Can you find distinct integers $a, b,$ and $c$ such that $a-b$ divides $b-c$, $b-c$ divides $c-a$, and $c-a$ divides $a-b$?
This problem is from "The Art and Craft of Problem Solving" page 74.
The author suggests approaching this problem using invariants. Any ideas?
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Can you find distinct integers $a, b,$ and $c$ such that $a-b$ divides $b-c$, $b-c$ divides $c-a$, and $c-a$ divides $a-b$?
The short answer is "No".
Let us state the following
Lemma. $\,\,$ If $P(x)$ is a polynomial having all integral coefficients, then $a-b$ divides $P(a) - P(b)$, being $a$ and $b$ integer numbers. In symbols: $$\underbrace{P(x) \in \mathbb{Z}[x]}_{\color{orange}{\displaystyle \text{Hypothesis}}} \implies \underbrace{a-b \mid P(a) - P(b) \qquad \forall \, a, b \in \mathbb{Z}}_{\color{red}{\displaystyle \text{Thesis}}}. \tag{0}\label{eq:0}$$
Proof. $\,\,$ $\color{orange}{\text{Hypothesis}} \Rightarrow$ Consider the $n^{\text{th}}$ degree polynomial $$P(x) = c_0 x^0 + c_1 x^1 + c_2 x^2 + \ldots + c_n x^n,$$ with $c_0, c_1, c_2, \ldots, c_n \in \mathbb{Z}$ (constant integer coefficients).
By evaluating the polynomial at $x = a$: $$P(a)= c_0 a^0 + c_1 a^1 + c_2 a^2 + \ldots + c_n a^n.$$ By evaluating the polynomial at $x = b$: $$P(b)= c_0 b^0 + c_1 b^1 + c_2 b^2 + \ldots + c_n b^n.$$ The difference between the two evaluations is: $$\begin{align} P(a)-P(b) &= c_0 a^0 + c_1 a^1 + c_2 a^2 + \ldots + c_n a^n - (c_0 b^0 + c_1 b^1 + c_2 b^2 + \ldots + c_n b^n) \\ &= c_0 a^0 + c_1 a^1 + c_2 a^2 + \ldots + c_n a^n - c_0 b^0 - c_1 b^1 - c_2 b^2 - \ldots - c_n b^n \\ &= c_0 (\underbrace{a^0}_{1} - \underbrace{b^0}_{1}) + c_1 (a^1 - b^1) + c_2 (a^2 - b^2) + \ldots + c_n (a^n - b^n) \\ &= c_0 (\underbrace{\cancel{1}-\cancel{1}}_{0}) + c_1 (a^1 - b^1) + c_2 (a^2 - b^2) + \ldots + c_n (a^n - b^n) \\ &= c_1 (a^1 - b^1) + c_2 (a^2 - b^2) + \ldots + c_n (a^n - b^n) \\ &= \sum_{k=1}^{n} c_k (a^k - b^k) \\ &= c_k \sum_{k=1}^{n} (a^k - b^k) \qquad \text{(by linearity of scaling of summations)} \tag{$\ast$}\label{eq:ast}. \end{align}$$ Recall that $$(a^k - b^k) = (a-b)\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right). \tag{$\ast \ast$}\label{eq:astast}$$ Substituting $\eqref{eq:astast}$ in $\eqref{eq:ast}$, we have $$\begin{align} P(a)-P(b) &= c_k \sum_{k=1}^{n}(a-b)\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right) \\ &= c_k (a-b)\sum_{k=1}^{n}\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right) \qquad \text{(by linearity of scaling of summations)}. \end{align}$$ Finally: $$\begin{align} P(a)-P(b) &= c_k (a-b)\sum_{k=1}^{n}\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right) \\ \iff \frac{P(a)-P(b)}{a-b} &= \frac{c_k \cancel{(a-b)}\large \sum\limits_{k=1}^{n}\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right)}{\bcancel{(a-b)}}, \end{align}$$ where in the last step, both the left-hand side and the right-hand side were divided by $(a-b)$, with $a \neq b \in \mathbb{Z}$. So: $$\frac{P(a)-P(b)}{a-b} = c_k \sum_{k=1}^{n}\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right).$$ Since $a, b, \{c_0, c_1, c_2, \ldots, c_n\} \in \mathbb{Z}$ (integer numbers), $Q := c_k \sum\limits_{k=1}^{n}\left(a^{k-1}\! +\! a^{k-2}b\!+\ldots+\! b^{k-2}a\!+\!b^{k-1}\right)$ is a constant polynomial with integer coefficients. From this, it follows that also $\dfrac{P(a)-P(b)}{a-b}$ belongs to $\mathbb{Z}$, so $a-b$ divides $P(a)-P(b)$ by the Polynomial Factor Theorem: $$\color{red}{\text{Thesis}} \Rightarrow \quad a-b \mid P(a) - P(b) \qquad \forall \, a, b \in \mathbb{Z}. \qquad \qquad \blacksquare$$
So:
${\color{blue}{\text{Statement 1.}}} \qquad$ Let $a, b$, and $c$ denote three distinct integers.
It is impossible that $a -b \mid b - c, \quad b-c \mid c - a, \quad c-a \mid a-b$. In symbols: $$\underbrace{a \neq b \neq c}_{\color{Goldenrod}{\displaystyle \text{Hypothesis}}} \implies \underbrace{a -b \nmid b - c, \quad b-c \nmid c - a, \quad c-a \nmid a-b}_{\color{SkyBlue}{\displaystyle \text{Thesis}}} \tag{1.1}\label{eq:1.1} $$ or $$\underbrace{a -b \mid b - c, \quad b-c \mid c - a, \quad c-a \mid a-b \quad \text{for} \quad a, b, c \in \mathbb{Z}}_{\color{magenta}{\displaystyle \text{Hypothesis}}} \implies \underbrace{a = b = c}_{\color{Grey}{\displaystyle \text{Thesis}}}\tag{1.2}\label{eq:1.2} $$
is equivalent to
${\color{blue}{\text{Statement 2.}}} \qquad$ Let $a$, $b$, and $c$ denote three distinct integers, and let $P$ denote a polynomial having all integral coefficients.
It is impossible that $P(a)=b$, $P(b)=c$, and $P(c)=a$, i.e. $a = b =c$ holds. $$ \underbrace{P \in \mathbb{Z}[x] \quad \text{such that} \quad P(a)=b, P(b)=c, P(c)=a}_{\color{Orchid}{\displaystyle \text{Hypothesis}}} \implies \underbrace{a = b = c}_{\color{Pink}{\displaystyle \text{Thesis}}}. \tag{2}\label{eq:2}$$
We will prove the $\color{blue}{\text{Statement}}$ (in the form $\color{blue}{1.}$ or $\color{blue}{2.}$) through two methods.
Assume for the sake of contradiction that there can be three distinct integers $a, b, c \in \mathbb{Z}, \quad a \neq b \neq c$ ($\color{Pink}{\displaystyle \text{Thesis}}$ denial) such that $P(a) = a, P(b) = c, P(c) = a$, with P polynomial with integral coefficients (maintenance of $\color{Orchid}{\displaystyle \text{Hypothesis}}$).
