Difference between revisions of "2007 USAMO Problems/Problem 5"
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− | Prove that for every [[nonnegative]] [[integer]] <math>n</math>, the number <math>7^{7^n}+1</math> is the [[product]] of at least <math> | + | Prove that for every [[nonnegative]] [[integer]] <math>n</math>, the number <math>7^{7^n}+1</math> is the [[product]] of at least <math>2n+3</math> (not necessarily distinct) [[prime]]s. |
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
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Revision as of 13:42, 4 July 2013
Problem
Prove that for every nonnegative integer , the number
is the product of at least
(not necessarily distinct) primes.
Contents
[hide]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
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.