3. Determine all functions $f$, defined on the set $\mathbb{Z}$ of integers and taking values in the set $\mathbb{R}$ of real numbers, that satisfy the following properties simultaneously:

- for every pair of integers $(m, n)$ with $m<n$ we have $f(m)<f(n)$;
- for every pair of integers $(m, n)$ there exists an integer $k$ such that $f(m)-f(n)=f(k)$.
SOLUTION: Ł̀ It is immediately verified that functions of the type $f(n)=\left(n-n_{0}\right) a$ with $n_{0}$ an integer and $a$ a positive real number satisfy the hypotheses: if $mf\left(n_{0}\right)=0$ for the first property, we can always divide by $a$), and suppose by contradiction that $b$ is a value of $f$. If $b>0$, consider the maximum $k$ natural number such that $k a$ is less than $b$; since we have already seen that $k a$ is assumed as a value by $f$, this must also happen for $b-k a$. But for our choice of $k$ we have $k a0$, and therefore, since $g(n+1)-g(n)$ is a value of the function, $g(n+1)-g(n) \geq g(1)$; on the other hand, $g(n+1)-g(1)<g(n+1)$ and thus, since $g(n+1)-g(n)$ is a value of the function, $g(n+1)-g(1) \leq g(n)$. In conclusion,

$$
g(n+1)=g(n)+g(1) \quad \text { for every integer } n \text {. }
$$

Now we prove by induction that $g(n)=n a$ for every integer $n$. We know that this is true for $n=0,1$. Suppose the thesis is true for a certain positive integer $n$: then the previous formula gives $g(n+1)=n a+a=(n+1) a$. Now suppose the thesis is true for a certain $m \leq 0$: setting $n=m-1$ in the previous formula we get $g(m)=g(m-1)+g(1)$, from which $g(m-1)=g(m)-g(1)=m a-a=(m-1) a$.