Prove that for every positive integer \(n\), \(n^3 + 2n\) is divisible by \(3\).
Solution
1. For every integer \(n\),\[ n^3+2n=n(n-1)(n+1)+3n. \]
Expanding the right-hand side gives \[ n(n^2-1)+3n=n^3-n+3n=n^3+2n. \]
2. The product \(n(n-1)(n+1)\) is divisible by \(3\).
The integers \(n-1,n,n+1\) are three consecutive integers, so exactly one of them is congruent to \(0\) modulo \(3\).
3. Hence \(n^3+2n\) is divisible by \(3\).
By 2, the first term on the right-hand side of 1 is divisible by \(3\), and \(3n\) is also divisible by \(3\). Therefore their sum is divisible by \(3\).