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.

Alright, I have this problem:

**Problem Statement:**
Find all functions \( f: \mathbb{R} \to \mathbb{R} \) such that for all real numbers \( x \) and \( y \),
\[ f(x)f(y) - f(xy) = x + y. \]

I need to find the sum of all possible values of \( f(7) \).

### Step 1: Understand the Functional Equation
First, I need to understand the given functional equation:
\[ f(x)f(y) - f(xy) = x + y. \]

This looks like a combination of multiplicative and additive terms. It's similar to Cauchy-like functional equations but more complex because it involves both \( f(xy) \) and \( f(x)f(y) \).

### Step 2: Assume a Polynomial Form
Functional equations often have polynomial solutions. Let's assume \( f \) is a polynomial. The equation involves \( f(xy) \) and \( f(x)f(y) \), which suggests a quadratic form.

#### General Form
Let's suppose \( f \) is quadratic:
\[ f(x) = ax^2 + bx + c. \]

Now, I need to plug this into the original equation and see if it works.

### Step 3: Expand \( f(x)f(y) \) and \( f(xy) \)
First, expand \( f(x)f(y) \):
\[ f(x)f(y) = (ax^2 + bx + c)(ay^2 + by + c) = a^2x^2y^2 + abx^2y + acx^2 + abxy^2 + b^2xy + bcx + acy^2 + bcy + c^2. \]

Next, expand \( f(xy) \):
\[ f(xy) = a(xy)^2 + b(xy) + c = ax^2y^2 + bxy + c. \]

Now, substitute into the original equation:
\[ f(x)f(y) - f(xy) = (a^2x^2y^2 + abx^2y + acx^2 + abxy^2 + b^2xy + bcx + acy^2 + bcy + c^2) - (ax^2y^2 + bxy + c) \]
\[ = (a^2 - a)x^2y^2 + abx^2y + abxy^2 + (b^2 - b)xy + acx^2 + acy^2 + bcx + bcy + (c^2 - c). \]

But according to the original equation, this must equal \( x + y \).

### Step 4: Compare Coefficients
The right-hand side is \( x + y \), which can be written as \( 0x^2y^2 + 0x^2y + 0xy^2 + 0xy + 0x^2 + 0y^2 + 1x + 1y + 0 \).

Now, compare coefficients term by term:

1. **\( x^2y^2 \) term**: \( a^2 - a = 0 \)
2. **\( x^2y \) term**: \( ab = 0 \)
3. **\( xy^2 \) term**: \( ab = 0 \)
4. **\( xy \) term**: \( b^2 - b = 0 \)
5. **\( x^2 \) term**: \( ac = 0 \)
6. **\( y^2 \) term**: \( ac = 0 \)
7. **\( x \) term**: \( bc = 1 \)
8. **\( y \) term**: \( bc = 1 \)
9. **Constant term**: \( c^2 - c = 0 \)

### Step 5: Solve the System of Equations
Let's solve these equations one by one.

1. **From \( a^2 - a = 0 \)**:
   \[ a(a - 1) = 0 \implies a = 0 \text{ or } a = 1. \]

2. **From \( ab = 0 \)**:
   - If \( a = 0 \), then \( b \) can be anything.
   - If \( a = 1 \), then \( b = 0 \).

3. **From \( b^2 - b = 0 \)**:
   \[ b(b - 1) = 0 \implies b = 0 \text{ or } b = 1. \]

But from the previous point, if \( a = 1 \), then \( b \) must be 0. So, \( b \) can only be 0 in this case.

4. **From \( ac = 0 \)**:
   - If \( a = 0 \), \( c \) can be anything.
   - If \( a = 1 \), \( c = 0 \).

5. **From \( bc = 1 \)**:
   - If \( a = 0 \), \( b \) can be 0 or 1.
     - If \( b = 0 \), \( bc = 0 \neq 1 \). Invalid.
     - If \( b = 1 \), \( c = 1 \).
   - If \( a = 1 \), \( b = 0 \), \( c = 0 \).
     - Then \( bc = 0 \neq 1 \). Invalid.

6. **From \( c^2 - c = 0 \)**:
   \[ c(c - 1) = 0 \implies c = 0 \text{ or } c = 1. \]

But from the previous analysis:
- For \( a = 0 \), \( b \) must be 1, and \( c \) must be 1 (from \( bc = 1 \)).
- For \( a = 1 \), \( b \) must be 0, and \( c \) must be 0, but \( bc = 0 \neq 1 \). So, \( a = 1 \) leads to a contradiction.

Thus, the only consistent solution is:
- \( a = 0 \)
- \( b = 1 \)
- \( c = 1 \)

### Step 6: Verify the Solution
Given \( f(x) = x + 1 \), let's verify it in the original equation:
\[ f(x)f(y) - f(xy) = (x + 1)(y + 1) - (xy + 1) = xy + x + y + 1 - xy - 1 = x + y. \]

This matches the right-hand side \( x + y \). So, \( f(x) = x + 1 \) is indeed a solution.

### Step 7: Check for Non-Polynomial Solutions
The problem doesn't specify that \( f \) is polynomial. Are there other solutions?

#### Consider Constant Function
Let's try \( f(x) = k \) for some constant \( k \).

Substitute into the original equation:
\[ k \cdot k - k = x + y \implies k^2 - k = x + y. \]

But the right-hand side depends on \( x \)