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 $n$ = 7$a$ + $b$, where $b$ = 1,2,3,4,5,6.

$n$³ = (7$a$ + $b$)(7$a$ + $b$)(7$a$ + $b$)

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

$b$³ 1,1,-1,1,-1,-1 mod 7, respectively.

Hence $n$³ 1 mod 7, where $n$ is not a multiple of 7.

Corollary

$n$³ -1,0,1 mod 7, where $n$N.