Prove that n3-n is divisible by 6 for all n 1.

Here is the proof using the Principle of Mathematical Induction.

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

Let $P(n)$ be the statement "$n^3-n$ is divisible by 6".

• Base Case (n=1): $P(1): 1^3-1 = 1-1 = 0$. Since 0 is divisible by 6 ($0 = 6 \times 0$), $P(1)$ is true.

• Inductive Hypothesis: Assume $P(m)$ is true for some positive integer $m$. That is, assume $m^3-m$ is divisible by 6. This means $m^3-m = 6k$ for some integer $k$.

• Inductive Step: We need to prove that $P(m+1)$ is true, i.e., $(m+1)^3-(m+1)$ is divisible by 6. Consider the expression $(m+1)^3-(m+1)$:

  $$(m+1)^3-(m+1)$$

  Expand the term $(m+1)^3$:

  $$(m^3+3m^2+3m+1) - (m+1)$$

  Distribute the negative sign:

  $$m^3+3m^2+3m+1 - m - 1$$

  Combine like terms:

  $$m^3+3m^2+2m$$

  Rearrange the terms to use the inductive hypothesis:

  $$(m^3-m) + 3m^2+3m$$

  By the Inductive Hypothesis, $m^3-m = 6k$. Substitute this into the expression:

  $$6k + 3m^2+3m$$

  Factor out $3m$ from the last two terms:

  $$6k + 3m(m+1)$$

  We know that $6k$ is divisible by 6. Now consider the term $3m(m+1)$. The product of two consecutive integers, $m(m+1)$, is always an even number (because either $m$ is even or $m+1$ is even). Therefore, $m(m+1)$ can be written as $2j$ for some integer $j$. Substitute this into the term:

  $$3(2j) = 6j$$

  So, $3m(m+1)$ is divisible by 6. Since both $6k$ and $3m(m+1)$ are divisible by 6, their sum is also divisible by 6:

  $$6k + 6j = 6(k+j)$$

  This shows that $(m+1)^3-(m+1)$ is divisible by 6. Thus, $P(m+1)$ is true.

By the Principle of Mathematical Induction, $P(n)$ is true for all $n \ge 1$.
The proof is complete.