, 2014

Let $f$ be any function satisfying
$$f(xy - 1) + f(x)f(y) = 2xy - 1 \quad (1)$$
for all $x, y \in \mathbb{R}$. Plug in $x = 0$ in (1), we get $f(-1) + f(x)f(0) = -1$ for all $x \in \mathbb{R}$.
If $f(0) \neq 0$, then $f$ is a constant function which does not satisfy the equation (1) for all $x, y \in \mathbb{R}$ and hence we get a contradiction. So $f(0) = 0$. Plug $x = y = 1$ in (1) to get $f(1)^2 = 1$. Thus $f(1) = 1$ or $f(1) = -1$.

Case 1: $f(1) = 1$. Substitute $x = xy$ and $y = 1$ in (1) to obtain $f(xy-1)+f(xy) = 2xy - 1$ for all $x, y \in \mathbb{R}$. Thus, by (1), we have
$$f(xy) = f(x)f(y) \qquad (2)$$
for all $x, y \in \mathbb{R}$. Next, we substitute $y = 1$ and $x = x + 1, y = 1$ in (1) to get
$$f(x-1) = 2x-1-f(x) \quad \text{and} \quad f(x+1) = 2x+1-f(x) \qquad (3)$$
for all $x \in \mathbb{R}$. Now letting $y = x$ in (1) and using (2) and (3), we then have
$$\begin{aligned} 2x^2 - 1 &= f(x^2 - 1) + f(x)^2 \\ &= f(x-1)f(x+1) + f(x)^2 \\ &= (2x-1-f(x))(2x+1-f(x)) + f(x)^2 \\ &= 2f(x)^2 - 4xf(x) + 4x^2 - 1 \end{aligned}$$
and hence $2(f(x) - x)^2 = 0$ which implies that $f(x) = x$ for all $x \in \mathbb{R}$.

Case 2: $f(1) = -1$. By the same substitutions as in case 1, we get $f(xy) = -f(x)f(y)$, $f(x-1) = 2x-1+f(x)$ and $f(x+1) = -(2x+1)+f(x)$ for all $x, y \in \mathbb{R}$.
Now plugging $y = x$ in (1) and using the previous three equations, we get
$$\begin{aligned} 2x^2 - 1 &= f(x^2 - 1) + f(x)^2 \\ &= -f(x-1)f(x+1) + f(x)^2 \\ &= -(2x-1+f(x))(-(2x+1)+f(x)) + f(x)^2 \\ &= 2f(x) + 4x^2 - 1 \end{aligned}$$
and thus $f(x) = -x^2$ for all $x \in \mathbb{R}$.

We can easily check that both functions obtained from those two cases satisfy (1) for all $x, y \in \mathbb{R}$. Hence, all functions satisfying (1) for all $x, y \in \mathbb{R}$ are $f(x) = x$ for all $x \in \mathbb{R}$ or $f(x) = -x^2$ for all $x \in \mathbb{R}$. $\square$