Difference between revisions of "2005 IMO Problems/Problem 4"
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+ | ==Problem== | ||
+ | |||
Determine all positive integers relatively prime to all the terms of the infinite sequence <cmath> a_n=2^n+3^n+6^n -1,\ n\geq 1. </cmath> | Determine all positive integers relatively prime to all the terms of the infinite sequence <cmath> a_n=2^n+3^n+6^n -1,\ n\geq 1. </cmath> | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=-rRPkQrmzJw | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | Let <math>k</math> be a positive integer that satisfies the given condition. | ||
+ | |||
+ | 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== | ||
+ | |||
+ | {{IMO box|year=2005|num-b=3|num-a=5}} |
Latest revision as of 15:43, 28 August 2024
Contents
Problem
Determine all positive integers relatively prime to all the terms of the infinite sequence
Video Solution
https://www.youtube.com/watch?v=-rRPkQrmzJw
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 |