# Thread: Proof by Induction (n^4 - 4n^2)

1. ## [SOLVED] Proof by Induction (n^4 - 4n^2)

Hi.

I need to prove that n^4 - 4n^2 is divisible by 3.

The induction hypnosis would be k^4 - 4k^2 is indeed divisible by 3, for k >= 0.

What I don't understand is after we expand out (k+1) -4(k+1)^2, why do we need to subtract (n^4-4n)?

I know that n^4 - 4n = 3t for some integer t (divisible by 3).

http://courses.cs.vt.edu/~cs4104/heath/Spring2005/resources/induction.pdf

The answer is given somewhere in the middle.

Hence (n+1)^4 - 4(n+1)2 is divisible by 3 if (n+1)^4 - 4(n+1)^2-(n^4 -4n^2)
is divisible by 3.
Thanks!

2. ## Re: Proof by Induction (n^4 - 4n^2)

Originally Posted by jwxie
Hi.

I need to prove that n^4 - 4n^2 is divisible by 3.

The induction hypnosis would be k^4 - 4k^2 is indeed divisible by 3, for k >= 0.

What I don't understand is after we expand out (k+1) -4(k+1)^2, why do we need to subtract (n^4-4n)?
Here is the trick:
$\displaystyle (n+1)^4-4(n+1)^2=n^4-4n^2+6n^2+4n(n^2-1)-3$

Each of $\displaystyle \color{blue}~(n^4-4n^2),~\&~(6n^2-3)$ is divisible by 3.

Look at $\displaystyle 4n(n^2-1)=4\color{blue}n(n-1)(n+1)$

But $\displaystyle n(n-1)(n+1)$ is the product of three consecutive integers.
Therefore it must be divisible by 3.

3. ## Re: Proof by Induction (n^4 - 4n^2)

Hi Plato,
At first I thought you were showing me what they did in the solution. Then I started questioning myself "what trick?" and when I worked on the example I actually worked through the example, and found out that we did the same thing.
Yes. Indeed. A nice trick Although I still don't get the reason why they would subtract anything in the first place.

Thanks!

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# prove by induction n^4-4n^2 is divisible by 3 for all n=>1

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