Question

Prove that "There exists no polynomial $f\left(x\right)$ with integral coefficients which satisfy $f\left(a\right)=b,f\left(b\right)=c,f\left(c\right)=a,$ where $a,b,c$ are distinct integers."

Answer 1

Let $f\left(x\right)=a_0+a_{1}x+a_2{x}^2+...+a_n{x}^n$, where $a_iϵ$ integer ($i=0,1,2,...,n$)

Now, $f\left(a\right)=a_0+a_{1}a+a_2{a}^2+...+a_n{a}^n=b$ (i)
$f\left(b\right)=a_0+a_{1}b+a_2{b}^2+...+a_n{b}^n=c$ (ii)
$f\left(c\right)=a_0+a_{1}c+a_2{c}^2+...+a_n{c}^n=a$ (iii)
which gives $f\left(a\right),f\left(b\right),f\left(c\right)$ $ϵ$ integer

$\therefore$ $f\left(a\right)-f\left(b\right)=(a-b)\times$ A function in terms of $a$ and $b$.
$\Rightarrow b-c=(a-b)f_1(a,b)$, where $f_1(a,b)$ is an integer.

Similarly, $(b-c)f_1(b,c)=c-a$
and $(c-a)f_1(c,a)=a-b$
Multiplying above expressions, we get
$f_1(a,b)f_1(b,c)f_1(c,a)=1$
$\Rightarrow$ $f_1(a,b)=1$, $f_1(b,c)=1$, $f_1(c,a)=1$ (as product of integers is 1, if each is one)
$\Rightarrow$ $|a-b|=|b-c|=|c-a|$
$\Rightarrow$ $a-b=b-c$ or $a-b=c-b$
$\Rightarrow$ $a=c$ which is not possible (as $a$, $b$, $c$ are distinct)
Similarly, we get other cases.
Hence, no polynomial exist.