2.19 - Algebraic proof

Algebra can establish a pattern for every allowed integer at once. You will turn expressions into visible multiples, choose between expansion and factorisation, and represent consecutive, even and odd integers accurately enough to prove general statements.

Proof and counterexample

Choose an integer, square the number one above it and subtract the square of the number one below it. Why is the answer always divisible by $4$ , even if you start with a negative integer? We will build an argument that covers every choice.

Checking a few numbers helps you guess a rule, but it cannot establish an unlimited claim. A pattern can last a surprisingly long time before it breaks: $n^2 + n + 41$ is prime at every integer $n$ from $0$ to $39$ , but at $n = 40$ it gives $1681 = 41^2$ . Forty successful checks still miss the failure.

Proof

A proof is a logical argument establishing a statement for every value allowed by its conditions. An algebraic proof uses a general variable and valid algebra, rather than a list of chosen numerical examples.

The conditions belong to the argument. A claim about every integer includes negative integers and zero; a claim about positive integers does not. Your letter must represent the full set the question names.

Counterexample

A counterexample is one allowed case that makes a general statement false. For “every integer $n$ makes $n^2 + n + 41$ prime”, $n = 40$ is a counterexample. A case outside the stated conditions cannot disprove the claim.

A proof of divisibility needs more than a factor that looks convenient. To show that $E$ is divisible by a non-zero integer $m$ , show that $E = mk$ for an integer $k$ . For instance, $6n + 12 = 6(n + 2)$ is divisible by $6$ when $n$ is an integer, because $n + 2$ is also an integer. Writing $6n + 12 = 6(n + 2)$ with $n = \frac12$ instead gives $15$ , which is not divisible by $6$ .

Key idea

Examples can suggest a universal claim; one counterexample can defeat it; a general argument is needed to prove it.

Focus on what checking examples can establish in this diagnostic question.

Expanding to expose a multiple

If expanding makes the highest-power terms cancel, the remaining expression may reveal the required factor immediately.

You already know

Expand every term and keep brackets round a subtracted expression. The identities $(a + b)^2 = a^2 + 2ab + b^2$ and $(a - b)^2 = a^2 - 2ab + b^2$ include their middle terms.

You are simplifying an expression, not solving an equation. Keep the variable general; do not replace it by a favourite value or set the expression equal to zero.

Showing divisibility

For integers $E$ and $m$ , with $m \ne 0$ , $E$ is divisible by $m$ exactly when

E = mk \quad \text{for some integer } k

In a proof, show the factor $m$ and explain why the remaining factor is an integer.
Not on the formulae sheet: learn it.

The following proof starts with an arbitrary integer $n$ , so every algebraic step applies to every allowed case.

Worked example 1: subtracting a square

Question. Prove that $(n + 6)^2 - n^2$ is divisible by $12$ for every integer $n$ .

Plan. Expand the squared bracket, cancel $n^2$ and take out $12$ .

\begin{aligned} (n + 6)^2 - n^2 &= n^2 + 12n + 36 - n^2 && \text{expand the square} \\ &= 12n + 36 && \text{cancel equal terms} \\ &= 12(n + 3) && \text{expose the required factor} \end{aligned}

Why this step? Factoring out $12$ changes “looks divisible” into a product of $12$ and an integer.

Answer. Since $n + 3$ is an integer, the expression is divisible by $12$ for every integer $n$ .

Check. At $n = -4$ , it gives $2^2 - (-4)^2 = -12$ , a multiple of $12$ . This checks the working at one value; the algebra gives the proof.

Change only the sign inside the bracket, from $n + 6$ to $n - 6$ . A negative coefficient after expansion does not prevent divisibility.

Worked example 2: a negative multiple

Question. Prove that $(n - 6)^2 - n^2$ is divisible by $12$ for every integer $n$ .

Plan. Expand, simplify and keep the positive factor $12$ outside.

\begin{aligned} (n - 6)^2 - n^2 &= n^2 - 12n + 36 - n^2 && \text{expand with the middle term} \\ &= -12n + 36 && \text{cancel the squared terms} \\ &= 12(3 - n) && \text{factor out }12 \end{aligned}

Why this step? The remaining integer may be negative or zero. Both are allowed: $-24 = 12(-2)$ and $0 = 12(0)$ .

Answer. Since $3 - n$ is an integer, the expression is divisible by $12$ for every integer $n$ .

Check. At $n = 4$ , the original gives $(-2)^2 - 4^2 = -12$ , and $12(3 - 4) = -12$ .

When two non-trivial squares are subtracted, carry the second expansion in brackets until you distribute the minus. For example,

(n + 5)^2 - (n + 3)^2 = n^2 + 10n + 25 - (n^2 + 6n + 9) = 4(n + 4)

This proves the expression is divisible by $4$ for every integer $n$ , because $n + 4$ is an integer.

Watch out

Wrong: $(n - 6)^2 = n^2 + 36$ . Right: $n^2 - 12n + 36$ . A missing middle term destroys the expression you were asked to prove something about.

Use a new even offset in the matched proof. Your final factor must establish divisibility for an arbitrary integer.

Using a shared factor in a proof

A product already on the page can sometimes reveal the required multiple more clearly than a full expansion.

For $AB - AC$ , the common factor $A$ gives $A(B - C)$ . This is the distributive law in reverse, and it is not supplied on the formulae sheet. Keep the subtraction inside brackets until both signs have been handled.

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2.19 - Algebraic proof

Proof and counterexample

Expanding to expose a multiple

Using a shared factor in a proof

Continue this lesson.

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