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

Revision as of 06:40, 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 $k$ .

@@ Line 10: / Line 10: @@
 So <math>7-b^2 \geq 0 \implies b=1,2</math>
-Testing for <math>b=1</math>
+Testing for <math>b=1</math> we find that <math>a+8 \mid 7a-1 \implies a+8 \mid 57</math>
+Therefore, <math>a=11, 49</math>, and we can easily check these.
+Testing for <math>b=2</math> and applying the division algorithm we find that <math>4a+9 \mid 79</math>, having no solutions in natural <math>a</math>.
+Hence, the only solutions are:
+<math>(a,b) = (11, 1), (49,1), (7k^2, 7k</math> for all natural <math>k</math>.

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

Revision as of 06:40, 10 April 2023

Solution