Let n be an odd integer. Prove that n^3/n is divisible by 24.

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- Oct 2nd 2008, 10:03 AMdolphinloverProof of divisibility
Let n be an odd integer. Prove that n^3/n is divisible by 24.

- Oct 2nd 2008, 10:09 AMicemanfan
$\displaystyle \frac{n^3}{n} = n^2$, which is not necessarily divisible by 24 if n is odd. Examples: $\displaystyle 3^2 = 9, 5^2 = 25$. Perhaps you wrote the problem incorrectly?

- Oct 2nd 2008, 10:13 AMdolphinlover
OOps! Yes I did. The correct one is: Prove (n^3)-n is divisible by 24.(Smirk)

- Oct 2nd 2008, 10:20 AMicemanfan
Ok. $\displaystyle n^3 - n = n(n^2 - 1) = n(n+1)(n-1)$.

Since the product is composed of three consecutive integers, one of them is divisible by 3. Also, n-1 and n+1 are consecutive even integers. Among every two consecutive even integers, one of them is divisible by 4, and the other is divisible by 2. Hence the two multiplied together are divisible by 8. Since 8 and 3 are both factors of the product and they are relatively prime, the product is divisible by 24. - Oct 2nd 2008, 10:34 AMdolphinlover
Thank you so much Icemanfan!