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September 14, 2016 at 1:47 am (This post was last modified: September 14, 2016 at 1:48 am by Whateverist.)
This wasn't presented as an occasion for producing a proof but I found it a fun problem anyway. I wonder what you think of my argument in favor of the proposition. (I'll hide it in case you want to have a go at it yourself first.)
Proposition: The cube of any odd number ≥ 3 decreased by that same odd number will always be divisible by 24
The prime factorization of 24 is 2^3 • 3. So to show n^3 – n (where n is an odd number ≥ 3) must be divisible by 24, we must show it has at least three factors of 2 and at least one factor of 3.
It is easy enough to show that n^3 – n must have three factors of 2,
since n^3 – n = n(n^2 – 1) = n(n – 1)(n + 1) = (n – 1)n(n + 1)
If n is odd, then it can be written as 2a – 1 and
then (n – 1) = 2a – 1 – 1 = 2a – 2 = 2(a – 1), an even number
and (n + 1) = 2a – 1 + 1 = 2a, also an even number
In fact, 2a and 2(a – 1) are consecutive even numbers and that means one of them is also a multiple of 4, or 2^2
Thus n^3 – n has at least 3 factors of 2 and so is divisible by 2^3.
But why must it have a factor of 3? In the first allowable case, where n = 3, n itself is a multiple of 3.
However, in the very next case, where n = 5, n lacks a fact of 3. In this case, n – 1 = 4, and, n + 1 = 6. So it is 6 which carries the factor of three. Since every third counting number is a multiple of three, and since (n – 1)n(n + 1) is the product of three consecutive counting numbers, (n – 1)n(n + 1) will always contain exactly one factor of three.
Since n^3 – n must always contain at least 3 factors of 2 and exactly 1 factor of 3, the cube of any odd number (3 or larger) decreased by that same odd number will always be divisible by 3.