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Difference between revisions of "1998 IMO Problems/Problem 4"

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Determine all pairs <math>(a, b)</math> of positive integers such that <math>ab^{2} + b + 7</math> divides
 
Determine all pairs <math>(a, b)</math> of positive integers such that <math>ab^{2} + b + 7</math> divides
 
<math>a^{2}b + a + b</math>.
 
<math>a^{2}b + a + b</math>.
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===Video Solution(In Chinese)===
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https://youtu.be/TAfXdhndY5M
 
===Solution===
 
===Solution===
 
We use the division algorithm to obtain <math>ab^2+b+7 \mid 7a-b^2</math>
 
We use the division algorithm to obtain <math>ab^2+b+7 \mid 7a-b^2</math>
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Written by dabab_kebab
 
Written by dabab_kebab
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==Video Solution by Flammable Maths==
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https://www.youtube.com/watch?v=MjyJzQeX6ec
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{{IMO box|year=1998|num-b=3|num-a=5}}

Latest revision as of 00:50, 28 August 2024

Determine all pairs $(a, b)$ of positive integers such that $ab^{2} + b + 7$ divides $a^{2}b + a + b$.

Video Solution(In Chinese)

https://youtu.be/TAfXdhndY5M

Solution

We use the division algorithm to obtain $ab^2+b+7 \mid 7a-b^2$ Here $7a-b^2=0$ is a solution of the original statement, possible when $a=7k^2$ and $b=7k$ where $k$ is any natural number. This is easily verified.

Otherwise we obtain the inequality (by basic properties of divisiblity): $7a-b^2 \geq ab^2+b+7 \implies 7a+7-ab^2-b^2 \geq b+14 \implies (7-b^2)(a+1) \geq b+14$ So $7-b^2 \geq 0 \implies b=1,2$

Testing for $b=1$ we find that $a+8 \mid 7a-1 \implies a+8 \mid 57$ Therefore, $a=11, 49$, and we can easily check these.

Testing for $b=2$ and applying the division algorithm we find that $4a+9 \mid 79$, having no solutions in natural $a$.

Hence, the only solutions are: $(a,b) = (11, 1), (49,1), (7k^2, 7k)$ for all natural $k$.

Written by dabab_kebab

Video Solution by Flammable Maths

https://www.youtube.com/watch?v=MjyJzQeX6ec

1998 IMO (Problems) • Resources
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 5
All IMO Problems and Solutions
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