Solution Set 1

1 (a)

An equilibrium 2-wheel is a quadrilateral with vertex numbers that satisfy the fact that the sum of any two adjacent numbers is equal to the sum of the two opposite numbers. Let the vertices be $A,B,C,D$ and the equilibrium condition be:

$$\{\begin{array}{l}A+B=C+D\\ B+C=D+A\end{array}$$

The solution is $A=C$ and $B=D$, i.e., the vertices are arranged as $A,B,A,B$. Thus, all balanced 2-wheels are of the alternating permutation form $A,B,A,B$, where $A,B$ are arbitrary real numbers.
All balanced 2-wheels are of alternating permutation form $A,B,A,B$, where $A$ and $B$ are arbitrary real numbers.

1 (b)

The primal n-wheel is defined as each number equal to the opposite number. Let the vertices be $a_1,a_2,\dots ,a_{2n}$ satisfying $a_i=a_{i+n}$. For any two adjacent numbers $a_i+a_{i+1}$, the two opposite numbers are $a_{i+n}+a_{i+n+1}$. Since $a_i=a_{i+n}$ and $a_{i+1}=a_{i+n+1}$, it follows:

$$a_i+a_{i+1}=a_{i+n}+a_{i+n+1}.$$

Thus, the primal n-wheel necessarily balances.

1 (c)

Case 1: When n is even

For each $i$, the equilibrium condition can be expanded as:

$$a_i+a_{i+1}=a_{i+n}+a_{i+n+1}.$$

There are a total of $2n$ equations, but due to circularity, the number of independent equations is $n$. The system of equations is:

$$\{\begin{array}{l}{a}_1+a_2=a_{1+n}+a_{2+n}\\ a_2+a_3=a_{2+n}+a_{3+n}\\ \dots \\ a_{n-1}+a_n=a_{2n-1}+a_{2n}\\ a_n+a_{n+1}=a_{2n}+a_1\end{array}$$

Put the 1st, 3rd, 5th… .n-1 equations and add the 2nd, 4th, 6th … .n equations are added to get

$$\{\begin{array}{l}{a}_1+a_2+a_3+a_4+\dots +a_{n-1}+a_n=a_{1+n}+a_{2+n}+a_{3+n}+a_{4+n}+\dots +a_{2n-1}+a_{2n}\\ a_2+a_3+a_4+a_5+\dots +a_n+a_{n+1}=a_{2+n}+a_{3+n}+a_{4+n}+a_{5+n}+\dots +a_{2n}+a_1\end{array}$$

Subtract this to get

$$\begin{array}{rl}{a}_{n+1}-a_1& =a_1-a_{n-1}\\ a_1& =a_{n-1}\end{array}$$

From this we are left with, for all i

$$a_i+a_{i+1}=a_{i+n}+a_{i+n+1}$$

So for even n there is no non-primal equilibrium $n$-wheel

Case 2: When n is odd

There are a total of $2n$ equations, but due to circularity, the number of independent equations is $n$. The system of equations is:

$$\{\begin{array}{l}{a}_1+a_2=a_{1+n}+a_{2+n}\\ a_2+a_3=a_{2+n}+a_{3+n}\\ \dots \\ a_{n-1}+a_n=a_{2n-1}+a_{2n}\\ a_n+a_{n+1}=a_{2n}+a_1\end{array}$$

When n is odd, we need to show that the given system of equations still has a solution in the case $a_1\ne a_{n+1}$.

Divide the variables into the first n variables $A=[a_1,a_2,\dots ,a_n]$ and the last n variables $B=[a_{n+1},a_{n+2},\dots ,a_{2n}]$. Each equation can be expressed as:
For $i=1,2,\dots ,n-1$:

$$A_i+A_{i+1}=B_i+B_{i+1}$$

When $i=n$:
$A_n+B_1=B_n+A_1$
Suppose there exists a constant $c$ such that $A_i=B_i+c(-1{)}^i$. Substituting into the first $n-1$ equations, the left and right sides are equal, so these equations are satisfied.
For the last equation, substituting $A_i=B_i+c(-1{)}^i$ gives:

$$\begin{array}{rl}{A}_n+B_1& =B_n+A_1\\ [B_n+c(-1{)}^n]+B_1& =B_n+[B_1+c(-1{)}^1]\end{array}$$

When n is odd, $(-1{)}^n=-1$, so the equation becomes:
$B_n-c+B_1=B_n+B_1-c$
i.e., the equation holds.
When $c\ne 0$, since $A_1=B_1-c\ne B_1$ i.e., $a_1\ne a_{n+1}$, such a solution exists.
Therefore

$$\boxed{\text{ exists a nonprimitive equilibrium if and only if }n\text{ is odd }n\text{-wheel}}$$

1 (d)

For all odd numbers $n$, envision a construction that fills the series $[1,2n,2,2n-1,3,2n-2,\dots ,n,n+1]$ into the positions $[1,2n,2n-1,2,3,2n-2,\dots ,n,n+1]$ respectively.
That is, for all $i$ (where $1\le i\le n-1$):

• If $i$ is odd, position $i$ is filled with the number $i$;
• If $i$ is even, the position $2n-i+1$ is filled with the number $i$.
  For all $i$ (where $n\le i\le 2n$):
• If $i$ is even, position $i$ is filled with the number $i$;
• If $i$ is odd, the position $2n-i+1$ is filled with the number $i$.
  Thus for any position $j$, if $j$ is even

$$\begin{array}{rl}& (a_j+a_{j+1})-(a_{j+n}+a_{j+n+1})\\ =& (2n-j+1+j+1)-(2n-(j+n)+1+j+n+1)\\ =& 0\end{array}$$

If $j$ is an odd number of pages, we have $a_i+a_{i+1}=a_{i+n}+a_{i+n+1}$.
So

$$\boxed{\text{ when and only when }n\text{ is odd there is a classical balanced }n\text{-wheel}}$$

1 (e)

There is a special case. Every pair of neighboring numbers produces a sum that is different except for the faces. This case has a specific placement.
For all odd numbers $n$, imagine a construction in which the series $[1,2,3,\dots ,n]$ is filled in positions $[1,3,5,\dots ,2n-1]$ respectively; and the series $[n+1,n+2,n+3,\dots ,2n]$ is filled in positions $[n,n+2,n+4,\dots ,n-2]$ respectively. This way the sum of each pair of numbers is completely different.

2 (a)

Proof using mathematical induction:

• Base Case: $a_1=F_0{a}_0+F_1{a}_1=0\cdot a_0+1\cdot a_1=a_1$ when $n=1$, holds.
• Induction Assumptions: Assume that the formula holds for $n=k$ and $n=k-1$.
• Induction step:

$$\begin{array}{rl}{a}_{k+1}& =a_k+a_{k-1}\\ & =\left(F_{k-1}{a}_0+F_k{a}_1\right)+\left(F_{k-2}{a}_0+F_{k-1}{a}_1\right)\\ & =(F_{k-1}+F_{k-2})a_0+(F_k+F_{k-1})a_1\\ & =F_k{a}_0+F_{k+1}{a}_1\end{array}$$

Therefore, Eq. holds for all $n\ge 1$.

2 (b)

It is known that $a_4=2a+3a_1=b$, solving for $a_1=\frac{b-2a}{3}$.
and $a_2=a+a_1$ and $a_3=a+2a_1$.
Substituting for $a_1$ gives:

$$a_1=\frac{b-2a}{3},a_2=\frac{a+b}{3},a_3=\frac{2b-a}{3}$$

2 (c)

We need to fulfill the following condition.
$a_1=\frac{b-2a}{3}$ requires $b-2a\equiv 0(\mathrm{mod}3)$
$a_2=\frac{a+b}{3}$ requires $a+b\equiv 0(\mathrm{mod}3)$
$a_3=\frac{2b-a}{3}$ requires $2b-a\equiv 0(\mathrm{mod}3)$

for the first and second conditions:
From $b-2a\equiv 0(\mathrm{mod}3)$ requires $b\equiv 2a(\mathrm{mod}3)$.
From $a+b\equiv 0(\mathrm{mod}3)$ we get $b\equiv -a(\mathrm{mod}3)$.
These two conditions are equivalent because $2a\equiv -a(\mathrm{mod}3)$ holds for any integer $a$.

