For given $ab\leq n$, do there exist $a'\geq a$ and $b'\geq b$ such that $a'b'=n$? For example, given $a=3$, $b=3$, and $n=14$, no such $a',b'$ exists. On the other hand, for $a=3$, $b=3$, and $n=12$, we can use $a'=3$ and $b'=4$.
Is there a simple formula that can help determine the answer to this question, rather than simply searching through possible combinations of $a'$ and $b'$?

In reverse, it's related to the formula for markup and margin. No possibility of such a factorization exists, with $ab=n$.  In fact: $$\sqrt{n}<a\land \sqrt{n}<b \implies n<ab<a'b'$$
Which is then a contradiction. One way would be:$${n\over ab}=(1+b_\text{markup})(1+a_\text{markup})$$ where you use percentage markup that can be integers when multiplied by b and a, respectively. But that's still more of a brute force approach.
