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.

I need to find all functions $f: \mathbb{R} \to \mathbb{R}$ that satisfy the functional equation $f(x)f(y) - f(xy) = x + y$ for all real numbers $x$ and $y$.

Let me denote the functional equation as:
$$f(x)f(y) - f(xy) = x + y \quad (1)$$

First, I'll find $f(0)$ by setting $x = 0$ in equation (1):
$$f(0)f(y) - f(0) = 0 + y$$
$$f(0)(f(y) - 1) = y$$

If $f(0) \neq 0$, then:
$$f(y) - 1 = \frac{y}{f(0)}$$
$$f(y) = 1 + \frac{y}{f(0)}$$

This would mean $f$ is linear. Let me check if this works.

If $f(y) = 1 + \frac{y}{f(0)}$, then $f(0) = 1 + \frac{0}{f(0)} = 1$.

So if $f(0) = 1$, then $f(y) = 1 + y$.

Let me verify this satisfies the original equation:
$$f(x)f(y) - f(xy) = (1 + x)(1 + y) - (1 + xy)$$
$$= 1 + x + y + xy - 1 - xy$$
$$= x + y$$

This works! So $f(x) = 1 + x$ is a solution.

Now, let me check if $f(0) = 0$ leads to other solutions.

If $f(0) = 0$, then from $f(0)(f(y) - 1) = y$, we have $0 = y$ for all $y$, which is impossible.

Therefore, $f(0) = 1$ and $f(x) = 1 + x$ is the only solution.

Let me double-check by using a different approach.

Setting $y = 1$ in equation (1):
$$f(x)f(1) - f(x) = x + 1$$
$$f(x)(f(1) - 1) = x + 1$$

If $f(1) - 1 \neq 0$, then:
$$f(x) = \frac{x + 1}{f(1) - 1}$$

Setting $x = 1$ in equation (1):
$$f(1)f(1) - f(1) = 1 + 1$$
$$f(1)^2 - f(1) = 2$$
$$f(1)^2 - f(1) - 2 = 0$$
$$(f(1) - 2)(f(1) + 1) = 0$$

So $f(1) = 2$ or $f(1) = -1$.

Case 1: If $f(1) = 2$, then $f(1) - 1 = 1$, so:
$$f(x) = \frac{x + 1}{1} = x + 1$$

Case 2: If $f(1) = -1$, then $f(1) - 1 = -2$, so:
$$f(x) = \frac{x + 1}{-2} = -\frac{x + 1}{2}$$

Let me verify Case 2: $f(x) = -\frac{x + 1}{2}$

$$f(x)f(y) - f(xy) = \left(-\frac{x + 1}{2}\right)\left(-\frac{y + 1}{2}\right) - \left(-\frac{xy + 1}{2}\right)$$
$$= \frac{(x + 1)(y + 1)}{4} + \frac{xy + 1}{2}$$
$$= \frac{xy + x + y + 1}{4} + \frac{2xy + 2}{4}$$
$$= \frac{xy + x + y + 1 + 2xy + 2}{4}$$
$$= \frac{3xy + x + y + 3}{4}$$

This should equal $x + y$, so:
$$\frac{3xy + x + y + 3}{4} = x + y$$
$$3xy + x + y + 3 = 4x + 4y$$
$$3xy + 3 = 3x + 3y$$
$$xy + 1 = x + y$$
$$xy - x - y + 1 = 0$$
$$(x-1)(y-1) = 0$$

This only holds when $x = 1$ or $y = 1$, not for all real $x, y$. So Case 2 doesn't work.

Therefore, the only solution is $f(x) = x + 1$.

The sum of all possible values of $f(7)$ is:
$$f(7) = 7 + 1 = 8$$

$\boxed{8}$