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Cute problem $f(xy-1) + f(x)f(y) = 2xy-1$ ....$P(x, y)$ If $f(0) \neq 0$ : $P(x, 0)$ : $f(x) = \frac{-1 - f(-1)}{f(0)}, \forall x \in \mathbb{R}$, but the constant function doesn't satisfy the problem So, $f(0) = 0$ By $P(xy, 1)$ and $P(x, y)$ : $f(xy)f(1)=f(x)f(y), \forall x$ and $y \in \mathbb{R}$ ....$(\alpha_2)$ $P(1, 1)$ : $f(1)^{2} = 1 \rightarrow f(1) \in$ {$1, -1$} Case 1) $f(1)=-1$ By $(\alpha_2)$ : $-f(xy)=f(x)f(y), \forall x$ and $y \in \mathbb{R}$ $P(x, 1)$ : $f(x-1)-= 2x+f(x)-1$ ....$(\alpha_3)$ $P(x+1, 1)$ : $-f(x+1) = 2x + f(x)-1$ ....$(\alpha_4)$ Multiplying $(\alpha_3)$ and $(\alpha_4)$ : $-f(x+1)f(x-1) = (2x-(f(x)-1))(2x+(f(x)-1)) \rightarrow f((x+1)(x-1))=(2x)^2 - (f(x)-1)^2 \rightarrow f(x^2 -1) + f(x)^2 = 4x^2 +2f(x)-1$ Replacing it in $P(x, x)$ : $4x^2 +2f(x)-1 = 2x^2 -1 \rightarrow f(x) = -x^2, \forall x \in \mathbb{R}$, and it satisfy the problem Case 2) $f(1)=1$ By $(\alpha_2)$ : $f(xy)=f(x)f(y), \forall x$ and $y \in \mathbb{R}$ $P(x, 1)$ : $f(x-1)-= 2x-f(x)-1$ ....$(\alpha_5)$ $P(x+1, 1)$ : $f(x+1) = 2x - f(x)-1$ ....$(\alpha_6)$ Multiplying $(\alpha_5)$ and $(\alpha_6)$ : $f(x+1)f(x-1) = (2x-f(x)+1))(2x-f(x)-1)) \rightarrow f((x+1)(x-1))=(2x-f(x))^2 - 1^2 \rightarrow f(x^2 -1) = 4x^2 -4xf(x)+ f(x)^2 -1$ Replacing it in $P(x, x)$ : $4x^2 -4xf(x)+ f(x)^2 -1 + f(x)^2 = 2x^2 -1 \rightarrow 2x^2 -4xf(x) + 2f(x)^2 =0 \rightarrow x^2 -2xf(x) + f(x)^2 =0 \rightarrow (f(x)-x)^2 = 0 \rightarrow f(x)=x, \forall x \in \mathbb{R}$, and it satisfy the problem Therefore, the functions are $f(x)=-x^2$ and $f(x)=x$