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
$$a + m \ge ma \iff am - a - m + 1 \le 1 \iff (a - 1)(m - 1) \le 1.$$
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 $b$.