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Difference between revisions of "2005 IMO Problems/Problem 4"

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==Problem==
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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>
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==Video Solution==
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https://www.youtube.com/watch?v=-rRPkQrmzJw
  
 
==Solution==
 
==Solution==
  
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For all primes <math>p</math> greater than <math>3</math>, by Fermat's last theorem, <math>n^{p-1} = 1</math> mod <math>p</math> if <math>n</math> and <math>p</math> are relatively prime. This means that <math>n^{p-3} = \frac{1}{n^2}</math> mod <math>p</math>. Plugging <math>n = p-3</math> back into the equation, we see that the value mod <math>p</math> is simply <math>\frac{2}{9} + \frac{3}{4} + \frac{1}{36} - 1 = 0</math>. Thus, the expression is divisible by <math>p</math>. Because the expression is clearly never divisible by <math>2</math> or <math>3</math>, our answer is all numbers of the form <math>2^a3^b</math>.
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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>
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==See Also==
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{{IMO box|year=2005|num-b=3|num-a=5}}

Latest revision as of 15:43, 28 August 2024

Problem

Determine all positive integers relatively prime to all the terms of the infinite sequence \[a_n=2^n+3^n+6^n -1,\ n\geq 1.\]

Video Solution

https://www.youtube.com/watch?v=-rRPkQrmzJw

Solution

Let $k$ be a positive integer that satisfies the given condition.

For all primes $p>3$, by Fermat's Little Theorem, $n^{p-1} \equiv 1\pmod p$ if $n$ and $p$ are relatively prime. This means that $n^{p-3} \equiv \frac{1}{n^2} \pmod p$. Plugging $n = p-3$ back into the equation, we see that the value$\mod p$ is simply $\frac{2}{9} + \frac{3}{4} + \frac{1}{36} - 1 = 0$. Thus, the expression is divisible by all primes $p>3.$ Since $a_2 = 48 = 2^4 \cdot 3,$ we can conclude that $k$ cannot have any prime divisors. Therefore, our answer is only $1.$

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
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