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.

To find the sum of all possible values of $f(7)$, we first determine the function $f(x)$ by analyzing the given functional equation:
$$f(x)f(y) - f(xy) = x + y$$
for all $x, y \in \mathbb{R}$.

**Step 1: Determine the value of $f(0)$.**
Let $x = 0$ and $y = 0$ in the equation:
$$f(0)f(0) - f(0 \cdot 0) = 0 + 0$$
$$f(0)^2 - f(0) = 0$$
$$f(0)(f(0) - 1) = 0$$
This implies that $f(0)$ can be either $0$ or $1$.

**Step 2: Test the case $f(0) = 0$.**
Substitute $y = 0$ into the original equation:
$$f(x)f(0) - f(x \cdot 0) = x + 0$$
$$f(x) \cdot 0 - f(0) = x$$
$$0 - 0 = x$$
$$0 = x$$
This result implies that the equation only holds for $x = 0$, but the problem states it must hold for **all** real numbers $x$. Thus, $f(0) = 0$ leads to a contradiction and is not a valid solution.

**Step 3: Test the case $f(0) = 1$.**
Substitute $y = 0$ into the original equation again, using $f(0) = 1$:
$$f(x)f(0) - f(x \cdot 0) = x + 0$$
$$f(x) \cdot 1 - f(0) = x$$
$$f(x) - 1 = x$$
$$f(x) = x + 1$$
This gives us a candidate function $f(x) = x + 1$.

**Step 4: Verify the candidate solution.**
We must check if $f(x) = x + 1$ satisfies the original equation for all $x, y$.
Left Hand Side (LHS):
$$f(x)f(y) - f(xy) = (x + 1)(y + 1) - (xy + 1)$$
$$= (xy + x + y + 1) - xy - 1$$
$$= x + y$$
Right Hand Side (RHS):
$$x + y$$
Since LHS = RHS, $f(x) = x + 1$ is a valid solution. Since the derivation in Step 3 showed that $f(0)=1$ uniquely determines $f(x)$ for all $x$, this is the only solution.

**Step 5: Calculate $f(7)$.**
Using the function $f(x) = x + 1$:
$$f(7) = 7 + 1 = 8$$

Since there is only one possible value for $f(7)$, the sum of all possible values is 8.

\boxed{8}