# Functional issues: Find all functions such as \$ f (x) f (y) = f (xy + 1) + f (x – y) – \$ 2

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 $$y = 1$$get $$g (x + 1) -g (x) = g (x) -g (x-1)$$. 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?