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

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.

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