1984 IMO Problems/Problem 6
Contents
Problem
Let be odd integers such that and . Prove that if and for some integers and , then .
Solution 1
Let . As , we infer that ; in particular, .
Now, , hence . It is easy to see that for , if , then . If , , so , which is in contradiction with the fact that . Thereby, .
Write . If , contradiction; so and , or equivalently ( otherwise or , contradiction )
Substituting back,
As and is odd, we get that . Furthermore, , hence . Now which together with the fact that , we get that , so the family of the solutions is described by the set
This solution was posted and copyrighted by TheFunkyRabbit. The original thread for this problem can be found here: [1]
Solution 2
Note that, if , then using , and , we get , which is clearly impossible. Thus, must hold.
Next, observe that, and implies . Keeping this in mind, we now observe that, , implies . Thus, . Now, note also that, if , then , and thus, the largest power of dividing either or is exactly . Clearly, , and thus, . Moreover, if , then we again have a contradiction, as . Thus, . This yields, , which brings us, . Now, using , we immediately obtain . This immediately establishes (since is odd), that .
This solution was posted and copyrighted by grupyorum. The original thread for this problem can be found here: [2]
See Also
1979 IMO (Problems) • Resources | ||
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