Problem
ALG-B1-M01-P009 When an Expression Is Definitely Composite
#9
★★☆☆☆ Level 2 of 5
Prove that for every integer \(n>2\), the number \(n^2-1\) is composite.
Factor \(n^2-1\) as a difference of squares.
We have \(n^2-1=(n-1)(n+1)\). For \(n>2\), both factors are greater than \(1\): \(n-1>1\), \(n+1>1\).
Therefore \(n^2-1\) is a product of two integers greater than \(1\), so it is composite.
Even a simple problem should include checking the factors.