Difference between revisions of "2005 IMO Problems/Problem 4"
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Let <math>k</math> be a positive integer that satisfies the given condition. | Let <math>k</math> be a positive integer that satisfies the given condition. | ||
− | For all primes <math>p>3</math>, by FLT, <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 | + | For all primes <math>p>3</math>, by FLT, <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 <math>p.</math> Since <math>a_2 = 48 = 2^4 \cdot 3,</math> <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}} |
Revision as of 22:10, 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 FLT, 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 Since 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 |