## Question

**Using mathematical induction, prove that 6 divides n3 − n whenever n is a nonnegative integer.**

## Answer

**ere’s the proof using mathematical induction:**

**1. Base Case (n = 0):**

- n³ – n = 0³ – 0 = 0
- 0 is divisible by 6, so the statement holds for n = 0.

**2. Inductive Hypothesis:**

- Assume the statement is true for some arbitrary nonnegative integer k, meaning 6 divides k³ – k.

**3. Inductive Step:**

- Prove the statement holds for k + 1, meaning 6 divides (k + 1)³ – (k + 1).
- Expand (k + 1)³: (k + 1)³ = k³ + 3k² + 3k + 1
- Rewrite (k + 1)³ – (k + 1): (k + 1)³ – (k + 1) = k³ + 3k² + 2k
- Factor out k: k³ + 3k² + 2k = k(k² + 3k + 2)
- Factor the quadratic: k(k² + 3k + 2) = k(k + 1)(k + 2)
- Observe that this expression is a product of three consecutive integers.
- Among any three consecutive integers, at least one must be even and one must be divisible by 3.
- Therefore, their product is divisible by 2 × 3 = 6.

**4. Conclusion:**

- The statement holds for the base case (n = 0).
- The inductive step proves that if the statement holds for k, it must also hold for k + 1.
- Therefore, by mathematical induction, the statement “6 divides n³ – n whenever n is a nonnegative integer” is proven to be true.