elementary number theory - Prove that there do not exist distinct integers $a,b,c$ and polynomial $P$, with integer coefficients, such that $P(a)=b, P(b)=c, P(c)=a$.

Let $P(x)$ be a polynomial of degree $n$ with integer coefficients. Assume that there exist three distinct integers $a,b,c$ such that $$P(a)=b, P(b)=c, P(c)=a.$$

Since the integers $a,b,c$ are all different, there must be a least among them. Without loss of generality, we can write $a

Note that, for any $x,y \in{\mathbb{R}}$:

$$P(x) -P(y)=a_n(x^n-y^n)+a_{n-1}(x^{n-1}-y^{n-1})+ \cdots + a_1(x-y).$$

Thus, using the identity

$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}),$$

it follows that

$$P(x) -P(y)=(x-y)Q(x,y),$$

where $Q(x,y)$ is a polynomial in $x,y$ of degree $(n-1)$. Evaluating $P(a)$ and $P(b)$ this way gives us the following equation:

$$P(a)-P(b)=(a-b) Q(a,b)=b-c$$

By our construction, the polynomial $Q$ must also have integer coefficients and so $Q(a,b)=m_1$, for some integer $m_1$. Arguing similarly as above, we get the following three equations:

$$(a-b)m_1=b-c$$

$$(b-c)m_2=c-a$$
$$(a-c)m_3=b-a$$

Clearly $m_i\neq0$ for any $i$. Dividing the first of these equations by the second:

$$\frac{a-b}{c-a}m_1=\frac{b-c}{(b-c)m_2}$$
$$\frac{a-b}{c-a}m_1=\frac{1}{m_2} \qquad (1)$$

A few algebraic manipulations of the last equation gives

$$\frac{1}{c-a}=\frac{m_3}{a-b},$$

which we can substitute into $(1)$ and get

$$m_1 m_3= \frac{1}{m_2}.$$

Since the $m_i$ are integers we must have $$m_i=\pm1.$$

We need only consider the two cases for $m_3$. If $m_3=-1$ this implies $c-a=b-a \implies c=b$. If $m_3= 1$ this implies $a-c=b-a$, which is not possible, since $a$ is less than both $c$ and $b$. Thus, since both possible values for $m_3$ lead to a contradiction, there is no polynomial $P(x)$ and distinct integers $a,b,c$ which satisfy the condition $P(a)=b, P(b)=c, P(c)=a.$ $\quad \square$