Divisibility of n cubed minus n by 6

Statement

Prove that for every $n\ge 1$ the number $n^3-n$ is divisible by $6$.

Solution

Base ($n=1$): $1^3-1=0$, divisible by $6$.

Step: suppose that $6\mid k^3-k$. Then $(k+1)^3-(k+1) = (k^3-k) + 3k^2+3k = (k^3-k) + 3k(k+1).$ The first summand is divisible by $6$ by hypothesis. In the second, $k(k+1)$ is the product of two consecutive integers, hence even: therefore $3k(k+1)$ is divisible by $6$. The sum is divisible by $6$, that is $6\mid (k+1)^3-(k+1)$. ∎

Links

Topics: Set theory
Concepts: Principle of induction
Methods: Induction
Skills: Proving
Exercise type: Proof