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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===Solution===
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We use the division algorithm to obtain <math>ab^2+b+7 \mid 7a-b^2</math>
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Here <math>7a-b^2=0</math> is a solution of the original statement, possible when <math>a=7k^2</math> and <math>b=7k</math> where <math>k</math> is any natural number. This is easily verified.
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Otherwise we obtain the inequality (by basic properties of divisiblity):
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<math>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</math>
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So <math>7-b^2 \geq 0 \implies b=1,2</math>
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Testing for <math>b=1</math>

Revision as of 05:38, 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 $b=1$