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Prove that n^3-n is divisible by 3

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- Dec 9th 2009, 12:45 PMwantanswersmathematical induction
Can you please help me with

Prove that n^3-n is divisible by 3 - Dec 9th 2009, 01:06 PMcraig
$\displaystyle n^3 - n = 3a$ for some $\displaystyle a$.

First we see if it holds for $\displaystyle n=0$

$\displaystyle 0^3 - 0 = 0$, therefore true for $\displaystyle n=0$

Now we use what's called the induction step, we assume that it's true for $\displaystyle n=k$ where k is some natural nubmer.

This gives us $\displaystyle k^3 - k = 3a$

We then prove this for $\displaystyle n=k+1$, giving us

$\displaystyle (k+1)^3 - k - 1$.

Expanding the brackets gives $\displaystyle k^3 + 3k^2 + 3k + 1 - k - 1$

$\displaystyle k^3 + 3k^2 + 2k$ which can be written as $\displaystyle k^3 + 3k^2 + 3k - k$

$\displaystyle (k^3-k) + 3(k^2 + k) = 3a + 3(k^2 + k)$.

Therefore it is true for $\displaystyle k+1$ if true for $\displaystyle k$.

You know it is true for 0, therefore true for 1. True for one, therefore true for 2....

Hope this helps