
Solution 1. Let $a-b=t$. Due to $a^{3}-b^{3} \geq 11 a b$ we conclude that $a>b$ so $t$ is a positive integer and the condition can be written as

$$
11 b(b+t) \leq t\left[b^{2}+b(b+t)+(b+t)^{2}\right] \leq 12 b(b+t)
$$

Since

$$
t\left[b^{2}+b(b+t)+(b+t)^{2}\right]=t\left(b^{2}+b^{2}+b t+b^{2}+2 b t+t^{2}\right)=3 t b(b+t)+t^{3}
$$

the condition can be rewritten as

$$
(11-3 t) b(b+t) \leq t^{3} \leq(12-3 t) b(b+t)
$$

We can not have $t \geq 4$ since in that case $t^{3} \leq(12-3 t) b(b+t)$ is not satisfied as the right hand side is not positive. Therefore it remains to check the cases when $t \in\{1,2,3\}$. If $t=3$, the above condition becomes

$$
2 b(b+3) \leq 27 \leq 3 b(b+3)
$$

If $b \geq 3$, the left hand side is greater than 27 and if $b=1$ the right hand side is smaller than 27 so there are no solutions in these cases. If $b=2$, we get a solution $(a, b)=(5,2)$.

If $t \leq 2$, we have

$$
(11-3 t) b(b+t) \geq(11-6) \cdot 1 \cdot(1+1)=10>t^{3}
$$

so there are no solutions in this case.

In summary, the only solution is $(a, b)=(5,2)$.


Problem 1. Find all pairs $(a, b)$ of positive integers such that

$$
11 a b \leq a^{3}-b^{3} \leq 12 a b
$$
