Saudi Arabian Mathematical Competitions, 2015

Let $f: \mathbb{Z} \rightarrow \mathbb{R}$ be such a function. We prove that $f(n+1)-f(n)$ is a positive constant $a$ independent of the integer $n$.

Indeed, assume that there exist two integers $n_{0}, n_{1}$ such that
$$f\left(n_{0}+1\right)-f\left(n_{0}\right)<f\left(n_{1}+1\right)-f\left(n_{1}\right) .$$
Let $k_{0}, k_{1}$ be integers such that $f\left(k_{0}\right)=f\left(n_{0}+1\right)-f\left(n_{0}\right)$ and $f\left(k_{1}\right)= f\left(n_{1}+1\right)-f\left(k_{0}\right)$. Because $f$ is strictly increasing, $f\left(k_{0}\right)>0$. We deduce that $f\left(n_{1}\right)<f\left(k_{1}\right)=f\left(n_{1}+1\right)-f\left(k_{0}\right)<f\left(n_{1}+1\right)$, which is impossible since $f$ is strictly increasing and $n_{1}, n_{1}+1$ are consecutive integers.

Moreover, there exists an integer $n_{0}$ such that $f\left(n_{0}\right)=f(0)-f(0)=0$.
Let $n>n_{0}$. We have
$$f(n)=f\left(n_{0}\right)+\sum_{m=0}^{n-n_{0}-1}\left(f\left(n_{0}+m+1\right)-f\left(n_{0}+m\right)\right)=a\left(n-n_{0}\right) .$$
We obtain a similar formula for $n<n_{0}$ and therefore
$$f(n)=a\left(n-n_{0}\right), \quad \text{ for all } n \in \mathbb{Z} .$$
Conversely, let $a>0$ be a positive real number and $n_{0}$ an integer and define $f(n)=a\left(n-n_{0}\right)$ for all $n \in \mathbb{Z}$. The function $f$ is strictly increasing and for any integers $m, n \in \mathbb{Z}$ we have $f(m)-f(n)=f\left(m-n+n_{0}\right)$.