Difference between revisions of "2005 IMO Problems/Problem 4"
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==Solution== | ==Solution== | ||
− | For all primes <math>p | + | 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 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>. |
==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 23:02, 25 November 2023
Problem
Determine all positive integers relatively prime to all the terms of the infinite sequence
Solution
For all primes , by FLT,
if
and
are relatively prime. This means that
. 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
.
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 |