1972 IMO Problems/Problem 3
Let and be arbitrary non-negative integers. Prove that is an integer. (.)
Contents
Solution
If , then which is integral.
Solution 2
Denote the given expression as . We intend to show that is integral for all . To start, we would like to find a recurrence relation for . First, let's look at : Second, let's look at : Combining, Therefore, we have found the recurrence relation
Note that is just , which is an integer for all . Then so is an integer, and therefore must be an integer, etc.
By induction, is an integer for all .
Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html
Solution 3
Let p be a prime, and n be an integer. Let be the largest positive integer such that
WTS: For all primes ,
We know
Lemma 2.1: Let be real numbers. Then
Proof of Lemma 2.1: Let and
On the other hand,
It is trivial that (Triangle Inequality)
Apply Lemma 2.1 to the problem: and we are pretty much done.
Note: I am lazy, so this is only the most important part. I hope you can come up with the rest of the solution. This is my work, but perhaps someone have come up with this method before I did.
See Also
1972 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |