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September 1st, 2009, 12:45 AM   #1
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A problem from 31st International Mathematical Olympiad

3. Determine all integers such that is an integer.
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September 1st, 2009, 06:48 AM   #2
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Re: A problem from 31st International Mathematical Olympiad

Well, for starters, n must be odd. Clearly, n = 3 works. I think that's the only solution, but I'm not sure how to show it. (There are no other small solutions.)

I can show worthless facts like 'such n are not divisible by 5' with modular arithmetic, but that's no help.
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September 1st, 2009, 10:49 AM   #3
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Re: A problem from 31st International Mathematical Olympiad

Let , where 3 and q are relatively prime, and q is odd
Then (1)
Clearly is divisible by if , and is divisible by if.
Did I miss something?

Yes, I did. In the expression (1) figure in brackets is not integer (prove or disprove).
Statement about s=1 is still correct if we move q inside the brackets - there are enough 3s there to compensate, but statement about q still need to be proved.
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September 1st, 2009, 05:50 PM   #4
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Right, CRGreathouse. n = 3 is clearly a solution to the problem. There could possibly be other solutions as well, but again we have to prove whether there is or not an another set of solutions.
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September 2nd, 2009, 06:06 AM   #5
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Re: A problem from 31st International Mathematical Olympiad

Let's take another way.
If n = p, p - odd prime (n is odd, obviously), then:
, so if n - prime it is 3.

Let , where q - smallest prime, divided n.
Then exists smallest r:. It is true, because q is a divisor of n.
Let , then:
If a - even: r - is not the smallest - contradiction
if a - odd: : - for some m and : - r is not the smallest.
So r divides n. Let's show that :
If : , then , so , and r is not smallest again.
So n cannot be composite and n = 3 is the only solution.

Is it messy?
Personally, I do not like this solution. Can somebody come up with the more elegant one?
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