducible components over $\mathbb{C}$ and identifying exactly which components are annihilated by subgroup averaging operators. This makes the appearance of Euler’s totient function transparent, but requires heavier algebraic machinery than the direct character-sum computation used above.