Is the function complicated?

Given that $f(2xy-1) = f(x)f(y) - f(x) - 2y -1$ and $f(0) = 1$ , we have:

$\begin{aligned} f(2xy-1) & = f(x)f(y) - f(x) - 2y -1 & \small \color{#3D99F6} \text{Putting }x=y=0 \\ f(-1) & = 1\cdot 1 - 1 -0-1 = -1 \end{aligned}$

Putting $x=0$ :

$\begin{aligned} f(-1) & = f(y) - 1 - 2y -1 \\ - 1 & = f(y) - 1 - 2y -1 \\ \implies f(y) & = 2y+1 & \small \color{#3D99F6} \text{Replace }y \text{ with }y \\ f(x) & = 2x+1 \end{aligned}$

Therefore,

• $f(5) = 11$ not a composite number.
• Since $f(-1) = -1$ , $f(x) \not \ge 0$ for all real $x$ .
• $f(7) = 13$ not an even number.
• $f(12) = 25$ is a perfect square.