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.

Let $P(x,y)$ be the assertion $f(x)f(y) - f(xy) = x + y$.
The domain and codomain of $f$ are $\mathbb{R}$.

Let's test some specific values for $x$ and $y$.

$P(0,0)$:
$f(0)f(0) - f(0) = 0 + 0 = 0$
$f(0)^2 - f(0) = 0$
$f(0)(f(0) - 1) = 0$
So $f(0) = 0$ or $f(0) = 1$.

Case 1: $f(0) = 0$.
Let's plug $y=0$ into the original equation.
$P(x,0)$:
$f(x)f(0) - f(0) = x + 0$
$f(x) \cdot 0 - 0 = x$
$0 = x$
This must hold for all $x \in \mathbb{R}$, which is clearly false.
So $f(0) \neq 0$.

Thus, we must have $f(0) = 1$.

Let's check $P(x,0)$ with $f(0)=1$.
$f(x) \cdot 1 - f(0) = x + 0$
$f(x) - 1 = x$
$f(x) = x + 1$.

Let's verify if $f(x) = x + 1$ is a solution.
LHS: $f(x)f(y) - f(xy) = (x+1)(y+1) - (xy+1)$
$= xy + x + y + 1 - xy - 1$
$= x + y$.
RHS: $x + y$.
LHS = RHS.
So $f(x) = x + 1$ is a solution.
In this case, $f(7) = 7 + 1 = 8$.

Is this the only solution?
Let's re-evaluate the deduction.
We found $f(0) \in \{0, 1\}$.
If $f(0)=0$, we got a contradiction $0=x$ for all $x$.
So $f(0)=1$ is necessary.
Then $P(x,0)$ gave $f(x) - f(0) = x$, so $f(x) - 1 = x$, so $f(x) = x+1$.
This derivation seems to imply uniqueness.

Wait, did I miss anything?
Let's check the steps carefully.
1. $P(0,0) \implies f(0)^2 - f(0) = 0 \implies f(0) \in \{0, 1\}$.
2. If $f(0)=0$, $P(x,0) \implies f(x)f(0) - f(0) = x \implies 0 - 0 = x \implies x=0$. Contradiction.
3. So $f(0)=1$.
4. $P(x,0) \implies f(x)f(0) - f(0) = x \implies f(x) \cdot 1 - 1 = x \implies f(x) = x+1$.
5. Check solution: $(x+1)(y+1) - (xy+1) = xy+x+y+1-xy-1 = x+y$. Correct.

Is there any constraint I missed?
The problem statement says "for all real numbers x and y".
The derivation holds for all $x$.
So $f(x) = x+1$ is the unique solution.
Then $f(7) = 8$.
The sum of all possible values is 8.

Let me double check if I made any assumption.
Maybe $f$ is not defined at 0? No, domain is $\mathbb{R}$.
Maybe $f(0)$ is not unique? No, $f(0)^2 - f(0) = 0$ is a quadratic equation.
Maybe $P(x,0)$ is not valid? It is valid for all $x, y$.

Let's try to find if there are other solutions.
Maybe I should check $P(1,1)$.
$f(1)^2 - f(1) = 1 + 1 = 2$.
$f(1)^2 - f(1) - 2 = 0$.
$(f(1) - 2)(f(1) + 1) = 0$.
So $f(1) = 2$ or $f(1) = -1$.

If $f(x) = x+1$, then $f(1) = 2$. This matches.
What if $f(1) = -1$?
Let's see if we can construct a solution with $f(1) = -1$.
We established $f(0)=1$.
$P(x,1) \implies f(x)f(1) - f(x) = x + 1$.
If $f(1) = -1$, then $f(x)(-1) - f(x) = x + 1$.
$-2f(x) = x + 1$.
$f(x) = -\frac{x+1}{2}$.
Let's check if this function works.
LHS: $f(x)f(y) - f(xy) = \left(-\frac{x+1}{2}\right)\left(-\frac{y+1}{2}\right) - \left(-\frac{xy+1}{2}\right)$
$= \frac{(x+1)(y+1)}{4} + \frac{xy+1}{2}$
$= \frac{xy + x + y + 1}{4} + \frac{2xy + 2}{4}$
$= \frac{3xy + x + y + 3}{4}$.
RHS: $x + y$.
Is $\frac{3xy + x + y + 3}{4} = x + y$?
$3xy + x + y + 3 = 4x + 4y$
$3xy - 3x - 3y + 3 = 0$
$3(xy - x - y + 1) = 0$
$3(x-1)(y-1) = 0$.
This must hold for all $x, y$. It does not.
So $f(1) = -1$ is not possible.

Wait, why did $P(x,1)$ imply $f(x) = -\frac{x+1}{2}$?
$P(x,1)$ is $f(x)f(1) - f(x) = x+1$.
This is $f(x)(f(1)-1) = x+1$.
If $f(1) \neq 1$, then $f(x) = \frac{x+1}{f(1)-1}$.
We found $f(1) \in \{2, -1\}$.
If $f(1)=2$, $f(x) = \frac{x+1}{2-1} = x+1$. This works.
If $f(1)=-1$, $f(x) = \frac{x+1}{-1-1} = -\frac{x+1}{2}$. This failed.
So $f(1)$ must be 2.

Is it possible that $f(1)=1$?
From $f(1)^2 - f(1) = 2$, $f(1)$ cannot be 1.
So $f(1)$ is uniquely determined as 2.
And $f(x)$ is uniquely determined as $x+1$.

Let's check if there are any other constraints.
Maybe $f(x)$ is not linear?
Let's try to derive $f(x)$ without assuming $f(0)=1$ first, although we proved it.
$f(0) \in \{0, 1\}$.
If $f(0)=0$, $f(x) \cdot 0 - 0 = x \implies x=0$, impossible.
So $f(0)=1$ is solid.
Then $f(x) - 1 = x \implies f(x) = x+1$.
This seems extremely straightforward. Is there a catch?

Let's re-read the problem carefully.
$f: \mathbb{R} \to \mathbb{R}$.
$f(x)f(y) - f(xy) = x + y$.
Find sum of all possible values of $f(7)$.

Let's