Induction divisibility (1 Viewer)

[enigma_1]

Examine the divisors of n^3−n for low odd values of n, make a judgement about the largest
integer divisor, and prove your result by induction.

What process do I need to use for this?

 

[Trebla]

Examine the divisors of n^3−n for low odd values of n, make a judgement about the largest
integer divisor, and prove your result by induction.

What process do I need to use for this?

n³ - n = n(n - 1)(n + 1)

This is the product of three consecutive integers. Since you have three consecutive integers, at least one of the integers must be divisible by 3 so you can factorise 3 out of the expression since one of the products has that factor.

Also, since n is an odd integer, (n - 1) and (n + 1) must be even integers, so you can factorise 2 out of both expressions (i.e. can factorise 4 out).

However, when you factorise 2 out of (n - 1) and (n + 1) the two products become consecutive integers (e.g. 10 x 12 factorises to 2 x 5 x 2 x 6 = 4 x 5 x 6). The product of two consecutive integers must be even as one of them must be an even integer. This means that we can factorise another 2 out of the expression.

From this, you can deduce that you can factorise 3, 4 and 2 out of the expression which leaves you with 24 x (some integer). Hence, the expression is divisible by 24.

Now you can proceed to prove it more formally by induction.