Problem 54: IMO Shortlist 1997, Number Theory

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by henderson, Nov 30, 2016, 6:39 PM

${\color{red}\bf{Problem \ 54}}$ Let $b, m, n$ be positive integers such that $b > 1$ and $m \neq n.$ Prove that if $b^m - 1$ and $b^n - 1$ have the same prime divisors, then $b + 1$ is a power of $2.$
(IMO Shortlist 1997)
${\color{red}\bf{Solution}}$ Without loss of generality, suppose $m < n.$ If $b+1$ is not a power of $2$ and $n \neq 6,$ then by Zsigmondy's theorem we can find some prime $p$ such that $p\mid b^n - 1$ but $p \nmid b^m - 1,$ which is impossible. Furthermore, $2^1 - 1 = 1,$ $2^2 - 1 = 3,$ $2^3 - 1 = 7,$ $2^4 - 1 = 3 \cdot 5,$ $2^5 - 1 = 31$ and $2^6 - 1 = 3^2 \cdot 7,$ so $2^6 - 1$ does not have the same set of prime factors as any of $2^i - 1$ with $1 \leq i \leq 5.$ Hence, $n \neq 6,$ so we may conclude that $b+1$ is a power of 2.

Note that Zsigmondy's theorem tells us that $b+1$ must be a power of two and that $(m,n) = (1,2)$ or $(m,n) = (2,1).$

This post has been edited 6 times. Last edited by henderson, Jan 4, 2017, 2:42 PM

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Problem 54: IMO Shortlist 1997, Number Theory

by henderson, Nov 30, 2016, 6:39 PM

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