1998 IMO Problems/Problem 4

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 $b=1$

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1998 IMO Problems/Problem 4

Solution