Prove that n3 -n is divisible by 3 ifn 2. Solution n3 - n = n (n+1) (n-1) n-1, n, n+1 are three consecutive numbers where n >=2 when multiplying 3 consecutive numbers, the product isdivisible by 3 see the link: http://mathforum.org/library/drmath/view/61347.htm the following is from the above link: Suppose the first of our three consecutive numbers is N, sothat they are N, N+1, and N+2. Now, our number N must leave a remainder of either 0, 1, or 2when divided by 3, so it can be written as either 3k, 3k+1, or 3k+2 for some k(the quotient). So we have three cases: N = 3k, N+1 = 3k+1, N+2 = 3k+2 ==> only N is a multiple of 3 N = 3k+1, N+1 = 3k+2, N+2 = 3k+3 ==> only N+2 is a multiple of 3 N = 3k+2, N+1 = 3k+3, N+2 = 3k+4 ==> only N+1 is a multiple of 3 So there is always exactly one multiple of 3 among them..

1. Prove that n3 -n is divisible by 3 ifn 2.
Solution
n3 - n = n (n+1) (n-1)
n-1, n, n+1 are three consecutive numbers where n >=2
when multiplying 3 consecutive numbers, the product isdivisible by 3
see the link: http://mathforum.org/library/drmath/view/61347.htm
the following is from the above link:
Suppose the first of our three consecutive numbers is N, sothat they are N, N+1, and N+2.
Now, our number N must leave a remainder of either 0, 1, or 2when divided by 3,
so it can be written as either 3k, 3k+1, or 3k+2 for some k(the quotient). So we have three cases:
N = 3k, N+1 = 3k+1, N+2 = 3k+2 ==> only N is a multiple of 3
N = 3k+1, N+1 = 3k+2, N+2 = 3k+3 ==> only N+2 is a multiple of 3
N = 3k+2, N+1 = 3k+3, N+2 = 3k+4 ==> only N+1 is a multiple of 3
So there is always exactly one multiple of 3 among them.