2007 IMO Problems/Problem 5
(Kevin Buzzard and Edward Crane, United Kingdom) Let and be positive integers. Show that if divides , then .
Lemma. If there is a counterexample for some value of , then there is a counterexample for this value of such that .
Proof. Suppose that . Note that , but . It follows that . Since it follows that can be written as , with . Then is a counterexample for which .
Now, suppose a counterexample exists. Let be a counterexample for which is minimal and . We note that and Now, Thus is a counterexample. But , which contradicts the minimality of . Therefore no counterexample exists.
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
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