75th Romanian Mathematical Olympiad

The hypothesis implies that $\sqrt{r+1}-r=a$, where $a\in \mathbb{Z}$.

We get: $r+1=r^2+2ra+a^2$.

Let $r=\frac{m}{n}$, with $m,n\in {\mathbb{N}}^{\ast }$ and $\text{gcd}(m,n)=1$. Then, $$mn+n^2=m^2+2mna+a^2{n}^2,$$ and therefore $$m^2=n(m+n-2ma-a^{2}n).$$ Hence, $m^2$ is divisible by $n$, which is only possible if $n=1$, thus $r=m\in \mathbb{N}$.