Introductory Induction proof that $n(n^2 +5)$ is divisible by $6$

Question

I am in currently in a discrete mathematics class, and I've done well on every problem I've encountered. Unfortunately, I find myself weak at some of the seemingly straight forward induction problems. As is the case of the following corollary. I will attempt to show as far as I've gotten in the problem and where I have hit a road block.

Theorem: $n(n^2 +5)$ is divisible by 6 for every integer $n\ge0$

Proof: Let the property $P(n)$ be the sentence "$n(n^2 +5)$ is divisible by 6."

For our base case we will show that $P(0)$. That is, $0(0^2 +5)$ is divisible by 6. But, $0(0^2 +5)=0$ so $P(0)$ says that $0$ is divisible by 6 which is clearly true because $0$ is divisible by all integers as was previously shown in our text.

Now, for our inductive step we must show that $P(k) \Rightarrow P(k+1)$. Now, by definition of divisibility, $P(k)$ says that $k(k^2 +5) =6r$ for some $r\in \mathbb Z$ (EDIT AFTER HINT). Also, $P(k+1)$ says that $(k+1)[(k+1)^2 +5]=6p$ for some $p\in \mathbb Z$. After expansion and substitution $(k+1)[(k+1)^2 +5 = k^3 + 3k^2 +8k +6 = (k^3 +5k)+3k^2 +3k +6 = 6r +3k^2 +3k + 6= 3(2r +k^2 + k +2)=3(2r+k(k+1)+2)$

However, note that $k(k+1)$ represents the pruduct of two consecutive integers and is thus even. That is, $k(k+1)=2q$ for some $q\in \mathbb Z$. Now, $2r+k(k+1)+2 = 2r+2q+2$, but this is simply the sum of three even numbers and is thus an even number itself. That is, $2r+2q+2 = 2p$ for some $p\in\mathbb Z$. Therefore, $(k+1)[(k+1)^2 +5] =6q$.

EDIT:PROBLEM SOLVED - This is where I get stuck I've tried to expand $(k+1)[(k+1)^2 +5]$ which gives me $k^3 + 3k^2 +8k +6$. I've tried to manipulate it so that $k^3 + 3k^2 +8k +6=(k^3 +5k)+3k^2 +3k +6 = 6r +3k^2 +3k + 6= 3(2r +k^2 + k +2)$. However, this only shows divisibility by 3.

Any help would be much appreciated. Thank you!

Answer 1

$\displaystyle k^2+k=k(k+1)$ is a product of two consecutive integers, hence divisible by $2$

W/O induction, $$n(n^2+5)=\underbrace{n(n-1)(n+1)}_{\text{product of } 3 \text{ consecutive integers }}+6n$$

Answer 2

Method 1:

$$0\left(0^2+5\right)=0$$ $$1\left(1^2+5\right)=6$$ $$2\left(2^2+5\right)=18$$ $$3\left(3^2+5\right)=42$$ $$4\left(4^2+5\right)=84$$ $$5\left(5^2+5\right)=150$$

Since all of these are $0 \pmod 6$, every other number must be $0 \pmod 6$ as well.

Answer 3

In the general setting of divisibility of a polynomial function by a constant I would expand the polynomial around the constant in a Taylor series, in this case $n(n^2+5)=246+113(n-6)+18(n-6)^2+(n-6)^3$. The induction then basically needs to consider the divisibility of the constant term in the Taylor series.