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.

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


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.

  • $\begingroup$ 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

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.