Problem 6.
 Let Q be the set of rational numbers. A function f : Q → Q is called aquaesulian if
the following property holds: for every x, y ∈ Q,
f(x + f(y)) = f(x) + y or f(f(x) + y) = x + f(y).
Show that there exists an integer c such that for any aquaesulian function f there are at most c different rational numbers of the form f(r) + f( − r) for some rational number r, and find the smallest possible value of c.

 

Solution

Claim 1. $0 \in f(\mathbb{Q})$

Proof. By $P(x, -f(x))$ we have that either $f(x + f(-f(x))) = 0$ or $f(-f(x)) + x = f(0)$ for each $x$. If the former ever holds then clearly $f$ attains $0$ as a value. Otherwise taking $x = f(0)$ yields $0$ as a value as well.

Claim 2. $f$ is injective

Assume $f(a) = f(b) = C$ for some $a$, $b$. By $P(a, a)$ and $P(b, b)$ we have $f(a + C) = a + C$ and $f(b + C) = b + C$. Now $P(a, b)$ yields either $f(a + C) = b + C$ or $f(b + C) = a + C$, both of which imply $a = b$.

Claim 3. $f(0) = 0$

Proof. Assume otherwise. By Claim 1 there exists a nonzero $t$ such that $f(t) = 0$. By $P(0, t)$ we have that one of $0$ and $t + f(0)$ maps to the other. Since $t$ is nonzero this is actually forced to be $f(t + f(0)) = 0 = f(t)$. Injectivity yields $t + f(0) = t$ but then $f(0) = 0$, contradicting that $t$ is nonzero.

Now I'm out of ideas so we'll try to bruteforce this. Let $a, b \in \mathbb{Q}$ be such that $f(a) + f(-a) \neq 0$ and $f(b) + f(-b) \neq 0$. I claim that these two values are actually equal. Assume that for some rational $x$ we have $f(x + f(a)) = f(x) + a$. By $P(x + f(a), -a)$ we obtain that two of the numbers

\[f(x) = f(x + f(a)) - a \quad \text{and} \quad x + f(a) + f(-a)\]
Is mapped to the other by $f$. If $f(x + f(a) + f(-a)) = f(x)$ then injectivity gives $f(a) + f(-a) = 0$ contradicting our choice of $a$. Therefore we must have $f(f(x)) = x + f(a) + f(-a)$ and $f(a) + f(-a) = f(f(x)) - x$. A similar relation holds for $b$ if $f(x + f(b)) = f(x) + b$, so it suffices to show that some value of $x$ satisfies this for $a$ and $b$ simultaneously. Notice that by $P(a, a)$ and $P(b, b)$, $x = a$ is a valid choice for $a$ and $x = b$ is a valid choice for $b$. Finally, by $P(a, b)$ we obtain that either $x = a$ is suitable for $b$ or $x = b$ is suitable for $a$, so in either case a suitable $x$ exists. It follows that $c = 2$ is suitable for all aquaesulian functions.

To finish we exhibit an aquaesulian function with two distinct values of $f(r) + f(-r)$. We claim that $f(x) = x - 2\{x\}$ works. This achieves $f(r) + f(-r) = 0$ for all integer $r$ and $f(r) + f(-r) = -2$ for all non-integer $r$. Consider two rational $x$, $y$ and assume WLOG that $\{x\} \ge \{y\}$, then

\begin{align*} f(x + f(y)) &= x + y - 2\{y\} - 2\{x + y - 2\{y\}\} \\ &= x + y - 2\{y\} - 2\{\{x\} + \{y\} - 2\{y\}\} \\ &= x + y - 2\{y\} - 2\{\{x\} - \{y\}\} \\ &= x + y - 2\{y\} - 2(\{x\} - \{y\}) \\ &= x + y - 2\{x\} = f(x) + y \end{align*}
Similarly $f(y + f(x)) = f(y) + x$ if $\{y\} \ge \{x\}$, which proves that $f$ is indeed aquaesulian.