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 September 1st, 2009, 12:56 AM #1 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 A problem from 32nd International Mathematical Olympiad 2. Let $n\,>\,6$ be an integer and $a_1,\,a_2,\,\ldots,\,a_k$ be all the natural numbers less than $n$ and relatively prime to $n.$ If $a_2\,-\,a_1\,=\,a_3\,-\,a_2\,=\,\cdots\,=\,a_k\,-\,a_{k\,-\,1}\,>\,0,=$ prove that $n$ must be either a prime number or a power of $2.$
 September 2nd, 2009, 07:55 AM #2 Senior Member   Joined: Dec 2008 Posts: 160 Thanks: 0 Re: A problem from 32nd International Mathematical Olympiad Clearly, $a_i$ form arithmetical progression with $a_1$ as base and fixed $d= a_{i + 1} - a_i$ If $n$ - prime, $d= 1, a_1 = 1, (a_i)$ - set of all numbers < n. If $n= 2^m, d = 2, a_1 = 1, (a_i)$ - set of all odd numbers < n If n has prime divider $q$ other than 2 - progression should have a 'dents' - numbers that multiple that prime divider, that is impossible, because, otherwise, it will not cover all numbers < n. Like: $q$ should not belong to the set, while $q - d * k$ for some k should.

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