Rioplatense Mathematical Olympiad

Let $P(x, y)$ denote the assertion
$$f(x + f(y + 1)) + f(xy) = f(x + 1)(f(y) + 1).$$
If $f$ is a constant $c$, we have $2c = c(c+1)$. This implies $c = 0$ or $c = 1$. Hence, there are two constant solutions, $f \equiv 0$ and $f \equiv 1$.

$$P(0, y) : \quad f(f(y + 1)) + f(0) = f(1)(f(y) + 1), \qquad (1)$$
$$P(x - 1, 0) : \quad f(x + f(1) - 1) = f(x)(f(0) + 1) - f(0). \qquad (2)$$
Assume $f(1) = 1$. Then (2) becomes $f(x) = f(x)(f(0) + 1) - f(0)$, or equivalently, $(f(x) - 1)f(0) = 0$. Hence $f(0) = 0$. Therefore, (1) becomes
$$f(f(y + 1)) = f(y) + 1. \qquad (3)$$
Setting $y = -1, -2, -3, \dots$ in (3), we conclude inductively that $f(x) = x$ for all $x \le 0$. In addition,
$$P(x, -1) : \quad f(x) + f(-x) = 0,$$
so $f(x) = x$ for all $x \in \mathbb{Z}$, which is clearly a solution. Assume now that $f(1) \neq 1$ and set $c = f(1) - 1 \neq 0$. If $f(0) + 1 = 0$, then (2) would imply that $f$ is constant. Therefore, it must be $h = f(0) + 1 \neq 0$. Equation (2) can now be rewritten as
$$f(x + c) = hf(x) - h + 1. \qquad (4)$$
Next we consider
$$P(x, 1) : \quad f(x + f(2)) + f(x) = f(x + 1)(f(1) + 1) \qquad (5)$$
$$P(x + c, 1) : \quad f(x + c + f(2)) + f(x + c) = f(x + c + 1)(f(1) + 1) \qquad (6)$$
Using (4), we find that (6) is equivalent to
$$h(f(x + f(2)) - h + 1 + hf(x) - h + 1 = (hf(x + 1) - h + 1)(f(1) + 1).$$
Subtracting $h$ times (5) we get
$$2(1 - h) = (1 - h)(f(1) + 1),$$
hence, either $h = 1$ or $f(1) + 1 = 2$. The latter is impossible due to our initial assumption that $f(1) \neq 1$. So $h = 1$, i.e., $f(0) = 0$. Now equation (4) becomes $f(x + c) = f(x)$ and, as $c \neq 0$, $f$ is periodic. Let $M = \max f$ and $-m = \min f$. Note $M, m \ge 0$ because $f(0) = 0$.
Choose $x, y \in \mathbb{Z}$ such that $f(x + 1) = f(y) = M$. Then, the right-hand side of $P(x, y)$ is equal to $M(M + 1)$, while the left-hand side is
$$f(x + f(y + 1)) + xy \le M + M = 2M.$$
So $M(M + 1) \le 2M$, and thus $M = 0$ or $M = 1$.

Similarly, if we now choose $x, y \in \mathbb{Z}$ such that $f(x+1) = f(y) = -m$, we find
$$-m(-m+1) = m^2 - m = f(x+f(y+1)) + xy \le 2M \le 2,$$
which implies $m \le 2$. So $f(x) \in \{-2, -1, 0, 1\}$ for all integers $x$. We analyze the possibilities for $f(1)$:
* $f(1) = -2$: setting $y = 1$ in (1) we get $f(f(2)) = 2$, absurd.
* $f(1) = 0$: in this case, $c = -1$ and $f$ is constant. Absurd.
* $f(1) = 1$: previously excluded.
* $f(1) = -1$: in this case, $c = -2$, so $f$ is 2-periodic, with $f(x) = f(0) = 0$ for all even $x$, and $f(x) = f(1) = -1$ for all odd $x$. However, $P(1, 1)$ gives $f(1+f(2)) + f(1) = f(2)(f(1)+1)$, so $-2=0$, a contradiction.
In conclusion, the only solutions are $f \equiv 0, f \equiv 1$, and $f(x) = x$ for all $x$.