IMO 2009 #1

by Wolstenholme, Nov 5, 2014, 9:29 PM

Let $ n$ be a positive integer and let $ a_1,a_2,a_3,\ldots,a_k$ $ \left( k\ge 2\right)$ be distinct integers in the set $ \left{ 1,2,\ldots,n\right}$ such that $ n$ divides $ a_i(a_{i + 1} - 1)$ for $ i = 1,2,\ldots,k - 1$. Prove that $ n$ does not divide $ a_k(a_1 - 1).$

Proof:

Assume the contrary, so that $ a_i \equiv a_ia_{i + 1} \pmod{n} $ for all $ i \in \{1, 2, \dots, k\} $ (with indices taken modulo $ k $).

Therefore we find that $ a_1 \equiv a_1a_2 \equiv \dots \equiv a_1a_2{\dots}a_k \equiv a_1a_2{\dots}a_{k - 2}a_k \equiv \dots \equiv a_1a_k \equiv a_k \pmod{n} $ but since $ a_1 $ and $ a_k $ are distinct elements of the set $ \{1, 2, 3, \dots, n\} $ this is clearly impossible, contradiction!

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  • Uh-oh ARML practice is Thursday... I should start the homework. :P

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  • Yes I am Sam, and Chebyshev polynomials aren't trivial, although they do make some problems trivial :P

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  • How are Chebyshev Polynomials trivial? :P

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  • Hi Wolstenholme did you actually use calc on that tstst problem :o

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