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>. |
Revision as of 06:40, 10 April 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
.