Question 349548: Prove the following statement.
If n=m^3-m for some integer m, then n is a multiple of 6.
My work so far:
Suppose n=m^3 - m. Thus n=m(m-1)(m+1). First if m=2k for some integer k. Thus n=(2k)(2k-1)(2k+1)= 2 (k(2k-1)(2k+1)). Also if m=2k+1 for some integer k, then n=(2k+1)(2k)(2k+2)=2(k(2k+1)(2k+2))
Thus n is divisible by 2.
I would like to find out how n is also divisible by 3 because if n is divisible by 3 as well then n would be divisible by 6.
Answer by jsmallt9(3758)   (Show Source):
You can put this solution on YOUR website! If n=m^3-m for some integer m, then n is a multiple of 6.
My work so far:
Suppose n=m^3 - m. Thus n=m(m-1)(m+1).
This is correct. From this point I would say- If m is an integer, then (m-1), m, and (m+1) are 3 consecutive integers
- In any set of 3 consecutive integers, there will be a multiple of 2 and a multiple of 3.
- Any number that is the product of a multiple of 2 and a multiple of 3 will be a multiple of 6.
This may not be particularly formal as a proof but the logic is perfectly sound.
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