Polynomial $P(a)=b,P(b)=c,P(c)=a$

Hint: If $P(a)=b$ and $P(b)=c$ then $a-b$ divides $b-c$.

Lemma:If $P(x)$ is a polynomial with integer coefficients then $a-b \mid P(a)-P(b)$

Proof:Take $P(x)=a_{n}x^n+a_{n-1}x^{n-1}+\dots+a_{1}x+a_0$

Then You cold easily prove the lemma by the identity $$a^k-b^k=(a-b)\sum_{\ell=0}^{k-1}a^\ell b^{k-\ell-1}$$

So we have:

$$b-c \mid a-b$$$$a-b \mid c-a$$ $$c-a \mid b-c$$

Which gives us $b-c \mid a-b \mid c-a \mid b-c$.

$$\Rightarrow \mid b-c \mid \ge \mid a-b \mid \ge \mid c-a \mid \ge \mid b-c \mid$$

then :

$\mid b-c \mid =\mid a-b \mid =\mid c-a \mid$

The problem is cyclic so without loss of generality we can take $a=max\{a,b,c\}$.So we have:

$a-b=a-c \Rightarrow b=c$

Which is wrong.

Hint $\ a\!-\!b\mid P(a)\!-\!P(b) = b\!-\!c.\,$ By symmetry $\,b\!-\!c\mid c\!-\!a,\ \ c\!-\!a\mid a\!-\!b.\ $ Chained, these yield a divisibility cycle $\ \color{#c00}j\mid k\mid n\mid \color{#c00} j,\ $ so $\ k,n = \pm j.\ $ But $\,j+k+n = 0\,\Rightarrow\, j=0\,\Rightarrow\,a=b\,\Rightarrow\!\Leftarrow$

Remark $ $ The divisibility is a specialization of the Factor Theorem $\,x-b\mid P(x)-P(b)\,$ or the Polynomial Congruence Rule $\bmod a\!-\!b\!:\ a\equiv b\,\Rightarrow\, P(a)\equiv P(b),\ $ for $\,P\in\Bbb Z[x],\,$ see here

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