Why n³ − n always divides by 6
Tonight's idea
Work out \(n^3 - n\) for \(n = 2, 3, 4, 5, 6\). You get 6, 24, 60, 120, 210. Every one divides by 6. After the last sheet you know that a run of successes proves nothing. We need a reason.
Factor. Take out \(n\), then use the difference of two squares:
\[n^3 - n = n(n^2 - 1) = (n - 1)\,n\,(n + 1).\]Three whole numbers in a row, multiplied. For \(n = 5\) that is \(4 \times 5 \times 6 = 120\).
Multiples of 3 come every third number, so any three in a row contain exactly one. Even numbers come every second number, so any two in a row contain one. The product therefore divides by 3 and by 2.
A number that divides by both 2 and 3 divides by 6, because 2 and 3 share no factor.
\(n^3 - n = (n - 1)\,n\,(n + 1)\): three in a row, so it divides by 6 for every whole number \(n\).
Watch one: \(n = 20\)
\(20^3 - 20 = 19 \times 20 \times 21\). Do not multiply out yet.
The multiple of 3 is \(21 = 3 \times 7\). An even one is \(20 = 2 \times 10\). Pull the 3 and the 2 to the front:
\[19 \times (2 \times 10) \times (3 \times 7) = 6 \times (19 \times 10 \times 7) = 6 \times 1330.\]Check: \(8000 - 20 = 7980\), and \(6 \times 1330 = 7980\).
Where you met the factoring: Expand and factor.
Your turn
- Write \(n^3 - n\) as three numbers in a row for (a) \(n = 4\); (b) \(n = 11\); (c) \(n = 100\). In each, say which of the three is the multiple of 3. Then divide the answers to (a) and (b) by 6.
- (a) Factor \(n^2 + n\) and explain why it is always even. (b) Is \(n^2 + n\) always a multiple of 3? Prove it or give a counterexample.
- (a) Work out \(n^3 + 5n\) for \(n = 1, 2, 3\). (b) Show that \(n^3 + 5n = (n^3 - n) + 6n\), and use this to prove that \(n^3 + 5n\) always divides by 6.
- When \(n\) is odd, \(n^3 - n\) divides by 24. (a) Check \(n = 3, 5, 7\). (b) If \(n\) is odd, then \(n - 1\) and \(n + 1\) are even numbers two apart. Explain why one of them is a multiple of 4, and why their product divides by 8. (c) Finish the proof. (d) Show that it can fail when \(n\) is even.
Stretch: \(n^5 - n\) and 30
Show that \(n^5 - n = (n - 1)\,n\,(n + 1)(n^2 + 1)\), and that \(n^2 + 1 = (n - 2)(n + 2) + 5\). Use these to prove that \(n^5 - n\) divides by 30 for every whole number \(n\).
Answers
- (a) \(3 \times 4 \times 5 = 60\); the multiple of 3 is 3. (b) \(10 \times 11 \times 12 = 1320\); it is 12. (c) \(99 \times 100 \times 101\); it is 99. \(60 \div 6 = 10\) and \(1320 \div 6 = 220\).
- (a) \(n^2 + n = n(n + 1)\), two whole numbers in a row. One of them is even, so the product is even. (b) No. \(n = 1\) gives 2, which is not a multiple of 3. Two in a row can miss the multiples of 3; three in a row cannot.
- (a) 6, 18, 42. (b) \((n^3 - n) + 6n = n^3 + 5n\). The first part divides by 6 by tonight's result, and \(6n\) clearly does. A sum of two multiples of 6 is a multiple of 6.
- (a) 24, 120, 336, and \(120 = 24 \times 5\), \(336 = 24 \times 14\). (b) Even numbers in order alternate between multiples of 4 and non-multiples: 2, 4, 6, 8, … So of two even numbers that are two apart, one is a multiple of 4. The other is still even, so the product divides by \(4 \times 2 = 8\). (c) One of the three numbers in a row is a multiple of 3. The product divides by 8 and by 3, which share no factor, so it divides by 24. (d) \(n = 2\) gives 6.
Stretch. \(n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = (n - 1)\,n\,(n + 1)(n^2 + 1)\), and \((n - 2)(n + 2) + 5 = n^2 - 4 + 5 = n^2 + 1\). So \[n^5 - n = (n - 2)(n - 1)\,n\,(n + 1)(n + 2) + 5(n - 1)\,n\,(n + 1).\] The first part is five in a row, which contains a multiple of 5. The second part has a factor 5. So \(n^5 - n\) divides by 5. It contains \((n - 1)\,n\,(n + 1)\), so it divides by 6. 5 and 6 share no factor, so it divides by 30.