TITLE: Polynomial Functional Equation: $f(x)f(y)+2=f(x)+f(y)+f(xy)$
QUESTION [5 upvotes]: Let $f(x)$ be a one-one, polynomial function such that  $f(x)f(y)+2=f(x)+f(y)+f(xy) , \ \forall x,y \in \mathbb R \setminus \{0\}$, $f(1) \ne 1$, $f'(1)=3$. Find $f(x)$.

I tried to find the degree of the polynomial from the equation by using suitable substitution, but it didn't work. Also, I found that $f(1)=2$ and then I substituted  $y=\frac{1}{x}$ to get $f(x)f\left(\frac{1}{x}\right)= f(x) + f\left(\frac{1}{x}\right)$. But I can't simplify further. Also, the answer given in my book is $f(x)=x^3+1$.
Any help will be appreciated.
Thanks.

REPLY [3 votes]: To add the solution by @Rory Daulton: Set $y=0$ and you get $$f(x)f(0)+2=f(x)+2f(0)$$ With $f(0)=2$ you get $$f(x)=2f(0)-2$$ which contradicts $f^\prime(1)=3$. Thus $f(0)=1$.
EDIT 1: Take $\partial_x$ from both sides in $$f(x)f(y)+2=f(x)+f(y)+f(xy)$$ which yields $$f^\prime(x)f(y)=f^\prime(x)+yf^\prime(xy)$$
Set $x=1$:$$3f(y)=3+yf^\prime(y)$$ Thus $$3f(y)-3=yf^\prime(y)$$ From This ODE you can derive, that the degree of the polynomial is 3. Now you can solve this ODE...