Difference between revisions of "2007 USAMO Problems/Problem 5"
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Revision as of 08:06, 7 August 2014
Contents
Problem
(Titu Andreescu) Prove that for every nonnegative integer , the number is the product of at least (not necessarily distinct) primes.
Hint 1 of 3
You may be stuck in "factoring" for . Keep trying! This is a #5 on a USAMO!
Hint 2 of 3
Believe it or not, is a difference of squares!
Final Hint
One of the perfect squares is .
Solution
Solution 1
We proceed by induction.
Let be . The result holds for because is the product of primes.
Now we assume the result holds for . Note that satisfies the recursion
Since is an odd power of , is a perfect square. Therefore is a difference of squares and thus composite, i.e. it is divisible by primes. By assumption, is divisible by primes. Thus is divisible by primes as desired.
Solution 2
Notice that . Therefore it suffices to show that is composite.
Let . The expression becomes
which is the shortened form of the geometric series . This can be factored as .
Since is an odd power of , is a perfect square, and so we can factor this by difference of squares. Therefore, it is composite.
See also
- <url>viewtopic.php?t=145849 Discussion on AoPS/MathLinks</url>
2007 USAMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.