Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
m (→‎Solution)
Line 17: Line 17:
 
Hence, the only solutions are:
 
Hence, the only solutions are:
 
<math>(a,b) = (11, 1), (49,1), (7k^2, 7k)</math> for all natural <math>k</math>.
 
<math>(a,b) = (11, 1), (49,1), (7k^2, 7k)</math> for all natural <math>k</math>.
 +
 +
Written by dabab_kebab

Revision as of 05:41, 10 April 2023

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_IMO_Problems/Problem_4&oldid=191957"