The function f : R -> R satisfies

f(x)f(y) = f(x + y) + f(xy)

for all real numbers x and y. Find all possible functions f.

Guest May 28, 2022

#1**0 **

When x = y = 0, \((f(0))^2 = 2f(0)\).

This implies f(0) = 0 or f(0) = 2.

Case 1: f(0) = 2.

When x = 0, 2 f(y) = f(y) + 2

f(y) = 2 for all real numbers y.

The function f(x) = 2 is a solution.

Case 2: f(0) = 0.

When x = 0, 0 = f(y) for all real numbers y.

The function f(x) = 0 is a solution.

Therefore, the possible functions are the constant functions f(x) = 0 and f(x) = 2.

MaxWong May 28, 2022