Problem 1694

1. Define the assertion: Let $P(x, y)$ be the assertion $f(x + y + xy) = f(x) + f(y) + f(xy)$.

2. **Evaluate $P(0, 0)$**:
$$P(0, 0) \implies f(0 + 0 + 0 \cdot 0) = f(0) + f(0) + f(0) \implies f(0) = 3f(0) \implies 2f(0) = 0 \implies f(0) = 0$$

3. **Evaluate $P(x, -1)$**:
$$P(x, -1) \implies f(x - 1 + x(-1)) = f(x) + f(-1) + f(-x) \implies f(-1) = f(x) + f(-1) + f(-x)$$
Since $f(-1) = f(-1)$, we get:
$$f(-x) = -f(x) \quad \text{(i.e., \( f \) is odd)}$$

4. **Evaluate $P(x, \frac{y}{x+1})$ for $x, y \neq -1$**:
$$P(x, \frac{y}{x+1}) \implies f(x + \frac{y}{x+1} + x \cdot \frac{y}{x+1}) = f(x) + f(\frac{y}{x+1}) + f(\frac{xy}{x+1})$$
Simplifying the left-hand side:
$$x + \frac{y}{x+1} + \frac{xy}{x+1} = x + \frac{y(1+x)}{x+1} = x + y$$
Thus:
$$f(x + y) = f(x) + f(\frac{y}{x+1}) + f(\frac{xy}{x+1})$$

5. **Evaluate $P(x, -\frac{y}{x+1})$**:
$$P(x, -\frac{y}{x+1}) \implies f(x - \frac{y}{x+1} + x \cdot -\frac{y}{x+1}) = f(x) + f(-\frac{y}{x+1}) + f(-\frac{xy}{x+1})$$
Simplifying the left-hand side:
$$x - \frac{y}{x+1} - \frac{xy}{x+1} = x - \frac{y(1+x)}{x+1} = x - y$$
Thus:
$$f(x - y) = f(x) - f(\frac{y}{x+1}) - f(\frac{xy}{x+1})$$

6. **Evaluate $P(y, \frac{x}{y+1})$**:
$$P(y, \frac{x}{y+1}) \implies f(y + \frac{x}{y+1} + y \cdot \frac{x}{y+1}) = f(y) + f(\frac{x}{y+1}) + f(\frac{xy}{y+1})$$
Simplifying the left-hand side:
$$y + \frac{x}{y+1} + \frac{xy}{y+1} = y + \frac{x(1+y)}{y+1} = y + x$$
Thus:
$$f(x + y) = f(y) + f(\frac{x}{y+1}) + f(\frac{xy}{y+1})$$

7. **Evaluate $P(y, -\frac{x}{y+1})$**:
$$P(y, -\frac{x}{y+1}) \implies f(y - \frac{x}{y+1} + y \cdot -\frac{x}{y+1}) = f(y) + f(-\frac{x}{y+1}) + f(-\frac{xy}{y+1})$$
Simplifying the left-hand side:
$$y - \frac{x}{y+1} - \frac{xy}{y+1} = y - \frac{x(1+y)}{y+1} = y - x$$
Thus:
$$f(y - x) = f(y) - f(\frac{x}{y+1}) - f(\frac{xy}{y+1})$$

8. Add the four equations:
$$f(x + y) = f(x) + f(\frac{y}{x+1}) + f(\frac{xy}{x+1})$$
$$f(x - y) = f(x) - f(\frac{y}{x+1}) - f(\frac{xy}{x+1})$$
$$f(x + y) = f(y) + f(\frac{x}{y+1}) + f(\frac{xy}{y+1})$$
$$f(y - x) = f(y) - f(\frac{x}{y+1}) - f(\frac{xy}{y+1})$$
Adding these four equations, we get:
$$2f(x + y) + 2f(x - y) = 2f(x) + 2f(y)$$
Dividing by 2:
$$f(x + y) + f(x - y) = f(x) + f(y)$$

9. **Remove the constraint $x, y \neq -1$**:
Let $x \neq 0$:
$$f((x-1) + 1) = f(x-1) + f(1) \implies f(x-1) = f(x) + f(-1)$$
Since $f(-1) = -f(1)$, we have:
$$f(x-1) = f(x) - f(1)$$
Thus, $f(x+y) = f(x) + f(y)$ for all $x \neq -1$ and $y$.

10. **Generalize to all $x, y$**:
Since $f(1+1) = f(1) + f(1)$ and $f(-1-1) = f(-1) + f(-1)$, we conclude:
$$f(x+y) = f(x) + f(y) \quad \forall x, y$$

$\blacksquare$