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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.

 May 28, 2022
 #1
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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.

 May 28, 2022

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