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 properties. It's similar to the Cauchy functional equation but more complex because it involves both \( f(xy) \) and \( f(x)f(y) \).

### Step 2: Assume a Polynomial Form
Given the complexity, I wonder if \( f \) is a polynomial. Let's assume \( f \) is a polynomial in \( x \). The equation involves \( f(xy) \) and \( f(x)f(y) \), which suggests a quadratic form.

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

Now, let's compute \( f(x)f(y) \) and \( f(xy) \):
\[ f(x)f(y) = (ax^2 + bx + c)(ay^2 + by + c) = a^2x^2y^2 + abx^2y + acx^2 + abxy^2 + b^2xy + bcx + acy^2 + bcy + c^2. \]

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

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

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

### Step 3: Compare Coefficients
The right-hand side is \( x + y \), which is a linear polynomial. Therefore, all other terms must vanish.

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

#### Case 1: \( a = 0 \)
If \( a = 0 \), the equation simplifies to:
\[ b^2xy + bcx + bcy + (c^2 - c) = x + y. \]

But \( b^2xy \) must be zero for all \( x, y \), so \( b = 0 \).

Now, the equation becomes:
\[ c^2 - c = x + y. \]

But \( c^2 - c \) is a constant, while \( x + y \) is not. This is a contradiction because the left side is constant and the right side varies with \( x \) and \( y \).

**Conclusion:** \( a = 0 \) leads to no valid solution.

#### Case 2: \( a = 1 \)
Now, set \( a = 1 \). The equation becomes:
\[ (1^2 - 1)x^2y^2 + b x^2 y + b x y^2 + (b^2 - b)xy + b c x + c y^2 + b c y + (c^2 - c) = x + y. \]

Simplify:
\[ b x^2 y + b x y^2 + (b^2 - b)xy + b c x + c y^2 + b c y + (c^2 - c) = x + y. \]

Again, we need to match coefficients for all terms.

1. **\( x^2 y \) and \( x y^2 \) terms**: \( b = 0 \).
   If \( b = 0 \), the equation reduces to:
   \[ c y^2 + (c^2 - c) = x + y. \]

   But \( c y^2 \) must be zero for all \( y \), so \( c = 0 \).

   Then, the equation becomes:
   \[ 0 = x + y. \]

   But this must hold for all \( x, y \), which is false.

**Conclusion:** \( b \) cannot be zero. Hmm, this suggests that our assumption of a quadratic form might be incomplete or incorrect.

### Step 4: Try Different Approach - Assume Linear Form
Since the quadratic approach didn't work, maybe \( f \) is linear. Let's try:
\[ f(x) = kx + d. \]

Compute \( f(x)f(y) \) and \( f(xy) \):
\[ f(x)f(y) = (kx + d)(ky + d) = k^2xy + kd x + kd y + d^2. \]

\[ f(xy) = kxy + d. \]

Substitute into the original equation:
\[ k^2xy + kd x + kd y + d^2 - (kxy + d) = x + y. \]

Simplify:
\[ (k^2 - k)xy + kd x + kd y + (d^2 - d) = x + y. \]

Again, compare coefficients:
1. **\( xy \) term**: \( k^2 - k = 0 \).
   \[ k(k - 1) = 0 \implies k = 0 \text{ or } k = 1. \]

#### Subcase 1: \( k = 0 \)
If \( k = 0 \), the equation becomes:
\[ d^2 - d = x + y. \]

Again, the left side is constant, and the right side is not. No solution here.

#### Subcase 2: \( k = 1 \)
If \( k = 1 \), the equation becomes:
\[ (1 - 1)xy + d x + d y + (d^2 - d) = x + y. \]

Simplify:
\[ d x + d y + (d^2 - d) = x + y. \]

Now, compare coefficients:
1. **\( x \) and \( y \) terms**: \( d = 1 \).
2. **Constant term**: \( d^2 - d = 0 \).

But if \( d = 1 \), then \( d^2 - d = 1 - 1 = 0 \), which matches the right-hand side since ther