Mathematical Induction

**xboxwee** · December 10th 2011, 02:47 PM

I was wondering if someone can show me step by step on how to do this question:
Prove that for all integers n >= 1, 6|n (n^2 + 5)
So far I got the bases step for when n = 1.
I am having troubles with the induction step after n = 1.

**Prove It** · December 10th 2011, 02:50 PM

Originally Posted by xboxwee

I was wondering if someone can show me step by step on how to do this question:
Prove that for all integers n >= 1, 6|n (n^2 + 5)
So far I got the bases step for when n = 1.
I am having troubles with the induction step after n = 1.

It's always more sophisticated to ask your questions in English instead of mathematical symbols.

Just so we're clear, you're asking to prove that for all positive integers, that 6 divides n(n^2 + 5)?

**xboxwee** · December 10th 2011, 02:52 PM

Yes.

**Prove It** · December 10th 2011, 03:05 PM

The base step is obvious.

For the inductive step, assume that 6 divides $\displaystyle \begin{align*} k(k^2 + 5) \end{align*}$ , in other words, write $\displaystyle \begin{align*} k(k^2 + 5) = 6m \end{align*}$ where $\displaystyle \begin{align*} m \end{align*}$ is some other positive integer.

Then we need to show that 6 divides $\displaystyle \begin{align*} (k + 1)\left[(k + 1)^2 + 5\right] \end{align*}$

$\displaystyle \begin{align*} (k + 1)\left[(k + 1)^2 + 5\right] &= (k + 1)\left(k^2 + 2k + 1 + 5\right) \\ &= k\left(k^2 + 5 + 2k + 1\right) + 1\left(k^2 + 5 + 2k + 1\right) \\ &= k\left(k^2 + 5\right) + k(2k + 1) + k^2 + 2k + 6 \\ &= 6m + 2k^2 + k + k^2 + 2k + 6 \\ &= 6m + 3k^2 + 3k + 6 \\ &= 6m + 3k(k + 1) + 6 \\ &= 6m + 3\cdot 2p + 6 \textrm{ where }p \textrm{ is some other positive integer, which we can do since }k(k + 1) \textrm{ is even...} \\ &= 6m + 6p + 6 \\ &= 6\left(m + p + 1\right) \textrm{ which is clearly divisible by }6 \end{align*}$

Q.E.D.

**xboxwee** · December 10th 2011, 03:15 PM

would this be right?

**xboxwee** · December 10th 2011, 03:17 PM

Thanks..

**Prove It** · December 10th 2011, 03:27 PM

Originally Posted by xboxwee

would this be right?

No it's not a complete proof, as you have not shown that your final line is divisible by 6. In fact, you have not used your assumption at all.

Also, the writing of your base step is atrocious.

$\displaystyle \begin{align*} 6|n\left(n^2 + 5\right) \end{align*}$ is a STATEMENT, that 6 divides $\displaystyle \begin{align*} n\left(n^2 + 5\right) \end{align*}$ . It can NOT be used in equations.

You would need to write something like

If $\displaystyle \begin{align*} n = 1 \end{align*}$ then $\displaystyle \begin{align*} n\left(n^2 + 5\right) = 1\left(1^2 + 5\right) = 6 \end{align*}$ which is divisible by 6.

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