Difference between revisions of "2005 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
| − | + | 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>. | |
Revision as of 22:51, 17 July 2023
Determine all positive integers relatively prime to all the terms of the infinite sequence ![\[a_n=2^n+3^n+6^n -1,\ n\geq 1.\]](http://latex.artofproblemsolving.com/a/3/f/a3f3c15ee9d92c2eb2747f800d908cfa7aa6a08b.png)
Solution
For all primes 
greater than 
, by Fermat's last theorem, 
mod if 
and are relatively prime. This means that 
mod . Plugging 
back into the equation, we see that the value mod is simply 
. Thus, the expression is divisible by . Because the expression is clearly never divisible by 
or , our answer is all numbers of the form 
.
