GCSE Algebra Proof: Proving an Expression is a Multiple of 4

Original exam question

Prove that (3n + 1)² - (3n - 1)² is a multiple of 4, for all positive integer values of n.

Proof questions are a key part of the GCSE Maths curriculum, especially for students aiming for the top grades. They require clear, logical steps to show that a statement is true for all cases. In this post, we'll walk through a classic example: proving that ((3n + 1)² - (3n - 1)²) is a multiple of 4 for all positive integer values of (n).

Understanding the Question

First, let's unpack what the question is asking.

We have an expression: ((3n + 1)² - (3n - 1)²).

(n) represents any positive integer (1, 2, 3, 4, ...).

We need to prove that the value of this expression is always a multiple of 4. This means that when you evaluate it, the result can be divided by 4 with no remainder, or written as (4 × k) where (k) is an integer.

A proof must work for every possible positive integer (n), not just a few examples we test.

Method 1: Expanding and Simplifying

The most straightforward approach is to expand each squared term, subtract, and simplify the result.

Step 1: Expand each bracket.
Recall that ((a + b)² = a² + 2ab + b²).

((3n + 1)² = (3n)² + 2(3n)(1) + (1)² = 9n² + 6n + 1)

((3n - 1)² = (3n)² + 2(3n)(-1) + (-1)² = 9n² - 6n + 1)

Step 2: Subtract the second expression from the first.

(3n + 1)² - (3n - 1)² = (9n² + 6n +
1) - (9n² - 6n + 1)

Step 3: Simplify carefully.
It's crucial to subtract all terms in the second bracket. A common mistake is to forget the parentheses.

= 9n² + 6n + 1 - 9n² + 6n - 1

Notice how (-(-6n)) becomes (+6n) and (-(+1)) becomes (-1).

Step 4: Combine like terms.

(9n² - 9n²) + (6n + 6n) + (1 -
1) = 0 + 12n + 0 = 12n

Step 5: Show the result is a multiple of 4.
We have simplified the original expression to (12n). Since (12n = 4 × 3n), and (3n) is an integer (because (n) is an integer), (12n) is clearly a multiple of 4.

Conclusion: ((3n + 1)² - (3n - 1)² = 12n = 4(3n)). Therefore, it is a multiple of 4 for all positive integer values of (n). ✅

Method 2: Using the Difference of Two Squares

This is a more elegant method that uses a key algebraic identity. The difference of two squares states: (a² - b² = (a - b)(a + b)).

Here, we can let:

(a = 3n + 1)

(b = 3n - 1)

Step 1: Apply the identity.

(3n + 1)² - (3n - 1)² = [(3n +
1) - (3n - 1)] × [(3n +
1) + (3n - 1)]

Step 2: Simplify each bracket.

First bracket: ((3n +
1) - (3n -
1) = 3n + 1 - 3n + 1 = 2)

Second bracket: ((3n + 1) + (3n - 1) = 3n + 1 + 3n - 1 = 6n)

Step 3: Multiply the results.

= 2 × 6n = 12n

Step 4: Factorise to show the multiple of 4.
As before, (12n = 4 × 3n), proving it's a multiple of 4.

This method is often quicker and involves less expansion, reducing the chance of arithmetic errors.

flowchart TD
    A["Start: Expression, a² - b²"] --> B["Choose Method"]
    B --> C["Method 1: Expand and Simplify"]
    B --> D["Method 2: Difference of Two Squares"]
    C --> E["Expand a² and b² separately"]
    E --> F["Subtract and simplify"]
    F --> G["Result: 12n"]
    D --> H["Apply identity: a²-b² = a-b a+b"]
    H --> I["Simplify each bracket"]
    I --> J["Multiply: 2 x 6n = 12n"]
    G --> K["Factorise: 12n = 4 3n"]
    J --> K
    K --> L["Conclusion: Multiple of 4 for all n"]

Writing Your Final Proof for the Exam

In an exam, you need to present your work clearly and logically. Here's how you might write it using Method 2, which is concise:

> Proof:
> Let (a = 3n +
1) and (b = 3n - 1).
> Then the expression is (a² - b²).
> Using the difference of two squares:
>
> a² - b² = (a - b)(a + b)
>
> Substituting back:
>
> = [(3n+1) - (3n-1)] × [(3n+1) + (3n-1)]
>
>
> = (3n+1-3n+1) × (3n+1+3n-1)
>
>
> = (2) × (6n) = 12n
>
> Since (12n = 4 × 3n), and (3n) is an integer, (12n) is a multiple of 4.
> Therefore, ((3n+1)² - (3n-1)²) is a multiple of 4 for all positive integer values of (n).

Common Mistakes to Avoid

Incorrect expansion: Forgetting that ((3n - 1)² = 9n² - 6n + 1), not (9n² - 6n - 1).

Sign errors when subtracting: Not applying the minus sign to every term in the second bracket when simplifying ((9n²⁺⁶ⁿ⁺¹) - (9n^2-6n+1)).

Forgetting the final step: Simplifying to (12n) is not enough. You must explicitly state that (12n = 4 × 3n) to conclude it's a multiple of 4.

Only showing examples: Substituting (n=1,2,3) shows it works for those cases, but it does not prove it works for all (n). A general algebraic proof is required.

Why This Question is Important

This question tests several key GCSE skills:

Algebraic manipulation (expanding brackets, simplifying expressions).

Knowledge of identities (difference of two squares).

Understanding of proof and mathematical reasoning.

Arithmetic with integers.

It's a typical style of question found on Edexcel, AQA, and OCR GCSE Maths papers, particularly in the higher tier.

Practice Idea

Try adapting the question yourself: what would you need to prove if the expression was ((5n+2)² - (5n-2)²)? Follow the same steps and see what multiple you get.

Final Tip: In the exam, if you're unsure which method to use, expanding and simplifying (Method 1) is always a safe, reliable option. Just be meticulous with your signs!

Good luck with your revision!