For any positive integer \(n\), prove that \(n^3 - n\) is divisible by 6.
Yes. For every positive integer \(n\), the number \(n^3 - n\) is divisible by \(6\).
Idea. Show that \(n^3 - n\) is divisible by \(2\) and by \(3\). Since \(\mathrm{lcm}(2,3) = 6\), divisibility by both \(2\) and \(3\) implies divisibility by \(6\).
Step 1: Factorise.
Write
\(n^3 - n = n(n^2 - 1)\)
\(= n(n - 1)(n + 1)\).
This is the product of three consecutive integers: \(n - 1\), \(n\), \(n + 1\).
Step 2: Divisible by \(2\).
Among any three consecutive integers, at least one is even. Therefore, \(n(n - 1)(n + 1)\) is divisible by \(2\).
Step 3: Divisible by \(3\).
Among any three consecutive integers, exactly one is a multiple of \(3\). Hence, \(n(n - 1)(n + 1)\) is divisible by \(3\).
Step 4: Combine the results.
Since the same number is divisible by \(2\) and by \(3\), it is divisible by \(6\).
Therefore, for every positive integer \(n\),
\(6 \mid (n^3 - n)\).
Optional check (small values).
\(n = 1\): \(1^3 - 1 = 0\), divisible by \(6\).
\(n = 2\): \(8 - 2 = 6\), divisible by \(6\).
\(n = 3\): \(27 - 3 = 24\), divisible by \(6\).
These examples agree with the proof.