147. IF N IS A POSITIVE INTEGER, IS N^3-N DIVISIBLE BY 4.
1. N = 2K + 1, WHERE K IS AN INTEGER.
WHY IS THIS SUFFICIENT -I GOT THIS WRONG :/
We can factor the expression in the question, using the difference of squares:
n^3 - n = n(n^2 - 1) = (n)(n-1)(n+1)
Arranging these factors in order, we want to know if (n-1)(n)(n+1) is divisible by 4. Now these three factors are consecutive integers since they are 1 apart. If n is odd, as Statement 1 tells us, then the other two factors, n-1 and n+1, must both be even, so they each must be divisible by 2, and their product must be divisible by 2*2 = 4.
While we don't need it for this question, it's actually true that (n-1)(n)(n+1) must be divisible by 8 if n is odd, since then n-1 and n+1 are consecutive even numbers, and whenever you have two consecutive even numbers, one of them must be a multiple of 4, and the other a multiple of 2.
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