Difference between revisions of "2000 IMO Problems/Problem 5"
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Let <math>N=2^n+1</math>. We will assume for the sake of contradiction that <math>n|N</math>. | Let <math>N=2^n+1</math>. We will assume for the sake of contradiction that <math>n|N</math>. | ||
Latest revision as of 14:39, 5 August 2024
Does there exist a positive integer such that
has exactly 2000 prime divisors and
divides
?
Solution
THIS SOLUTION IS WRONG.. ANOTHER SOLUTION IS NEEDED..
Let . We will assume for the sake of contradiction that
.
. So 2 does not divide
, and so
is odd.
Select an arbitrary prime factor of and call it
. Let's represent
in the form
, where
is not divisible by
.
Note that and
are both odd since
is odd. By repeated applications of Fermat's Little Theorem:
(mod
)
Continuing in this manner, and inducting on k from 1 to ,
(mod
)
(mod
)
So we have (mod
)
Since is relatively prime to
,
(mod
)
(mod
)
Since is odd,
is not divisible by
. Hence
is not divisible by
. So we have a contradiction, and our original assumption was false, and therefore
is still not divisible by
.
See Also
2000 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |