2022 Junior Balkan MO, Problem 1

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sarjinius

239 posts

sarjinius

#1 h Jun 30, 2022, 1:09 PM • 2 Y

Y by Stepinac, lian_the_noob12

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

This post has been edited 1 time. Last edited by sarjinius, Jun 30, 2022, 1:54 PM

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BarisKoyuncu

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BarisKoyuncu

#2 h Jun 30, 2022, 1:13 PM • 3 Y

Y by physicskiddo, farhad.fritl, Math_.only.

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

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bitrak

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bitrak

#3 h Jun 30, 2022, 9:25 PM • 2 Y

Y by Math_.only., togrul123

I will go step by step.

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JustARandomGuy__

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JustARandomGuy__

#4 h Jun 30, 2022, 9:33 PM • 2 Y

Y by KhayalAliyev, togrul123

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

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Weighted_Dirichlet

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Weighted_Dirichlet

#5 h Jul 1, 2022, 4:17 PM • 2 Y

Y by Nabilos, togrul123

@BarisKoyuncu exactly my solution in exam

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Nabilos

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Nabilos

#6 h Jul 1, 2022, 6:19 PM

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Weighted_Dirichlet wrote:

@BarisKoyuncu exactly my solution in exam

Congratulations

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Vulch

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Vulch

#7 h Jul 2, 2022, 1:36 AM

Y by

$11ab \le a^3 - b^3 \le 12ab\implies 11ab \le (a-b)((a-b)^2 +3ab)\le 12ab.$
$=11p\le d(d^2 +3p)\le 12p,$ where $p=ab,$ and $d=(a-b).$

Can anybody help me from here?

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Johnweak

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Johnweak

#8 h Jul 2, 2022, 2:48 AM

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bitrak wrote:

I will go step by step.

Nice solution !!

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mihaig

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mihaig

#9 h Jul 2, 2022, 3:31 AM

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sarjinius wrote:

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

Who was the author?

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mihaig

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mihaig

#10 h Jul 2, 2022, 2:12 PM

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Hm? ......

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Vulch

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Vulch

#11 h Jul 3, 2022, 8:01 AM

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Solution:

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parmenides51

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parmenides51

#12 h Jul 3, 2022, 8:03 AM

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proposed by Croatia

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triangle112

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triangle112

#13 h Jul 4, 2022, 11:34 PM • 4 Y

Y by ihavealotofquestions, Mango247, Mango247, Mango247

Author is Ivan Novak, Croatia

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Iora

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Iora

#14 h Jul 11, 2022, 12:07 AM

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I overcomplicated things again :maybe:

Solution

This post has been edited 1 time. Last edited by Iora, Jul 11, 2022, 12:25 AM

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sttsmet

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sttsmet

#15 h Jul 12, 2022, 6:38 AM

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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

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amogususususus

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amogususususus

#16 h Aug 26, 2022, 7:56 AM

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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 $(5,2)$ .

This post has been edited 2 times. Last edited by amogususususus, Oct 11, 2023, 12:45 PM

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john0512

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john0512

#17 h Feb 9, 2023, 7:54 PM

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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 $(5,2).$

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andrewthenerd

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andrewthenerd

#18 h Apr 29, 2023, 8:54 AM

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$a^3 - b^3 = (a-b)(a^2 + ab + b^2) \leq 12ab$
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.

This post has been edited 1 time. Last edited by andrewthenerd, Apr 29, 2023, 8:56 AM

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dancho

38 posts

dancho

#19 h Jun 21, 2023, 9:19 PM

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

This post has been edited 3 times. Last edited by dancho, Jun 21, 2023, 9:37 PM

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ehuseyinyigit

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ehuseyinyigit

#20 h Nov 14, 2023, 4:47 PM

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

$14ab\leq a^3-b^3\leq 15ab$

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ehuseyinyigit

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ehuseyinyigit

#21 h Nov 14, 2023, 4:48 PM

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

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ehuseyinyigit

780 posts

ehuseyinyigit

#22 h Nov 14, 2023, 4:55 PM • 1 Y

Y by Math_.only.

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 $\left(a,b\right)=\left(\dfrac{3b+\lambda -2}{3},b\right)$ .

This post has been edited 2 times. Last edited by ehuseyinyigit, Dec 5, 2023, 2:59 PM

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Math_.only.

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Math_.only.

#23 h May 24, 2024, 8:39 AM

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Good solution

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banananjin

4 posts

banananjin

#25 h Dec 2, 2024, 5:16 PM • 2 Y

Y by triangle112, Fibonacci_11235

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.

This post has been edited 1 time. Last edited by banananjin, Dec 2, 2024, 5:17 PM
Reason: hihi

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eg4334

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eg4334

#26 h Jan 9, 2025, 4:27 AM

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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.

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anudeep

96 posts

anudeep

#27 h Yesterday at 4:34 AM

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We claim $(5,2)$ is the only pair.

Claim 1. $0<a-b<4$ .
Proof. The lower bound is obvious and the upper bound is obtained as,
$12ab\ge a^3-b^3=(a-b)(a^2+ab+b^2)>(a-b)(3ab).$
And bashing the heck out of remaining cases yields the required. $\square$

"Life is too short to argue just say trivial by case bash and move on."

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