Difference between revisions of "1972 IMO Problems/Problem 3"
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Combining, | Combining, | ||
− | <math>4f(m,n-1)-f(m+1,n-1)=f(m,n) (\frac{4(m+n | + | <math>4f(m,n-1)-f(m+1,n-1)=f(m,n) (\frac{4(m+n)}{2(2n-1)}-\frac{2m+1}{2n-1})=f(m,n) \frac{2m+2n-2m-1}{2n-1}=f(m,n)</math>. |
Therefore, we have found the recurrence relation <math>f(m,n)=4f(m,n-1)-f(m+1,n-1)</math>. | Therefore, we have found the recurrence relation <math>f(m,n)=4f(m,n-1)-f(m+1,n-1)</math>. |
Revision as of 19:12, 21 November 2018
Let and be arbitrary non-negative integers. Prove that is an integer. (.)
Solution 1
Let . 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 .
We can see that is integral because the RHS is just , which we know to be integral for all .
So, must be integral, and then must be integral, etc.
By induction, is integral for all .
Borrowed from http://www.cs.cornell.edu/~asdas/imo/imo/isoln/isoln723.html
Solution 2
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
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.