Difference between revisions of "2005 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | + | Let <math>k</math> be a positive integer that satisfies the given condition. | |
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+ | For all primes <math>p>3</math>, by [[Fermat's Little Theorem]], <math>n^{p-1} \equiv 1\pmod p</math> if <math>n</math> and <math>p</math> are relatively prime. This means that <math>n^{p-3} \equiv \frac{1}{n^2} \pmod p</math>. Plugging <math>n = p-3</math> back into the equation, we see that the value<math>\mod p</math> is simply <math>\frac{2}{9} + \frac{3}{4} + \frac{1}{36} - 1 = 0</math>. Thus, the expression is divisible by all primes <math>p>3.</math> Since <math>a_2 = 48 = 2^4 \cdot 3,</math> we can conclude that <math>k</math> cannot have any prime divisors. Therefore, our answer is only <math>1.</math> | ||
==See Also== | ==See Also== | ||
{{IMO box|year=2005|num-b=3|num-a=5}} | {{IMO box|year=2005|num-b=3|num-a=5}} |
Latest revision as of 23:14, 25 November 2023
Problem
Determine all positive integers relatively prime to all the terms of the infinite sequence
Solution
Let be a positive integer that satisfies the given condition.
For all primes , by Fermat's Little Theorem,
if
and
are relatively prime. This means that
. Plugging
back into the equation, we see that the value
is simply
. Thus, the expression is divisible by all primes
Since
we can conclude that
cannot have any prime divisors. Therefore, our answer is only
See Also
2005 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |