$f(x)$ is a polynomial satisfying $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2)$), given $f(2)=5.$ 
If f(x) is a polynomial satisfying  $2 + f(x)f(y)=f(x)+f(y)+f(xy)$, find $f(f(2))$, given $f(2)=5.$

ATTEMPT:- $f(f(2))=f(5)$, We can find $f(0)$,$f(1)$ and $f(1/2)$ to be $1,2$ and $5/4$ respectively.
we can change the function to the form $g(x)*g(y)=g(xy)$ by basic transformation.
but how to get $f(5)$ without using transformations.

Substitute $y = 2$ to get $2+f(x)f(2) = f(x)+f(2)+f(2x)$, which simplifies to $f(2x)-1 = 4(f(x)-1)$ for all reals $x$. 
Now, suppose that $x_0 \neq 0$ is a zero of $f(x)-1$. Then $f(2x_0)-1 = 4(f(x_0)-1) = 4 \cdot 0 = 0$, and so, $2x_0$ is also a zero of $f(x)-1$. Continue this process indefinitely to get that $2^kx_0$ is a zero of $f(x)-1$ for all non-negative integers $k$. But $f(x)-1$ is a non-zero polynomial, and thus, can only have finitely many zeros, a contradiction. 
Therefore, the only zero of $f(x) - 1$ is $0$, i.e. $f(x)-1 = Cx^n$ for some constant $C$ and some non-negative integer $n$. 
Now, use the fact that $f(2x)-1 = 4(f(x)-1)$ to determine $n$, and then use the fact that $f(2) = 5$ to determine $C$. 
