---
title: "Induction: n⁴ − 4n² divisible by 3"
subject: "Theory of Computation"
chapter: 1
key: "semester_3/theory_of_computation:induction-n4-4n2"
source: https://www.bctengineer.com/q/semester_3/theory_of_computation/induction-n4-4n2
site: BCT Engineering
---

# Induction: n⁴ − 4n² divisible by 3

Theory of Computation · Chapter 1: Introduction

- Asked in: 2074 Asoj
- Video: [Mathematical Induction | Show that (n³+2n) is divisible by 3](https://www.youtube.com/watch?v=8NFnJQ3hxD0) — Tambuwal Maths Class
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/theory_of_computation/induction-n4-4n2
- Whole chapter: https://www.bctengineer.com/md/semester_3/theory_of_computation/chapter-1

## Question

Use the principle of mathematical induction to prove that n⁴ − 4n² is divisible by 3 for n ≥ 0.

## Short answer

Base n=0 holds. For n=k+1, (k+1)⁴–4(k+1)² = (k⁴–4k²) + 4(k³–k) + 3(2k²–1). Each term is divisible by 3: first by hypothesis; second because k³–k is a product of three consecutive integers; third is a multiple of 3.

## Answer

Let P(n): n⁴ − 4n² is divisible by 3. We prove P(n) for all n ≥ 0.

#### Basis (n = 0)
0⁴ − 4·0² = 0 = 3·0, so P(0) is true.

#### Induction hypothesis
Assume P(k): k⁴ − 4k² = 3m for some integer m.

#### Induction step
Expand (k+1)⁴ − 4(k+1)²:

```text
(k+1)⁴ − 4(k+1)² = (k⁴ + 4k³ + 6k² + 4k + 1) − 4(k² + 2k + 1)
                 = k⁴ − 4k² + 4k³ + 6k² − 4k − 3
                 = (k⁴ − 4k²) + 4(k³ − k) + 3(2k² − 1)
```

- (k⁴ − 4k²) = 3m by the hypothesis.
- 3(2k² − 1) is a multiple of 3.
- 4(k³ − k): we show k³ − k is divisible by 3 (small lemma Q(k)).
  - Q(0): 0 − 0 = 0 is divisible by 3.
  - If k³ − k = 3t, then (k+1)³ − (k+1) = k³ + 3k² + 3k + 1 − k − 1 = (k³ − k) + 3(k² + k) = 3(t + k² + k), divisible by 3.

Hence all three terms are multiples of 3, so (k+1)⁴ − 4(k+1)² is divisible by 3, i.e. P(k+1) holds.

#### Conclusion
By induction, n⁴ − 4n² is divisible by 3 for all n ≥ 0. ∎

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which grouping of (k+1)^4 − 4(k+1)^2 is used in the step?
   A. (k^4 − 4k^2) + 3(2k^2 − 1) only
   B. (k^4 − 4k^2) + 4k^3 + 6k^2 − 3
   C. (k^4 − 4k^2) · (k+1)
   D. (k^4 − 4k^2) + 4(k^3 − k) + 3(2k^2 − 1)
   Answer: D. Expanding gives k^4 − 4k^2 + 4k^3 + 6k^2 − 4k − 3, which regroups into three parts. (Key point: Expansion)

2. Why does the proof need the lemma that k^3 − k is divisible by 3?
   A. Because the basis case fails without it
   B. Because 3(2k^2 − 1) is not divisible by 3
   C. The term 4(k^3 − k) is not obviously a multiple of 3 and needs it
   D. Because the hypothesis is about k^3 − k
   Answer: C. The first and third groups are clearly multiples of 3; 4(k^3 − k) needs k^3 − k divisible by 3. (Key point: Why a lemma)

3. How is the lemma Q(k+1) shown from Q(k)?
   A. (k+1)^3 − (k+1) = (k^3 − k) + 3(k^2 + k)
   B. (k+1)^3 − (k+1) = 3(k^3 − k)
   C. (k+1)^3 − (k+1) = k^3 + 1 − k
   D. (k+1)^3 − (k+1) = (k^3 − k) + 3
   Answer: A. Both pieces are multiples of 3: k^3 − k = 3t by hypothesis and 3(k^2 + k) obviously. (Key point: Lemma proof)

## Close variants set in other exams

- [Induction: 5^n − 1 divisible by 4](https://www.bctengineer.com/question.md?key=semester_3%2Ftheory_of_computation%3Ainduction-five-power) — Theory of Computation, chapter 1, asked 2×

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