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No n-th Powers Here! (Posted on 2005-09-05) Difficulty: 3 of 5
Show that the product of three consecutive positive integers cannot be the n-th power of an integer, for any integer n>1.

See The Solution Submitted by McWorter    
Rating: 3.4000 (5 votes)

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Solution Solution? | Comment 1 of 7
We assume that it is possible, and prove it by contradiction. (integers mean positive integers, I'm just lazy :p)

Let the 3 consecutive integers be x-1, x, x+1. Clearly x shares no common factor with either x-1 or x+1. If the product is an n-th power, then x = a^n for some integer a.

Then we have (a^n - 1)(a^n)(a^n+1) = b^n for some integers a,b.
a^n (a^2n - 1) = b^n

Then a divides b, so b = ka for some integer k.
a^2n - 1 = k^n
(a^2)^n - k^n = 1
We know this is impossible for n > 1, as all the gaps between 2 n-th powers is always >1.

  Posted by Tan Kiat Chuan on 2005-09-05 14:55:00
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