Problem 1467

Solution:
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 $(a, b)=(5,2)$.

Solution:
First, from $a^{3}-b^{3} \geq 11 a b>0$ it follows that $a>b$, implying that $a-b \geq 1$. Note that
$$a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right)=(a-b)\left[(a-b)^{2}+3 a b\right] \geq(a-b)(1+3 a b)>3 a b(a-b)$$
Therefore $12 a b \geq a^{3}-b^{3}>3 a b(a-b)$, which implies that $a-b<4$ so $a-b \in\{1,2,3\}$. We discuss three possible cases:
- $a-b=1$

After replacing $a=b+1$, the condition $a^{3}-b^{3} \geq 11 a b$ reduces to $1 \geq 8 b^{2}+8 b$, which is not satisfied for any positive integer $b$.

- $a-b=2$

After replacing $a=b+2$, the condition $a^{3}-b^{3} \geq 11 a b$ reduces to $8 \geq 5 b^{2}+10 b$, which is also not satisfied for any positive integer $b$.

- $a-b=3$

After replacing $a=b+3$, the condition $a^{3}-b^{3} \geq 11 a b$ reduces to $27 \geq 2 b^{2}+6 b$. The last inequality holds true only for $b=1$ and $b=2$. For $b=1$ we get $a=4$ and for $b=2$ we get $a=5$. Direct verification shows that $a^{3}-b^{3} \leq 12 a b$ is satisfied only for $(a, b)=(5,2)$.

In summary, $(a, b)=(5,2)$ is the only pair of positive integers satisfying all conditions of the problem.