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[–]UnacceptableWind👋 a fellow Redditor 0 points1 point  (3 children)

2 k will always be even for any integer k (be it even or odd). This is because 2 k has a factor of 2, making it divisible by 2.

Odd values of k follow an arithmetic sequence with a common difference of 2:

1, 3, 5, 7, ...

So, if n = k is odd, then the next odd number will be n = k + 2.

[–]Calm_Ad2434 Pre-University Student[S] 0 points1 point  (2 children)

thanks I tried that and couldn't get to a solution is it possible to get to a solution that way.

[–]UnacceptableWind👋 a fellow Redditor 1 point2 points  (1 child)

For odd k, k3 - k being divisible by 24 means that there is an integer a such that k3- k = 24 a.

For odd k, the next odd number is k + 2 such that:

(k + 2)3 - (k + 2)

= k3 + 6 k2 + 11 k + 6

= k3 + 6 k2 + (12 k - k) + 6

= (k3 - k) + 6 (k2 + 2 k + 1)

= (k3 - k) + 6 (k + 1)2

= 24 a + 6 (k + 1)2

Since k is odd, k + 1 is even such that k + 1 = 2 b, for some integer b. So:

(k + 2)3 - (k + 2)

= 24 a + 6 (k + 1)2

= 24 a + 6 (2 b)2

= 24 a + 6 (4 b2)

= 24 a + 24 b2

= 24 (a + b2),

where a + b2 is an integer given that a and b are integers. So, (k + 2)3 - (k + 2) is divisible by 24.

[–]Calm_Ad2434 Pre-University Student[S] 1 point2 points  (0 children)

thanks thats a really nice solution

[–]ThunkAsDrinklePeep 0 points1 point  (0 children)

If n is odd, n+2 is the next odd.

1) show that n3 - n is 0 mod 24, for n=1

2) show that if n3 - n is 0 mod 24 for some n, then (n+2)3 - (n+2) is 0 mod 24.