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>. | ||
− | <math>2^n+1 \equiv 0 | + | <math>2^n+1 \equiv 0 \pmod{n} \Rightarrow 2^n \equiv -1 \pmod{n}</math>. So 2 does not divide <math>n</math>, and so <math>n</math> is odd. |
Select an arbitrary prime factor of <math>n</math> and call it <math>p</math>. Let's represent <math>n</math> in the form <math>p^am</math>, where <math>m</math> is not divisible by <math>p</math>. | Select an arbitrary prime factor of <math>n</math> and call it <math>p</math>. Let's represent <math>n</math> in the form <math>p^am</math>, where <math>m</math> is not divisible by <math>p</math>. |
Revision as of 01:04, 8 March 2024
Does there exist a positive integer such that
has exactly 2000 prime divisors and
divides
?
Solution
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 |