For the third condition:
When $b\equiv -a(\mathrm{mod}3)$, the substitution $2(-a)-a=-3a\equiv 0(\mathrm{mod}3)$ holds.
So the three conditions can be reduced to $a+b\equiv 0(\mathrm{mod}3)$, i.e., $3\mid (a+b)$.
For all integers a and b it is only necessary to satisfy

$$\boxed{3\mid (a+b)}$$

2 (d)

For positive integers $k$, the integers $a,b$ need to satisfy $F_k\mid (b-F_{k-1}a)$, i.e. $b\equiv F_{k-1}a(\mathrm{mod}{F}_k)$.
For any $n\ge 1$, when $n=k$, substituting the known conditions $a_0=a$ and $a_k=b$, we have:

$$\begin{array}{rl}b& =F_{k-1}a+F_k{a}_1\\ & \\ a_1& =\frac{b-F_{k-1}a}{F_k}\end{array}$$

To ensure that $a_1$ is an integer, $b\equiv F_{k-1}a(\mathrm{mod}{F}_k)$ must be satisfied. In other words there exists an integer $m$ such that $b=F_{k-1}a+F_{k}m$, i.e. $a_1=m$. At this point, the initial terms $a_0=a$ and $a_1=m$ are both integers. According to the recurrence relation $a_n=a_{n-1}+a_{n-2}$, all subsequent terms are sums of integers, so they are all integers.
In summary, all terms of the sequence are integers if and only if $F_k$ divides $(b-F_{k-1}a)$. Therefore, the integers $a,b$ must satisfy:

$$\boxed{F_k\mid (b-F_{k-1}a)}$$

2 (e)

For the extended case of problem (d), given two nonconsecutive terms $a_m$ and $a_n$ ($m<n$), what conditions need to be satisfied to require that all the terms be integers?

According to the conclusion of problem (a), for any $k\ge 0$, there are:

$$a_k=F_{k-1}{a}_0+F_k{a}_1$$

When given $a_m=p$ and $a_n=q$, the joint equation:

$$\{\begin{array}{l}{F}_{m-1}{a}_0+F_m{a}_1=p\\ \\ F_{n-1}{a}_0+F_n{a}_1=q\end{array}$$

The determinant of the above system of equations is:

$$D=\left|\begin{array}{cc}{F}_{m-1}& F_m\\ F_{n-1}& F_n\end{array}\right|=F_{m-1}{F}_n-F_m{F}_{n-1}$$

By the Cassini constant extension of the Fibonacci numbers, $D=(-1{)}^m{F}_{n-m}$. The solution is:

$$a_0=\frac{F_{n}p-F_{m}q}{D},a_1=\frac{F_{n-1}q-F_{m-1}p}{D}$$

For $a_0$ and $a_1$ to be integers, the denominator $D=(-1{)}^m{F}_{n-m}$ must divide the numerator:

• $F_{n-m}\mid (F_{n}p-F_{m}q)$
• $F_{n-m}\mid (F_{n-1}q-F_{m-1}p)$
  Since the Fibonacci numbers satisfy $\gcd (F_k,F_{k+1})=1$ and $\gcd (F_k,F_n)=F_{\gcd (k,n)}$, the conditions can be further simplified.
  By the linear combination property of the Fibonacci numbers, the above integro-division condition is equivalent to:

$$F_{n-m}\mid (p\cdot F_n-q\cdot F_m)$$

If all terms are integers, then $a_0$ and $a_1$ must be integers, so $F_{n-m}$ divides the numerator.
If $F_{n-m}\mid (F_{n}p-F_{m}q)$, then $a_0$ and $a_1$ are integers, and all terms of the recursively generated sequence are integers.
All terms of the f-sequence given the discontinuous terms $a_m=p$ and $a_n=q$ are integers if and only if $F_{n}p-F_{m}q$ is divisible by $F_{n-m}$.

$$\boxed{F_{n-m}\mid (F_{n}p-F_{m}q)}$$

3 (a)

Considering the letters $A,B,C$ as generating elements, the subword rule defines the relation:

• $AA\sim \epsilon$ (the empty string), i.e. $A^2=1$
• $BB\sim \epsilon$, i.e. $B^2=1$
• $CC\sim \epsilon$, i.e. $C^2=1$
• $ABAB\sim \epsilon$, i.e. $(AB{)}^2=1$
• $BCBC\sim \epsilon$, i.e. $(BC{)}^2=1$
• $CACA\sim \epsilon$, i.e. $(CA{)}^2=1$
  This defines a group that generates the elements $A,B,C$ satisfying:

$$A^2=B^2=C^2=1,(AB{)}^2=(BC{)}^2=(CA{)}^2=1$$

$(AB{)}^2=1$ implies that $AB=B{A}^{-1}=BA$ (since $A^2=1$), so $A$ and $B$ are exchangeable. Similarly, $BC=CB$ and $CA=AC$. Thus, $A,B,C$ are exchangeable, and the corresponding group is $(\mathbb{Z}/2\mathbb{Z}{)}^3$, i.e. the direct product of the three $\mathbb{Z}/2\mathbb{Z}$.
The elements of the group $(\mathbb{Z}/2\mathbb{Z}{)}^3$ can be written as $A^a{B}^b{C}^c$, where $a,b,c\in 0,1$ (since $A^2=1$, $A^1=A$, $A^2=\epsilon$, and other similar reasoning applies to $B,C$). The possible elements are $\epsilon ,A,B,C,AB,AC,BC,ABC$, totaling $2^3=8$.

