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

I will go step by step.

because $a > b$, from the inequality $a^3 - b^3 > 3ab(a-b)$ and $a^3 - b^3 \leq 12ab$, we can deduce that $a-b \leq 3$. The rest as above

@BarisKoyuncu exactly my solution in exam

Weighted_Dirichlet wrote: @BarisKoyuncu exactly my solution in exam Congratulations

$11ab \le a^3 - b^3 \le 12ab\implies 11ab \le (a-b)((a-b)^2 +3ab)\le 12ab.$ $=11p\le d(d^2 +3p)\le 12p,$ where $p=ab,$ and $d=(a-b).$ Can anybody help me from here?

bitrak wrote: I will go step by step. Nice solution !!

sarjinius wrote: Find all pairs of positive integers $(a, b)$ such that $$11ab \le a^3 - b^3 \le 12ab.$$ Who was the author?

Hm? ......

Solution:

proposed by Croatia

Author is Ivan Novak, Croatia

Vulch wrote: $11ab \le a^3 - b^3 \le 12ab\implies 11ab \le (a-b)((a-b)^2 +3ab)\le 12ab.$ $=11p\le d(d^2 +3p)\le 12p,$ where $p=ab,$ and $d=(a-b).$ Can anybody help me from here? Sure! You have $11p-3dp \leq d^3 \leq 12p-3dp$ but $d^3$ is positive, therefore $p(12-3d) > 0 \implies 12 > 3d \implies d \leq 3$ We take cases and the conclusion follows

Clearly $a>b$, let $a=b+x$ with $x\in\mathbb{Z^+}$ $11ab\le a^{3}-b^{3} \le 12ab$ $\Rightarrow 11b^{2}+11bx \le x^{3}+3bx^{2}+3b^{2}x \le12b^{2}+12bx$ For $x\ge4$, $3bx^{2}\ge12bx$ and $3b^{2}x\ge12b^{2}$ $\Rightarrow 3bx^{2}+3b^{2}x\ge12b^{2}+12bx$ which results in contradiction, hence $x\le3$ For $x=1$, $8b^{2}+8b-1\le0$. No positive integer solution exist. For $x=2$, $5b^{2}+10b-8\le0$. Again, no positive integer solution exist. For $x=3$, $b^{2}+3b-11\le0$. Only $b=2$ satisfies. Therefore, the only solution is $(5,2)$.

Note that clearly $a>b$ since $a^3>b^3$. Let $x=a-b$ and $y=ab$ so that they are both positive. Then, $$11y\leq x(x^2+3y)\leq 12y$$$$11\leq \frac{x^3}{y}+3x\leq 12.$$ Case 1: $x=1$. Then $$8\leq \frac{1}{y}\leq 9,$$but this is not satisfied by any positive integer $y$. Case 2: $x=2$. Then $$5\leq \frac{8}{y}\leq 6,$$but again this is not possible. Case 3: $x=3$. Then $$2\leq \frac{27}{y}\leq 3,$$and since $y$ is a positive integer, $9\leq y\leq 13.$ However, remember that $a=b+3$, so $y=ab=b(b+3)$ From $9\leq y\leq 13$, only 10 can be expressed as $b(b+3)$, so we must have $x=3,y=10$ so $$(a,b)=(5,2).$$ $x\geq4$ is clearly not possible since $x^3/y$ is positive, so the only solution is $(5,2).$

$a^3 - b^3 = (a-b)(a^2 + ab + b^2) \leq 12ab$ Since $a^2 + b^2 \geq 2ab,$ hence $(a-b)(3ab) \leq a^3 - b^3 \leq 12ab \implies a-b \leq 4$. (further noticing that $a^2 + b^2 = 2ab$ iff $a=b$ which is impossible, hence $a^2 + b^2 > 2ab \implies a-b \leq 3)$ Case bash $a-b =1,2,3$ gives the only solution $(a,b)=(5,2)$. QED.

Obviosly $a>b$. Let $a=b+x$ where $x \in \mathbb{N}$. We get $A=(3x-11)b^2+(3x^2-11x)b+x^3\geq0$ and $B=(3x-12)b^2+(3x^2-12x)b+x^3\leq0$. Now we will find the discriminant for $b$ in both equations. $\Delta_A=-x^2(3x^2+22x-121)$ which needs to be less than or equal to zero, because $A\geq0$. $\Delta_B=-3x^2(x+12)(x-4)$ which needs to be greater than or equal to zero, because $B\leq0$. Solving $\Delta_A\leq0$ and $\Delta_B\geq0$ in natural numbers we get $x=3$. Now we need to solve $-2b^2-6b+27\geq0$ and $-3b^2-3b+9\leq0$. The above has only one solution in natural numbers: $(a,b)=(5,2)$

Version 1 Find all pairs of positive integers $(a,b)$ such that $$14ab\leq a^3-b^3\leq 15ab$$

Let us realise that this version problem has an equality case which original problem don't.

Generalization 1 Prove that for $\lambda \in \left(2,\dfrac{13+3\sqrt{21}}{2}\right)$ reals, the $(a,b)$ positive integer pairs which holds the inequality $$\lambda ab\leq a^3-b^3\leq \left(\lambda +1\right)ab$$are $$b\in \left(\dfrac{\left(\lambda -2\right)^3}{27\left(3b+\lambda -2\right)},\dfrac{\left(\lambda -2\right)^3}{18\left(3b+\lambda -2\right)}\right)$$for $b$ between below is $\left(a,b\right)=\left(\dfrac{3b+\lambda -2}{3},b\right)$.

Good solution

triangle112 wrote: Author is Ivan Novak, Croatia This is false information. Authors of this problem are in fact Borna Banjanin and Tin Salopek. They gave Ivan Novak the idea when he was their university assistant.

We have $11ab \leq (a-b)(a^2+ab+b^2) \leq 12ab$, and $a > b$. By AMGM $(a-b)(a^2+ab+b^2) \geq 3(a-b)ab$ so it follows that $a-b \leq 4$ But equality occurs only when $a=b$ so $a-b=4$ is not possible. We now split into cases. For each case $a=b+1, b=2, b+3$ we obtain a quadratic in $b$ just looking at the left side of the inequality. If $a=b+1$, we have $b \in [ \frac14 (-2-\sqrt{6}), \frac14 (\sqrt{6}-2) ]$, no solutions. If $a=b+2$, we similarly get no solutions. If $a=b+3$, we get $b \in [\frac12(-3-3\sqrt{7}), \frac12 (3\sqrt{7}-3)]$ giving us $b=1, 2$. A finite check gives us only $\boxed{(5, 2)}$ as a solution.