Difference between revisions of "2010 IMO Problems/Problem 3"
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+ | == See Also == | ||
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Latest revision as of 22:53, 23 October 2010
Problem
Find all functions such that is a perfect square for all
Author: Gabriel Carroll, USA
Solution
Suppose such function exist then:
Lemma 1)
Assume for contradiction that
has to be a perfect square
but .
A square cannot be between 2 consecutive squares. Contradiction. Thus,
Lemma 2) (we have show that it can't be 0)
Assume for contradiction, that .
Then there must exist a prime number such that and are in the same residue class modulo .
If where is not divisible by .
If .
Consider an such that
, where is not divisible by
If .
Consider an such that
, where is not divisible by
At least one of , is not divisible by . Hence,
At least one of , is divisible by an odd amount of .
Hence, that number is not a perfect square.
If , then , .
, which is not perfect square because is never a perfect square.
If , then , .
Thus, ,
See Also
2010 IMO (Problems) • Resources | ||
Preceded by Problem 2 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 4 |
All IMO Problems and Solutions |