69th Belarusian Mathematical Olympiad

From the equality $n=a^2+b$ it follows that $b$ is divisible by $a$, since $a$ divides both $n$ and $a^2$. Let $b=ma$, where $m>1$. Then $n=a^2+ma=a(a+m)$, so $a+m$ divides $n$. Let us show that $a<a+m<b$. Suppose $a+m\ge b$, then $$a+m\ge ma⟺am-a-m+1\le 1⟺(a-1)(m-1)\le 1.$$ The latter inequality holds only if $a=m=2$ whence $b=4$, but then $n=2\cdot 4=8<10$, which contradicts with the conditions of the problem. Therefore, $a+m$ is the required divisor of $n$, located between $a$ and $b$.