$$\boxed{\text{The number of different words that can be formed is 8}}$$

3 (b)

Non-adjacent letters cannot be exchanged. For example, A and C are not “directly interchangeable”, nor are B and D. Thus four letters are only “directly interchangeable” in the sense of “directly interchangeable”.
So the four letters are only interchangeable with each other in “adjacent” rings (A↔B, B↔C, C↔D, D↔A).
Therefore, if “AC” is present, it cannot be exchanged for the string “CA”!
Consider the string: $(AC{)}^n$, i.e. $ACACACAC...$ (n for AC). When $n\ne m$, it is not possible to simplify $(AC{)}^n$ to $(AC{)}^m$ by existing rules.

$$\boxed{\text{An infinite number of words can be obtained using this type of element splicing}}$$

3 (c)

As with (b), non-adjacent letters cannot be exchanged. Consider also the string: $(AC{)}^n$, i.e. $ACACACAC...$ (n for AC). When $n\ne m$, it is not possible to simplify $(AC{)}^n$ to $(AC{)}^m$ by existing rules.

$$\boxed{\text{An infinite number of words can be obtained using this type of element splicing}}$$

Both the five-letter form and the four-letter form create an infinite number of strings that cannot be compared. However, five letters have more string arrangements than four letters for the same length. More in a sense.

3 (d)

The previous rule rule is shaped like this:

• Each letter satisfies $X^2=e$ (free to delete/add AA to indicate that the square of $X$ is equal to the identity element)
• For some pairs of letters X, Y, if deletion/addition of the substring XYXY is allowed, this corresponds to $e(XY{)}^2=e$.
  Since $X^2=Y^2=e$, $(XY{)}^2=e⟺XY=YX$, i.e. they are interchangeable with each other.
  Thus, when we add only the relations $X^2=e$ and $(XY{)}^2=e⟹X,Yexchange$, the resulting group is called the “right-corner Coxeter group”. Each letter is treated as a vertex, and an edge is connected between the two vertices if they are “exchangeable”. If the graph is clique, it means that all letters are exchangeable with each other. Thus, all generators are sequentially exchangeable, each of order 2, and the whole group is isomorphic to

$$\underset{\text{number of letters }n}{\underbrace{Z_2\times Z_2\times \cdots \times Z_2}}$$

The size of this group is $2^n$ and is a finite group.
If the graph has missing edges (indicating that a pair of letters is not exchanged), an infinite number of distinguishable elements are created, and it can generally be introduced that the group is infinite.
In (b), the fact that 4 letters form a quadrilateral (cycle) instead of being completely contiguous already leads to an infinite group; the 5 letters in (c) are also 5-cycles, and are similarly infinite.

4 (a)

When $m=1$, m-Hilbert numbers are all positive integers (since $1\cdot k+1=k+1$, which covers all positive integers). The 1-Hilbert prime is then almost identical to what we normally define as a “prime number” (greater than 1 and not factor in smaller positive integers). The prime factorization of positive integers in the usual sense satisfies the unique factorization theorem. Therefore, the m=1 case must be uniquely decomposable.
When $m=2$, the m-Hilbert numbers are all the odd numbers. A 2-Hilbert prime is a number in the set of odd numbers that cannot be multiplied by a smaller set of odd factors. In other words, it is an “irreducible element in the set of odd numbers”.
But we know that all odd numbers in the usual factorization of integers factor into only a few odd primes. Since 2 is not in this set, there is no problem of multiple factorization due to “missing 2”; within the set of odd numbers, the factorization still corresponds only to the same set of odd primes, and remains unique. Thus, the case m=2 is still uniquely factorized.

$$\boxed{\text{The unique factorization occurs in the m=1 and m=2 cases}}$$

4 (b)

The 4-Hilbert numbers and 5-Hilbert numbers do not factor uniquely. The 4-Hilbert numbers are “closer” to having unique factorization in the sense that the failure is more structured, involving primes from a single residue class (e.g., primes ≡ 3 mod 4). In contrast, the 5-Hilbert numbers might involve more complex interactions between different residue classes, leading to a higher degree of non-unique factorization. Higher values of m can exhibit even worse factorization with more distinct factorizations.