Difference between revisions of "1998 IMO Problems/Problem 4"
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So <math>7-b^2 \geq 0 \implies b=1,2</math> | 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>. | ||
+ | |||
+ | Written by dabab_kebab | ||
+ | |||
+ | ==See Also== | ||
+ | |||
+ | {{IMO box|year=1998|num-b=3|num-a=5}} |
Latest revision as of 23:48, 18 November 2023
Determine all pairs of positive integers such that divides .
Solution
We use the division algorithm to obtain Here is a solution of the original statement, possible when and where is any natural number. This is easily verified.
Otherwise we obtain the inequality (by basic properties of divisiblity): So
Testing for we find that Therefore, , and we can easily check these.
Testing for and applying the division algorithm we find that , having no solutions in natural .
Hence, the only solutions are: for all natural .
Written by dabab_kebab
See Also
1998 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |