24th Hellenic Mathematical Olympiad

It is necessary to have $n^2-9\ge 0\iff n\ge 3$ or $n\le -3$. Let $A-B=\sqrt{n^2+24}-\sqrt{n^2-9}=d\in \mathbb{Z}$. Then $d>0$ and $$\begin{array}{rl}\sqrt{n^2+24}& =\sqrt{n^2-9}+d\\ \Rightarrow n^2+24& =n^2-9+d^2+2d\sqrt{n^2-9}\\ \iff d^2& =33-2d\sqrt{n^2-9}<33\Rightarrow d\in \{1,2,3,4,5\}.\end{array}$$ Moreover $$\sqrt{n^2-9}=\frac{33-d^2}{2d}\iff n^2={\left(\frac{33-d^2}{2d}\right)}^2+9.(1)$$ If $d$ is even, that is $d=2$ or $4$, then from (1) we have $n^2\notin \mathbb{Z}$, absurd. For $d=3$ or $5$, also $n^2\notin \mathbb{Z}$, while for $d=1$, we find $n=\pm 5$. Therefore the integer we seek is $5$ or $-5$.