The polynomial function

It is given that: $\space f(x)f(y) = f(x) + f(y) + f(xy) - 2$ , where $\space f(x) = \displaystyle \sum_{k=0}^n a_kx^k$ .

$\begin{aligned} \text{Therefore, we have } \space f(0)\color{#3D99F6}{f(2)} & = f(0) + \color{#3D99F6}{f(2)} + f(0) - 2 \quad \quad \small \color{#3D99F6}{\text{Since } f(2) = 5} \\ \color{#3D99F6}{5}f(0) & = f(0) + \color{#3D99F6}{5} + f(0) - 2 \\ \Rightarrow f(0) & = 1 \\ \Rightarrow f(x) & = \sum_{k=1}^n a_kx^k + 1 \end{aligned}$

$\begin{aligned} \text{And, we have } \space f(1)\color{#3D99F6}{f(2)} & = f(1) + \color{#3D99F6}{f(2)} + \color{#3D99F6}{f(2)} - 2 \\ \color{#3D99F6}{5}f(1) & = f(1) +2(\color{#3D99F6}{5}) - 2 \\ \Rightarrow f(1) & = 2 \\ \Rightarrow f(x)f\left(\frac{1}{x}\right) & = f(x) + f\left(\frac{1}{x}\right) + f(1) - 2 \\ & = f(x) + f\left(\frac{1}{x}\right) + 2 - 2 \\ \Rightarrow f(x)f\left(\frac{1}{x}\right) & = f(x) + f\left(\frac{1}{x}\right) \end{aligned}$

Therefore, the polynomial must be of the form $\space f(x) = (x^n + 1)$ , so that:

$\begin{aligned} f(x)f\left(\frac{1}{x}\right) & = (x^n + 1) \left(\frac{1}{x^n} +1\right) \\ & = 1 + x^n + \frac{1}{x^n} + 1 \\ & = f(x) + f\left(\frac{1}{x}\right) \end{aligned}$

Since $f(2) = 5$ , $\Rightarrow f(x) = x^2 + 1 \quad \Rightarrow f(3) = 3^2+1 = \boxed{10}$

If f(x) is a polynomial, then it is differentiable over all real x. Differentiating the above functional equation with respect to x and y each gives:

$f'(x)f(y) = f '(x) + yf'(xy) \tag{i}$

$f(x)f'(y) = f'(y) + xf'(xy) \tag{ii}$

and equating (i) with (ii) yields:

$\frac{xf'(x)}{f(x)-1} = \frac{yf'(y)}{f(y)-1} = A \ \text{(real constant)} \tag{iii}$

and integrating (iii) with respect to $x$ gives:

$\ln[f(x)-1] = A\ln x + B$

or

$f(x) = e^Bx^A + 1 \tag{iv}$

Plugging in the boundary condition f(2) = 5 will produce

$4 = e^B \times 2^A \Rightarrow 2^2 = e^B \times 2^A$

which holds only for $A = 2, B = 0$ . Thus, the sought polynomial is $f(x) = x^2 + 1$ with $f(3) = \boxed{10}$ .

A Nice question!!:):)

$It's\quad given\quad that\quad f\left( x \right) .f\left( y \right) =f\left( x \right) +f\left( y \right) +f\left( xy \right) -2\\ Also,\quad f\left( 2 \right) =5.\quad \quad \quad \quad \longrightarrow eq.1\\ Substitute\quad x=2\quad in\quad first\quad equation\\ f\left( 2 \right) .f\left( y \right) =f\left( 2 \right) +f\left( y \right) +f\left( 2y \right) -2\\ 5f\left( y \right) =5+f\left( y \right) +f\left( 2y \right) -2\\ 4f\left( y \right) =3+f\left( 2y \right) \\ f\left( 2y \right) =4f\left( y \right) -3$

$Now,\quad put\quad y=1:\\ f\left( 2y \right)=4f\left( y \right)-3\\ f\left( 2 \right)=4f\left( 1 \right)-3\\ 5+3=4f\left( 1 \right)\\ f\left( 1 \right)=2\quad \quad \quad \quad \quad \longrightarrow eq.2$

$Similarly,\quad put\quad y=2\\ f\left( 2y \right) =4f\left( y \right) -3\\ f\left( 4 \right) =4(f\left( 2 \right) )-3\\ f\left( 4 \right) =20-3\\ f\left( 4 \right) =17\quad \quad \quad \quad \quad \longrightarrow eq.3$

$From\quad eq.1,\quad eq.2,\quad and\quad eq.3\\ we\quad can\quad deduce\quad that\quad \\ f\left( x \right)={ x }^{ 2 }+1\\ Therefore,\quad f\left( 3 \right)={ 3 }^{ 2 }+1\\ \Rightarrow f\left( 3 \right)=10$

Cheers!!:):)

Note that $\begin{aligned} f(x)f(y) &= f(x)+f(y)+f(xy) - 2 \iff (f(x)-1)(f(y)-1) = f(xy)-1 \\ &\implies f(x^n)-1 = \left(f(x)-1\right)^n,\quad \forall n \geq 1 \\ &\implies f(2^n) = 1 + \left(f(2)-1\right)^n = 1 + 4^n = 1 + \left(2^n\right)^2, \quad \forall n \geq 1 \end{aligned}$

Since $f(x) = 1 + x^2$ for the infinitely many values $x=2^n$ for positive integers $n$ and $f$ is a polynomial, it follows that $f(x) = 1 + x^2$ for all $x$ .

In particular, $f(3) = 1 + 3^2 = \boxed{10}$