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.

We are given a functional equation:

$$
f(x)f(y) - f(xy) = x + y \quad \text{for all } x, y \in \mathbb{R}.
$$

Our goal is to find **all possible values** of $ f(7) $, then sum them.

---

### Step 1: Try plugging in simple values.

#### Let $ x = 0 $:

$$
f(0)f(y) - f(0\cdot y) = 0 + y \Rightarrow f(0)f(y) - f(0) = y.
$$

So:
$$
f(0)(f(y) - 1) = y. \tag{1}
$$

This is very useful! It relates $ f(y) $ to $ f(0) $.

Let’s denote $ a = f(0) $. Then equation (1) becomes:

$$
a(f(y) - 1) = y \Rightarrow f(y) = 1 + \frac{y}{a}, \quad \text{provided } a \ne 0.
$$

So if $ f(0) \ne 0 $, then $ f $ must be of the form:

$$
f(x) = 1 + \frac{x}{a}. \tag{2}
$$

Let’s test whether such a function can satisfy the original equation.

Plug (2) into the functional equation:

Left-hand side:
$$
f(x)f(y) - f(xy) = \left(1 + \frac{x}{a}\right)\left(1 + \frac{y}{a}\right) - \left(1 + \frac{xy}{a}\right).
$$

Compute:

$$
= 1 + \frac{x}{a} + \frac{y}{a} + \frac{xy}{a^2} - 1 - \frac{xy}{a}
= \frac{x + y}{a} + xy\left(\frac{1}{a^2} - \frac{1}{a}\right).
$$

We want this to equal $ x + y $ for all $ x, y $. So:

$$
\frac{x + y}{a} + xy\left(\frac{1}{a^2} - \frac{1}{a}\right) = x + y.
$$

This must hold for all real $ x, y $. Compare coefficients.

- Coefficient of $ x + y $: $ \frac{1}{a} $ must equal 1 ⇒ $ a = 1 $.
- Coefficient of $ xy $: $ \frac{1}{a^2} - \frac{1}{a} = 0 $.

Check if $ a = 1 $ satisfies this:

$$
\frac{1}{1^2} - \frac{1}{1} = 1 - 1 = 0.
$$

Yes! So $ a = 1 $ works.

Thus, one possible function is:

$$
f(x) = 1 + x.
$$

Check: $ f(x)f(y) - f(xy) = (1+x)(1+y) - (1+xy) = 1 + x + y + xy - 1 - xy = x + y $. ✅

So $ f(x) = x + 1 $ is a solution, and then $ f(7) = 8 $.

But wait — we assumed $ a = f(0) \ne 0 $. What if $ f(0) = 0 $? Let's check that case.

---

### Step 2: Consider $ f(0) = 0 $

Recall from equation (1):

$$
f(0)(f(y) - 1) = y.
$$

If $ f(0) = 0 $, then left side is 0, so:

$$
0 = y \quad \text{for all } y,
$$

which is impossible. So **$ f(0) \ne 0 $**.

Thus, the only possible form is $ f(x) = 1 + \frac{x}{a} $, and we found that only $ a = 1 $ works.

Wait — but maybe there are other forms? Let's double-check.

We derived $ f(y) = 1 + \frac{y}{a} $ assuming $ a = f(0) \ne 0 $. That derivation was correct: from (1), $ a(f(y) - 1) = y $, so $ f(y) = 1 + \frac{y}{a} $.

So all solutions must be of that form, and only $ a = 1 $ works.

But wait — could there be other solutions not captured by this derivation? Let's double-check the derivation.

From $ x = 0 $:

$$
f(0)f(y) - f(0) = y \Rightarrow f(0)(f(y) - 1) = y.
$$

Yes, that's correct for all $ y $. So unless $ f(0) = 0 $, which leads to contradiction, we must have $ f(y) = 1 + \frac{y}{f(0)} $.

So only possible solution is $ f(x) = 1 + x $.

But the problem says **"sum of all possible values of $ f(7) $"**, implying possibly more than one solution.

Let’s double-check our algebra when plugging in the general form.

Let $ f(x) = 1 + \frac{x}{a} $. Then:

$$
f(x)f(y) = \left(1 + \frac{x}{a}\right)\left(1 + \frac{y}{a}\right) = 1 + \frac{x + y}{a} + \frac{xy}{a^2},
$$
$$
f(xy) = 1 + \frac{xy}{a}.
$$

So:

$$
f(x)f(y) - f(xy) = \left(1 + \frac{x + y}{a} + \frac{xy}{a^2}\right) - \left(1 + \frac{xy}{a}\right) = \frac{x + y}{a} + xy\left(\frac{1}{a^2} - \frac{1}{a}\right).
$$

Set equal to $ x + y $:

$$
\frac{x + y}{a} + xy\left(\frac{1}{a^2} - \frac{1}{a}\right) = x + y.
$$

This must hold for all real $ x, y $. So treat as a polynomial identity in $ x $ and $ y $:

Left side: linear terms in $ x $ and $ y $, plus a term proportional to $ xy $.

Right side: only linear terms, no $ xy $ term.

So coefficient of $ xy $ must be zero:

$$
\frac{1}{a^2} - \frac{1}{a} = 0 \Rightarrow \frac{1 - a}{a^2} = 0 \Rightarrow a = 1.
$$

Then coefficient of $ x + y $: $ \frac{1}{a} = 1 $, matches.

So only $ a = 1 $ works. So only one solution.

But maybe there's another solution not of this form? Let's try to find other possibilities.

Suppose $ f $ is constant: $ f(x) = c $. Then: