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February 19th, 2018, 11:34 AM   #1
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Using Mathematical Induction

Dear My Math Forum Community:

Prove: n^3 - n is divisible by 3 for every positive integer n.

[ Hint: use (k + 1)^3 - (k + 1) = (k^3 - k) + 3(k^2 + k). This proof is to
illustrate the concept of mathematical induction. ]
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February 19th, 2018, 11:51 AM   #2
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Base case is $n=1$ and 0 is clearly divisible by 3 so it holds.

Now assume inductively that it holds for $k^3 - k$. Applying the hint you have
\[ (k + 1)^3 - (k + 1) = (k^3 - k) + 3(k^2 + k) \]
so by your inductive assumption, the left summand is a multiple of 3. Clearly the right summand is as well and therefore, $(k+1)^3 - (k+1)$ is a multiple of which and by induction, it holds for all $k \in \mathbb{N}$.
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February 19th, 2018, 11:51 AM   #3
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Quote:
Originally Posted by Carl James Mesaros View Post
Dear My Math Forum Community:

Prove: n^3 - n is divisible by 3 for every positive integer n.
Yet another problem better solved without induction.

$n^3 - n = n (n^2 -1) = n (n + 1) (n - 1)$

One of $n$, $n + 1$, and $n -1$ is divisible by $3$; therefore their product is.
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February 19th, 2018, 11:54 AM   #4
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Quote:
Originally Posted by Maschke View Post
Yet another problem better solved without induction.

$n^3 - n = n (n^2 -1) = n (n + 1) (n - 1)$

One of $n$, $n + 1$, and $n -1$ is divisible by $3$; therefore their product is.
I think the point is not to prove this "theorem" which is trivial. Rather, it is to learn how to apply induction, which the OP failed to do since he just asked someone to do it for him.
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February 19th, 2018, 11:54 AM   #5
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$n^3-n=n(n-1)(n+1)=(n-1)n(n+1).$ Now it's easily seen that this is divisible by 3. If you need to solve this by induction, you can start by telling us what your initial approach would be.
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February 20th, 2018, 09:45 AM   #6
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If you are required to use induction, then you start, of course, by observing that when n= 1, this is 1- 1= 0 which is divisible by 6.

The next step is to show that "if, for some positive integer k, k^3- k is divisible by 6 then so is (k+1)^3- (k+1)".

To do that, calculate (k+1)^3- (k+1)= k^3+ 3k^2+ 3k+ 1- k- 1= k^3+ 3k^2+ 2k= (k^3- k)+ (3k^2+ 3k). Since k^3- k was divisible by 6 we can write this as 6n+ 3k(k+ 1) and we now can observe that, since k and k+ 1 are consecutive integers one of them is divisible by 2.
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