math-func-01

# A Multiplicative Functional Equation

Let $f:\mathbb{R}\to\mathbb{R}$ be a function satisfying
$$f(x)\,f(y) - f(xy) = x + y$$
for **all** real numbers $x$ and $y$.

Find the sum of all possible values of $f(7)$.

Give your final answer as a single integer.

We are given the functional equation:
$$f(x)f(y) - f(xy) = x + y$$
for all real numbers $x$ and $y$.

Let's analyze the properties of the function $f$.
First, let $x = 1$ and $y = 1$:
$$f(1)f(1) - f(1) = 1 + 1$$
$$f(1)^2 - f(1) - 2 = 0$$
$$(f(1) - 2)(f(1) + 1) = 0$$
So, $f(1) = 2$ or $f(1) = -1$.

Next, let $y = 1$:
$$f(x)f(1) - f(x) = x + 1$$
$$f(x)(f(1) - 1) = x + 1$$
This gives us a linear expression for $f(x)$ in terms of $x$.
Case 1: $f(1) = 2$.
$$f(x)(2 - 1) = x + 1 \implies f(x) = x + 1$$
Case 2: $f(1) = -1$.
$$f(x)(-1 - 1) = x + 1 \implies -2f(x) = x + 1 \implies f(x) = -\frac{1}{2}x - \frac{1}{2}$$

We must check if these candidates satisfy the original equation for all $x, y$.
Substitute $f(x) = x + 1$:
$$LHS = (x+1)(y+1) - (xy+1) = xy + x + y + 1 - xy - 1 = x + y = RHS$$
So $f(x) = x + 1$ is a solution.
Substitute $f(x) = -\frac{1}{2}x - \frac{1}{2}$:
$$LHS = \left(-\frac{1}{2}x - \frac{1}{2}\right)\left(-\frac{1}{2}y - \frac{1}{2}\right) - \left(-\frac{1}{2}xy - \frac{1}{2}\right)$$
$$= \frac{1}{4}(x+1)(y+1) + \frac{1}{2}xy + \frac{1}{2}$$
$$= \frac{1}{4}(xy + x + y + 1) + \frac{1}{2}xy + \frac{1}{2}$$
$$= \frac{1}{4}xy + \frac{1}{4}x + \frac{1}{4}y + \frac{1}{4} + \frac{2}{4}xy + \frac{2}{4}$$
$$= \frac{3}{4}xy + \frac{1}{4}x + \frac{1}{4}y + \frac{3}{4}$$
This is not equal to $x + y$ for all $x, y$ (e.g., if $x=0, y=0$, LHS is $3/4 \neq 0$).
Thus, the only solution is $f(x) = x + 1$.

We need to find the sum of all possible values of $f(7)$.
Since there is only one solution, there is only one possible value for $f(7)$.
$$f(7) = 7 + 1 = 8$$
The sum of all possible values is 8.

\boxed{8}