Proving results with identities

To show that a statement is true for every integer \(n\), write the numbers in terms of \(n\), expand and simplify, then write the result in a form that shows the property.

Why: trying a few values only gives examples. The table below suggests that \((n+1)^2-(n-1)^2\) is always a multiple of \(4\), but it cannot check every \(n\). Algebra does, because \(n\) stands for any integer.

Method

1. Write each number in terms of \(n\), for example two numbers either side of \(n\) as \(n-1\) and \(n+1\).
2. Form the expression that the statement describes.
3. Expand with the identities and collect like terms.
4. Take out the required factor, and state that the remaining bracket is an integer because \(n\) is an integer.
5. Finish with a sentence that answers the statement.

Example

Show that \((n+1)^2-(n-1)^2\) is a multiple of \(4\) for every integer \(n\).

\(\begin{aligned} (n+1)^2-(n-1)^2 &= (n^2+2n+1)-(n^2-2n+1) \\ &= 4n \end{aligned}\)

Since \(n\) is an integer, \(4n\) is \(4\) times an integer, so \((n+1)^2-(n-1)^2\) is a multiple of \(4\).

To show a result is odd, split \(1\) off the constant term so that the rest has a factor \(2\): \(4n^2+8n-1=4n^2+8n-2+1=2(2n^2+4n-1)+1\), which is \(2\times(\text{an integer})+1\).

Check: try one value, such as \(n=5\): \(6^2-4^2=20=4\times5\). A test like this is a check, not a proof. A common error is to stop at \(4n\) without saying why it is a multiple of \(4\).