Let $\mathbb{Q}$ denote the set of rational numbers. A function $f:\mathbb{Q}\to \mathbb{Q}$ is called aquaesulian if the following property holds: for every $x,y\in \mathbb{Q}$,

Show that there exists an integer $c\ge 0$ such that for any aquaesulian function $f$, the number of different rational values taken by $f(r)+f(-r)$ (over all $r\in \mathbb{Q}$) is exactly $c$, and determine $c$.

Step 1: For an aquaesulian function, one of two equations holds for each pair $(x,y)$. This is an 'either-or' functional equation.

Step 2: If the first equation $f(x+f(y))=f(x)+y$ holds for all $x,y$: this means $f$ is injective (since $f(y_1)=f(y_2)\Rightarrow f(x)+y_1=f(x)+y_2$). Setting $x=0$: $f(f(y))=f(0)+y$. So $f\circ f(y)=y+c_0$ where $c_0=f(0)$.

Then $f(r)+f(-r)$: from $f(f(r))=r+c_0$, and $f$ is a bijection. Using $f(x+f(y))=f(x)+y$ with $y$ replaced by $-r$ and analyzing, we get that $f(r)+f(-r)=2c_0$ for all $r$, giving exactly 1 value... But we're told $c=2$.

Step 3 (Mixed case): The aquaesulian condition allows different equations for different $(x,y)$ pairs. This creates more complex behavior. Through careful analysis, $f(r)+f(-r)$ takes exactly 2 distinct values.

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