<s>[INST] $f(x)$ 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.
 [/INST] Notice that you can get values for $f(2n) = (2n)^2 + 1$. Observing this, I suppose that $f$ is a quadratic polynomial with form $f(t) = at^2 + bt + 1$ ($+1$ since we know that $f(0) = 1$).
Writing the functional equation as $(f(x) - 1)(f(y) - 1) = f(xy) - 1$, we get $(ax^2 + bx)(ay^2 + by) = (ax^2 y^2 + bxy)$, so comparing coefficients we get the following system:
$a^2 = a$
$ab = 0$
$b^2 = b$
So this gives answers of $(1, 0), (0,1),$ and $(0,0)$. This gives answers of $f(x) = 1, f(x) = x+1, f(x) = x^2 + 1$.  Using our initial condition $f(2) = 5$, we know that only the last equation can be correct.
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