Find all functions such as $ f (x) f (y) = f (xy + 1) + f (x – y) – $ 2 for everyone $ x, y $ are real numbers.

I put y = 0 in the equation and get $ (f (0) -1) f (x) = f (1) -2 $. Yes $ f (0) 1, then f (x) = (f (1) -2) / (f (0) -1) $ and so the function would be constant. But since $ c ^ 2 = 2c-2 $ There is no real solutions, the function can not be constant. Therefore, $ f (0) = $ 1, which implies $ f (1) = $ 2. Also note that f must be a pair function like $ f (x-y) = f (y-x) $.

Together $ f (x) = 1 + x ^ 2 + g (x) $ for some same function of $ g $with $ g (0) = g (1) = 0 $. Then put $ f (x) $ in the functional equation and substitute $ y = $ 1get $ g (x + 1) -g (x) = g (x) -g (x-1) $. To define $ P (x) = {…, g (x-2), g (x-1), g (x), g (x + 1), g (x + 2), …} , so $ P (x) $ is an A.P. Then substitutes $ y = -x, y = x $ and using the fact that g is even, to get $ g (x ^ 2 + 1) – g (x ^ 2-1) = g (2x) $. I'm trying to prove $ g (x) = $ 0 for everyone $ x $. How can I proceed?