mathschallenge.net

Problem

Prove that for any number that is not a multiple of seven, then its cube will be one more or one less than a multiple of 7.

Solution

Given n is any number that is not evenly divisible by 7, let n = 7a + b, where b = 1,2,3,4,5,6.

n³ = (7a + b)(7a + b)(7a + b)

Clearly the only term under expansion that does not have at least one multiple of 7 will be b³ and as b = 1,2,3,4,5,6 we get b³ = 1,8,27,64,125,216.

b³ 1,1,-1,1,-1,-1 mod 7, respectively.

Hence n³ 1 mod 7, where n is not a multiple of 7.

Corollary

n³ -1,0,1 mod 7, where nN.