Given that f (x) is defined as R, for any x, y belongs to R, f (x + y) + F (X-Y) = 2F (x) (y), and f (0) is not equal to 0 If there is a constant C, Let f (C / 2) = 0. Proof: for any x belonging to R, f (x + C) = - f (x)

Given that f (x) is defined as R, for any x, y belongs to R, f (x + y) + F (X-Y) = 2F (x) (y), and f (0) is not equal to 0 If there is a constant C, Let f (C / 2) = 0. Proof: for any x belonging to R, f (x + C) = - f (x)

Let y = 0 substitute f (x + y) + F (X-Y) = 2F (x) (y)
Then 2F (x) = 2F (x) ^ 2, so f (x) = f (x) ^ 2, so f (x) = 0 or 1
Let y = C / 2 be replaced by F (x + y) + F (X-Y) = 2F (x) (y)
Then f (x + C / 2) + F (x-C / 2) = 2F (x) f (C / 2) = 0
So f (x + C / 2) ^ 2 + F (x-C / 2) ^ 2 = 0
So f (x + C / 2) = f (x-C / 2) = 0 {x is not equal to positive and negative C / 2}
Let x = x + C / 2, so f (x + C) = f (x) = 0 = - f (x) {x is not equal to 0 or - C}
We only need to prove that f (c) = - f (0) = - 1 and - f (- C) = f (0) = 1
That is to say, f (c) = f (- C) = - 1
Let x = y = C / 2, substitute f (x + y) + F (X-Y) = 2F (x) (y)
F (c) + F (0) = 2F (C / 2) ^ 2 = 0, because f (0) = 1, so f (c) = - 1
Let x = C, y = C, substitute f (x + y) + F (X-Y) = 2F (x) (y)
We get f (2C) + F (0) = 2F (c) ^ 2 = 2
Let x = C, y = - C, substitute f (x + y) + F (X-Y) = 2F (x) (y)
F (0) + F (2C) = 2F (c) f (- C) = 2, because f (c) = - 1, so f (- C) = - 1
So for any x belonging to R, f (x + C) = - f (x)