$${\Huge \neg} \,\, \color{Pink}{\displaystyle \text{Thesis}} \, + \, \color{Orchid}{\displaystyle \text{Hypothesis}} \Rightarrow \exists \, a \neq b \neq c \quad \text{such that} \quad \begin{cases} P(a) = b \\ \vphantom0 \\ P(b) = c \\ \vphantom0 \\ P(c) = a \end{cases} \quad \text{for} \quad a, b, c \in \mathbb{Z}.$$
By the Lemma \eqref{eq:0}, this is equivalent to asserting that $$\begin{cases} a - b \mid \underbrace{P(a)}_{\displaystyle b} - \underbrace{P(b)}_{\displaystyle c} \\ \vphantom0 \\ b - c \mid \underbrace{P(b)}_{\displaystyle c} - \underbrace{P(c)}_{\displaystyle a} \\ \vphantom0 \\ c - a \mid \underbrace{P(c)}_{\displaystyle a} - \underbrace{P(a)}_{\displaystyle b} \end{cases} \implies \begin{cases} a - b \mid b- c \\ \vphantom0 \\ b - c \mid c - a \\ \vphantom0 \\ c- a \mid a - b \end{cases} \Rightarrow {\Huge \neg} \,\, \color{SkyBlue}{\displaystyle \text{Thesis}},$$ i.e. denying the thesis of \eqref{eq:1.1}. Asserting that a variable $v$ divides a variable $w$ is equivalent to asserting that $|v| \leq |w|$. Thus, for the three divisibilities, the relations $$\begin{cases} a - b \mid b- c \\ \vphantom0 \\ b - c \mid c - a \\ \vphantom0 \\ c- a \mid a - b \end{cases} \implies \begin{cases} |a - b|\leq |b- c| \\ \vphantom0 \\ |b - c| \leq |c - a| \\ \vphantom0 \\ |c- a| \leq |a - b| \end{cases}$$ are obtained.
Putting these together gives: $$|a - b|\leq |b- c|\leq |c - a|\leq |a - b|,$$ a chain of inequalities whose furthest left and right sides return $$|a - b|\leq|a - b|,$$ which holds if and only if $$|a-b| = |a-b|,$$ trivially.
So, for the chain of inequalities to be valid, one must have: $$|a - b|= |b- c|=|c - a|=|a - b|,$$ i.e. $$|a-b| = |b-c| = |c-a|.$$
Analyzing only the first two members, we have: $$\begin{align} |a-b| &= |b-c| \\ (|a-b|)^2 &= (|b-c|)^2 \qquad \text{(squaring both members)} \\ (a-b)^2 &= (b-c)^2 \qquad \text{(property of absolute value:}\, |x|^2 = (x)^2 \text{)} \\ a^2 - 2ab + b^2 &= b^2 - 2bc +c^2 \qquad \text{(square of a binomial)} \\ a^2 - 2ab + \cancel{b^2} &= \cancel{b^2} - 2bc +c^2 \qquad \text{(simplify like terms)} \\ a^2-c^2+2bc-2ab &= 0 \qquad \text{(writing the equation in implicit form)} \\ (a-c)(a+c) - 2b(a-c) &= 0 \qquad \text{(partial factor out)} \\ (a-c)(a+c-2b) &= 0, \qquad \text{(total factor out)}\\ \end{align}$$ from which $$a-c = 0 \qquad \text{and} \qquad a+c - 2b = 0,$$ which gives $$\begin{align} a= c \qquad &\text{and} \qquad a+c - 2b = 0 \\ a= c \qquad &\text{and} \qquad c+c - 2b = 0 \qquad \text{(substituting} \, c=a \, \text{into the second equation)} \\ a= c \qquad &\text{and} \qquad 2c - 2b = 0 \\ a= c \qquad &\text{and} \qquad \cancel{2}c = \cancel{2}b \\ a=c \qquad &\text{and} \qquad c = b. \end{align}$$
Finally:
$$a = b = c \Rightarrow {\Huge \neg} \,\, \color{GoldenRod}{\displaystyle \text{Hypothesis}},$$
the negation of $\color{GoldenRod}{\displaystyle \text{Hypothesis}}$ of \eqref{eq:1.1}, so the negation of $\color{Orchid}{\displaystyle \text{Hypothesis}}$ of \eqref{eq:2}.
