Problem 1

Find all pairs of positive integers $(a, b)$ such that $$11ab \le a^3 - b^3 \le 12ab.$$
 
Solution
Clearly, $a>b$.

If $a\ge b+4$, then $a^3-b^3-12ab=3ab(a-b-4)+(a-b)^3>0$, contradiction.

If $a\leq b+2$, then $a^3-b^3-11ab=(a-b)^3-ab(11-3(a-b))\leq 8-5ab\leq -2<0$, contradiction.

Hence, $a=b+3$. This gives us $$11b(b+3)\leq (b+3)^3-b^3\leq 12b(b+3)\Leftrightarrow 9\leq b(b+3)\leq \frac{27}2\Rightarrow b=2$$
So the only solution is $(a,b)=(5,2)$.