Discrete Mathematics Question, Proof by Induction.

Question

anonymous · Answer

$$n(n^2 + 5)$$ Divisible by 6, for all n >= 0.

What I have so far is that my basis is when n = 0 then 0 = 6 * 0.

Next I have my Induced Hypothesis: 
$$p(k) = k(k^2 + 5)$$
So 
$$k^3 = 6j - 5k$$

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Now to the prove step: $$p(k+1) = (k+1)((k+1)^2 + 5)$$ is divisible by 6?

I did math all the way down to...
$$6j + 3k^2 + 3k + 6$$
The factor of three is throwing me off, how do I prove it's divisible by 6?

anonymous · Answer

I just did something that I've never done before and am a little uncomfortable turning in homework with it on it but... I separated
$$3(k^2 + k)$$
And I set k equal to both an odd in one instance by saying suppose k is odd and supposed k is even so I got
$$((2m + 1)^2 + 2m + 1) = 2(2n^2+n)$$
Where 2n^2 + n is some integer.

I did the same with evens and got:
$$((2n^2) + 2n) = 2(2n^2 + n)$$
Where 2n^2 + n is some integer. So my new equation is both these plugged in so

$$6j + 6 + 3(2m(or-n)) = 6(j + 1 + m(or-n))$$

Therefore it is divisible by 6?

I'll be back in a few hours to award medals for any advise on this problem. Thanks in advance!

freckles · Answer

let's look at 3k^2+3k
3k^2+3k
=3k(k+1)

what do you notice about the product of k and k+1?

freckles · Answer

like if you multiply two consecutive integers what can always say about the product

anonymous · Answer

https://www.youtube.com/watch?v=buZeAty1axM

anonymous · Answer

Ok let me see if I get this because it seems simpler. Basically that since (ignoring the 3 for now)
k times k+1, it's always an even value? So with that we can say that k(k+1) = 2m where m is some integer by closure of z by +, *. So then we end up with 6m.

anonymous · Answer

Thanks for linking the video too, will check it out soon.

freckles · Answer

exactly since k and k+1 are consecutive integers one of them has to be even
and therefore k(k+1) is always even :)