Question 72
Let $\mathbb{Z}$ be the set of integers. We consider functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ satisfying
$$
f(f(x+y)+y)=f(f(x)+y)
$$
for all integers $x$ and $y$. For such a function, we say that an integer $v$ is $f$-rare if the set
$$
X_{v}=\{x \in \mathbb{Z}: f(x)=v\}
$$
is finite and nonempty.
a. Prove that there exists such a function $f$ for which there is an $f$-rare integer.
b. Prove that no such function $f$ can have more than one $f$-rare integer.
$$
f(f(x+y)+y)=f(f(x)+y)
$$
for all integers $x$ and $y$. For such a function, we say that an integer $v$ is $f$-rare if the set
$$
X_{v}=\{x \in \mathbb{Z}: f(x)=v\}
$$
is finite and nonempty.
a. Prove that there exists such a function $f$ for which there is an $f$-rare integer.
b. Prove that no such function $f$ can have more than one $f$-rare integer.