Question

If $f$ is a polynomial function satisfying $2+f\left(x\right)\cdot f\left(y\right)=f\left(x\right)+f\left(y\right)+f\left(xy\right),\forall x,yϵR$ and if $f\left(2\right)=5$,then find $f\left(f\right(2\left)\right)$.

Answer 1

Given, $2+f\left(x\right)f\left(y\right)=f\left(x\right)+f\left(y\right)+f\left(xy\right)$
or $1-f\left(x\right)-f\left(y\right)+f\left(x\right)f\left(y\right)=f\left(xy\right)-1$
or $(1-f(x\left)\right)(1-f(y\left)\right)=f\left(xy\right)-1$
The above result holds if and only if,
$f\left(x\right)=1+x^n$
if $f\left(x\right)=a_n{x}^n+a_{n-1}{x}^{n-1}+...+a_0$
Then, consider $(1+f(x\left)\right)(1-f(y\left)\right)=f\left(xy\right)-1$
Compare constant term on either side, we have
$1-a_0=a_0-1\Rightarrow a_0=1$
Comparing coefficient of $x^n{y}^n$, we get
$a_n^2=a_n\Rightarrow a_n=1$ or otherwise polynomial would not be of n degree.
Comparing coefficient of $x,x^1,....,x^{n-1}$ on either sides, we have
$a_1=a_2=...=a_{n-1}=0$
$\Rightarrow a_n=1$ and $f\left(x\right)=x^n+1$
Given, $f\left(2\right)=5$ ie, $2^n+1=5$
$\Rightarrow n=2$
Thus, $f\left(x\right)=x^2+1$
$f\left(f\right(2\left)\right)=f\left(5\right)=5^2+1=26$