Let y = f (x) (x belongs to R) satisfy f (x1) + F (x2) = f (x1 * x2) for any real number x1, X2, and prove that (1) f (1) = f (- 1) = 0 (2) f (x) is even function

Let y = f (x) (x belongs to R) satisfy f (x1) + F (x2) = f (x1 * x2) for any real number x1, X2, and prove that (1) f (1) = f (- 1) = 0 (2) f (x) is even function

(1) Let X1 = x2 = 1, then f (x1) + F (x2) = f (1) + F (1) = f (1)
So 2F (1) = f (1)
F (1) = 0
Similarly, let X1 = x2 = - 1, then f (- 1) + F (- 1) = f (1) = 0
So f (- 1) = 0
(2) Let x2 = 1 / x1, then
f(x1)+f(1/x1)=f(x1*1/x1)=f(1)=0
f(-x1)+f(1/x1)=f(-x1*1/x1)=f(-1)=0
By subtracting the above two formulas, f (x1) - f (x2) = 0
So: F (x1) = f (x2)
So f (x) is an even function