Proving: 6 Divides (n^3-n) for All Integers n

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SUMMARY

The discussion centers on proving that 6 divides the expression \( n^3 - n \) for all integers \( n \). The expression can be factored as \( n(n-1)(n+1) \), which represents the product of three consecutive integers. It is established that among any three consecutive integers, at least one is even (ensuring divisibility by 2) and at least one is divisible by 3, thus confirming that \( n^3 - n \) is divisible by 6 for all integers \( n \).

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Homework Statement



prove:6 divides (n^3-n) for all integers n.

Homework Equations


n^3-n=(n)(n+1)(n-1)


The Attempt at a Solution


tried to use direct proof. Then used cases that involed n=2k for some integer k and n=2k+1 for some integer k. However, i could not get it so that 6 was factored from either odd/even of n^3-n i.e. n^3-n=6m for some integer m.
Just a hint please.
 
Physics news on Phys.org
What's the remainder on division by 3 of the 3 factors you have exhibited?
 
One of any two consecutive number is even. One of any three consecutive numbers is a multiple of 3.
 

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