MathDB
There exists P such that P|2^m -1 [Iran Second Round 1994]

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November 26, 2010
number theory proposednumber theory

Problem Statement

Let n>3n >3 be an odd positive integer and n=∏i=1kpiαin=\prod_{i=1}^k p_i^{\alpha_i} where pip_i are primes and αi\alpha_i are positive integers. We know that m=n(1−1p1)(1−1p2)(1−1p3)⋯(1−1pn).m=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3}) \cdots (1-\frac{1}{p_n}). Prove that there exists a prime PP such that P∣2m−1P|2^m -1 but P∤n.P \nmid n.
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