number cubic polynomials possible

Question

Let $p(x)$ be a cubic polynomial with integral coefficients , such that $p(a)=b$, $p(b)=c$, $p(c)=a$ for $a,b,c$ being distinct integers . find number of such possible polynomials.

Answer 1

Assume first that $a=0$. Then $(X-c)$ is a factor of $p(X)$ and the constant term is $b$, which implies that $c\mid b$. Also, $b\mid p(b)=c$, hence $c=\pm b$ and by distinctness, $b=-c$.

Now if we drop the assumption that $a=0$, we obtain from $p(X)$ the polynomial $q(X)=p(X+a)-a$ with the property $q(0)=b-a=:b'$, $q(b')=c-a=:c'$ and $q(c')=0$, i.e. a similar polynomial which does have $a=0$. From the first paragraph we conclude $b'+c'=0$, i.e. $a=\frac{b+c}{2}$. By cyclic symmetry, we may assume wlog. that $a$ is the smallest of the three numbers; contradiction.

Answer 2

As is well known about polynomials of integer coefficients, $ a-b \mid P(a) - P(b) $.

This gives us that $ a-b \mid P(a) - P(b) = b-c$ and $b-c \mid P(b) - P(c) = c-a$ and $ c-a \mid P(c) - P(a) = a-b$.

Hence, we must have $|a-b|=|b-c|=|c-a|$, or that two of them have the same value. Say $a-b = b-c = k$, then we have that $c-a = - [ (a-b) - (b-c) ] $, so $|k| = 2k$. This gives us $k=0$, or that $a=b=c$, which contradicts the assumption of distinct integers.