Functional Equation

The functional identity can be written $f(xy) - 1 \; = \; \big(f(x) - 1\big)\big(f(y) - 1\big) \;.$ Thus $f(x)-1 = (f(x)-1)(f(1)-1)$ for all $x$ . Since $f(2) = 5$ , $f$ is not identically equal to $1$ , and hence $f(1) = 2$ .

Since $f(2x)-1 = (f(2) - 1)(f(x) - 1) = 4(f(x) -1)$ , we deduce by induction that $f(2^y) \,=\, 4^y + 1$ for all integers $y \ge 0$ . Since the polynomial functions $f(x)$ and $x^2 + 1$ agree at an infinite number of points (the integer powers of $2$ ), they must be identical, and so $f(x) = x^2 + 1$ , and hence $f(3) = \boxed{10}$ .

Input $y=2$

$\begin{aligned} 5f(x)&=f(x)+5+f(2x)-2\\4f(x)&=3+f(2x)\end{aligned}$

Input $x=1$ in the equation obtained.

$\begin{aligned} 4f(1)&=3+5\\f(1)&=2\end{aligned}$

Input $y=\frac{1}{x}$ in the given equation.

$\begin{aligned} f(x).f\left(\frac{1}{x}\right)&=f(x)+f\left(\frac{1}{x}\right)+f(1)-2\\f(x).f\left(\frac{1}{x}\right)&=f(x)+f\left(\frac{1}{x}\right)\\f(x).f\left(\frac{1}{x}\right)-f(x)-\left(\frac{1}{x}\right)+1&=1\\\left(f(x)-1\right)\left(f\left(\frac{1}{x}\right)-1\right)&=1\end{aligned}$

Input $g(x)=f(x)-1$

$\Rightarrow g(x).g\left(\frac{1}{x}\right)=1$

Since $f$ is a polynomial function, $g$ is ought to be a polynomial function. Hence, for some whole number $n$

$\begin{aligned} g(x)&=\pm{x}^{n}\\f(x)&=1\pm{x}^n \end{aligned}$

Input $x=2$

$\begin{aligned}f(2)&=1\pm 2^{n}\\5&=1\pm 2^{n}\\n&=2\\\Rightarrow f(x)&=1+{x}^2\end{aligned}$

Input $x=3$

$\begin{aligned} f(3)&=1+{3}^2\\&\huge\color{#3D99F6}{=\boxed{10}}\end{aligned}$

---

This is a useful result which one can use on coming across functional equations. $\begin{aligned} f(x).f\left(\frac{1}{x}\right)&=f(x)+f\left(\frac{1}{x}\right)\\\Rightarrow f(x)&=1\pm {x}^n \end{aligned}$ Adding or subtracting a constant term from a polynomial yields another polynomial. Hence $g$ is a polynomial. Multiplicative identity holds good only for monomials.