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Prove that for integers $n > 0$, $n^3 + 5n$ is divisible by $6$.

Here is what I have done:

Base Step: $n=1$, $1^3+5(1)=6$

Inductive Step:

$p(k)=k^3 + 5k =6m$, $m$ is some integer

$p(k+1)=(k+1)^3+5(k+1)=6m$ $m$ is some integer

Since both are equal to $6m$ I set them equal to each other.

$k^3 + 5k=(k+1)^3+5(k+1)$

$k^3+5k=(k+1)[(k+1)^2+5]$

Proven that $p(k)=p(k+1)$

I do not think this is the correct way of proving this problem but I couldn't think of anything else.

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  • $\begingroup$ You want to expand $(k+1)^3+5(k+1)$ and then use that $k^3+5k=6m$ to show that $(k+1)^3+5(k+1)=6n$ for some integer n. $\endgroup$
    – user84413
    Feb 26 '14 at 1:44
  • $\begingroup$ When I expand, I get $6m+3k^2 +3k +6$. How would I use that to show $(k+1)^3+5(k+1)$? $\endgroup$
    – Kot
    Feb 26 '14 at 1:50
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Hint: $(k+1)^3+5(k+1) = (k^3 + 5k) + 3k^2 + 3k + 6$

Since $(k^3 + 5k)$ is divisible by 6, all you have to prove is $3k^2 + 3k + 6$ is divisible by 6 too. Since adding $6$ does not change divisibility, you just have to proove $3k(k+1)$ is divisible by 6. Think about even and odd numbers.

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  • $\begingroup$ Thank you! Figured it out. $\endgroup$
    – Kot
    Feb 26 '14 at 2:38

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