$f(x)\cdot f(y)=f(x)+f(y)+f(xy)-2$, $f(2)=5$, then $f(3)=?$
My attempt is as follows:-
Finding $f(1)$
$$f(1).f(2)=f(1)+f(2)+f(2)-2$$ $$5f(1)=f(1)+10-2$$ $$4f(1)=8$$ $$f(1)=2$$
Finding $f(0)$ $$f(0).f(2)=f(0)+f(2)+f(0)-2$$ $$5f(0)=2f(0)+5-2$$ $$3f(0)=3$$ $$f(0)=1$$
Finding $f(3)$ $$f(0).f(3)=f(0)+f(3)+f(0)-2$$ $$f(3)=f(0)+f(3)+f(0)-2$$ $$0=0$$
Let's take the help of $f(1)$ $$f(1).f(3)=f(1)+f(3)+f(3)-2$$ $$2f(3)=2f(3)+2-2$$ $$0=0$$
Let's take the help of $f(2)$ $$f(2).f(3)=f(2)+f(3)+f(6)-2$$ $$5f(3)=5+f(3)+f(6)-2$$ $$4f(3)=3+f(6)$$
But we don't $f(6)$
I am not seeing any way to calculate $f(3)$. Any suggestions?
Note that if $g(x)=f(x)-1$, then $g(2)=4$ and $$g(x)\cdot g(y)=g(xy)$$This relation implies that $g(0)=0$ and $g(1)=1$.
So, we know that for all $n\in\mathbb N\cup\{0\}$, $g(2^n)=2^{2n}$. There is only one polynomial that satisfies this, which is $g(x)=x^2$ (if two polynomials agree on infinitely many points, their difference is also a polynomial, which can't have infinitely many roots unless it is trivial). So, $$f(x)=x^2+1$$