$\perp \quad$ Contradiction. $\qquad \qquad \qquad \square$
$$\color{magenta}{\displaystyle \text{Hypothesis}} \Rightarrow a -b \mid b - c, \quad b-c \mid c - a, \quad c-a \mid a-b \quad \text{for} \quad a, b, c \in \mathbb{Z}.$$ By $\color{magenta}{\displaystyle \text{Hypothesis}}$, the three quantities $$\frac{b-c}{a-b} = \alpha, \qquad \frac{c-a}{b-c} = \beta, \qquad \frac{a-b}{c-a} = \gamma, \qquad \qquad \text{for} \quad \alpha, \beta, \gamma \in \mathbb{Z} \setminus \{0 \}$$ are all integers. Multiplying these three factors, they cancel out: $$\frac{{\color{red}{\cancel{b-c}}}}{{\color{blue}{\bcancel{a-b}}}} \cdot \frac{\cancel{c-a}}{{\color{red}{\bcancel{b-c}}}} \cdot \frac{{\color{blue}{\cancel{a-b}}}}{\bcancel{c-a}} = 1,$$ so their product $\alpha \cdot \beta \cdot \gamma = 1$. Analyzing the various cases:
Case 1. $$\alpha \cdot \beta \cdot \gamma = 1 \qquad \text{if} \qquad \boxed{\alpha = 1, \qquad \beta = 1, \qquad \gamma = 1}.$$
We have: $$\begin{cases} \dfrac{b-c}{a-b} = 1 \\ \vphantom0 \\ \dfrac{c-a}{b-c} = 1 \\ \vphantom0 \\ \dfrac{a-b}{c-a} = 1 \end{cases} \iff \begin{cases} b-c = a-b \\ \vphantom0 \\ c-a = b -c \\ \vphantom0 \\ a-b = c-a \end{cases} \iff \begin{cases} 2b = a + c \\ \vphantom0 \\ 2c = a +b \\ \vphantom0 \\ 2a = b+c \end{cases} \iff \begin{cases} b = \dfrac{a+c}{2} \\ \vphantom0 \\ c = \dfrac{a+b}{2} \\ \vphantom0 \\ a = \dfrac{b+c}{2} \end{cases}.$$ If all three numbers are equal, each number is the average of the other two. This is impossible if the quantities are distinct.
$\text{A})$ Suppose that $a < c$. Then, since $b$ is the average of $a$ and $c$, we get $a < b < c$. But $c$ is the average of $a$ and $b$, with the constraint $a < c < b$. Contradiction.
$\text{B})$ Suppose that $a < b$. Then, since $c$ is the average of $a$ and $b$, we get $a < c < b$. But $b$ is the average of $a$ and $c$, with the constraint $a < b < c$. Contradiction.
$\text{C})$ Suppose that $b < c$. Then, since $a$ is the average of $b$ and $c$, we get $b < a < c$. But $c$ is the average of $a$ and $b$, with the constraint $a < c < b$. Contradiction.
Case 2. Two out of three number are equal to $-1$.
If just one of the numbers is equal to $−1$, we get
$$\begin{align} \frac{b−c}{a-b} = -1 \qquad \text{or} \qquad &\frac{c-a}{b-c} = -1 \qquad \qquad \text{or} \qquad \frac{a-b}{c-a} = -1 \\ \cancel{b}- c = \cancel{b}-a \qquad \text{or} \qquad &\cancel{c} -a = \cancel{c}-b \qquad \,\, \text{or} \qquad \cancel{a}-b = \cancel{a} -c \\ -c = -a \qquad \text{or} \qquad &-a = - b \qquad \quad \quad \,\,\,\, \text{or} \qquad -b = -c \\ a = c \qquad \text{or} \qquad &a = b \qquad \qquad \qquad \quad \text{or} \qquad b = c, \end{align}$$ which contradicts the fact that $a,b,c$ are distinct.
So, $$a = b = c \Rightarrow \color{Grey}{\displaystyle \text{Thesis}}. \qquad \qquad \qquad \square$$
