Problem 6

A function

f:\mathbb Q\to\mathbb Q

is called aquaesulian if the following property holds: for every

x,y\in\mathbb Q

f(x+f(y))=f(x)+y

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

Step 5 of 5: At most two nonzero-compatible values coincide

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In plain words

If two different numbers both have a nonzero value of f(r) plus f(-r), applying the dichotomy from Step 4 to a direction that must hold between them shows their two values are forced to be equal, so there is really only one nonzero value possible, plus the value 0.

c=2

Detailed analysis

Suppose $a\ne b$ both have $f(a)+f(-a)\ne0$ and $f(b)+f(-b)\ne0$ . By hypothesis, without loss of generality $a\to b$ . Applying Step 4 with $s=r=a$ (using $a\to a$ from Step 1) and separately with $s=a,r=b$ (using $a\to b$ ), and ruling out the " $=0$ " branches by assumption, both give an expression for $f(f(a))-a$ , forcing $f(a)+f(-a)=f(b)+f(-b)$ . Hence all nonzero values of $f(r)+f(-r)$ coincide, so together with the possible value $0$ , at most $2$ distinct values occur in general; combined with the construction of Step 2, the smallest possible $c$ is $c=2$ .