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.

The problem asks for the sum of all possible values of $f(7)$ given the functional equation $f(x)f(y) - f(xy) = x + y$ for all $x, y \in \mathbb{R}$.

Let the given equation be
(1) $f(x)f(y) - f(xy) = x + y$.

First, let's try to find some specific values of $f(x)$.
Let $x=0$.
$f(0)f(y) - f(0) = 0 + y = y$.
So, $f(0)(f(y) - 1) = y$.
This implies that $f(0) \neq 0$ because if $f(0)=0$, then $0 = y$ for all $y$, which is impossible.
So $f(0) \neq 0$.
From $f(0)(f(y) - 1) = y$, we can express $f(y)$ in terms of $y$.
$f(y) - 1 = \frac{y}{f(0)}$.
$f(y) = 1 + \frac{y}{f(0)}$.
Let $c = f(0)$. Then $f(y) = 1 + \frac{y}{c}$.
Let's check if this form satisfies the original equation.
Substitute $f(x) = 1 + \frac{x}{c}$ and $f(y) = 1 + \frac{y}{c}$ and $f(xy) = 1 + \frac{xy}{c}$ into (1).
LHS: $(1 + \frac{x}{c})(1 + \frac{y}{c}) - (1 + \frac{xy}{c})$
$= 1 + \frac{x}{c} + \frac{y}{c} + \frac{xy}{c^2} - 1 - \frac{xy}{c}$
$= \frac{x}{c} + \frac{y}{c} + \frac{xy}{c^2} - \frac{xy}{c}$
$= \frac{x+y}{c} + xy(\frac{1}{c^2} - \frac{1}{c})$.
RHS: $x + y$.
So we need $\frac{x+y}{c} + xy(\frac{1}{c^2} - \frac{1}{c}) = x + y$ for all $x, y$.
Comparing coefficients of $x+y$ and $xy$:
Coefficient of $x+y$: $\frac{1}{c} = 1 \implies c = 1$.
Coefficient of $xy$: $\frac{1}{c^2} - \frac{1}{c} = 0$.
If $c=1$, then $\frac{1}{1} - \frac{1}{1} = 0$, which holds.
So $c=1$ is a solution.
This gives $f(x) = 1 + x$.
Let's check this solution.
$f(x)f(y) - f(xy) = (1+x)(1+y) - (1+xy) = 1 + x + y + xy - 1 - xy = x + y$.
This works. So $f(x) = x+1$ is a solution.
In this case, $f(7) = 7+1 = 8$.

However, I assumed $f(0) \neq 0$ to derive $f(y) = 1 + y/c$.
Is it possible that $f(0) = 0$?
If $f(0) = 0$, then from $f(0)(f(y) - 1) = y$, we get $0 = y$ for all $y$, which is a contradiction.
So $f(0) \neq 0$ is correct.
Wait, let me double check the step $f(0)(f(y) - 1) = y$.
The equation is $f(x)f(y) - f(xy) = x + y$.
Set $x=0$: $f(0)f(y) - f(0) = 0 + y$.
$f(0)(f(y) - 1) = y$.
If $f(0) = 0$, then $0 = y$ for all $y$, impossible.
So $f(0) \neq 0$.
Then $f(y) = 1 + \frac{y}{f(0)}$.
Let $k = f(0)$. Then $f(x) = 1 + \frac{x}{k}$.
Substituting this back into the original equation led to $k=1$.
So $f(x) = x+1$ seems to be the only solution.
But wait, the problem asks for the "sum of all possible values of $f(7)$". This phrasing suggests there might be more than one solution.
Let me re-read the derivation.
$f(0)(f(y) - 1) = y$.
This implies $f(y) = 1 + \frac{y}{f(0)}$.
Is it possible that $f(0)$ depends on $y$? No, $f(0)$ is a constant.
So $f(y)$ must be linear.
Let's check if there are any other possibilities.
Maybe I made an assumption somewhere.
Let's check $x=1$.
$f(1)f(y) - f(y) = 1 + y$.
$f(y)(f(1) - 1) = 1 + y$.
This implies $f(y) = \frac{1+y}{f(1)-1}$.
Let $A = f(1) - 1$. Then $f(y) = \frac{1}{A} + \frac{y}{A}$.
This is of the form $f(y) = \alpha + \beta y$.
Comparing with $f(y) = 1 + \frac{y}{f(0)}$, we have $\alpha = 1$ and $\beta = \frac{1}{f(0)}$.
Also from $f(y) = \frac{1+y}{A}$, we have $\alpha = \frac{1}{A}$ and $\beta = \frac{1}{A}$.
So $\alpha = \beta$.
Thus $f(y) = \alpha(1+y)$.
Let's substitute $f(x) = \alpha(1+x)$ into the original equation.
LHS: $\alpha(1+x)\alpha(1+y) - \alpha(1+xy) = \alpha^2(1+x+y+xy) - \alpha(1+xy)$.
$= \alpha^2 + \alpha^2 x + \alpha^2 y + \alpha^2 xy - \alpha - \alpha xy$.
$= (\alpha^2 - \alpha) + \alpha^2(x+y) + (\alpha^2 - \alpha)xy$.
RHS: $x+y$.
We need this to hold for all $x, y$.
Coefficient of $x+y$: $\alpha^2 = 1$.
Coefficient of $xy$: $\alpha^2 - \alpha = 0$.
Constant term: $\alpha^2 - \alpha = 0$.
From $\alpha^2 = 1$, we have $\alpha = 1$ or $\alpha = -1$.
If $\alpha = 1$, then $\alpha^2 - \alpha = 1 - 1 = 0$. This works.
If $\alpha = -1$, then $\alpha^2 - \alpha = 1 - (-1) = 2 \neq 0$. This does not work.
So $\alpha = 1$ is the only solution.
This leads to $f(x) = 1+x$.
So $f(7) = 8$.
Why would the problem ask for the sum of all possible values?
Maybe I missed something.
Let's re-evaluate $f(0)(f(y)-1) = y$.
This equation must h