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Prove the following statement by mathematical induction:
$n(n^2+5)$ is divisible by 6 for all integer $n \ge 1$

My attempt:
Let the given statement be p(n).
(1) $1(1^2+5)$=6 Hence, p(1) is true.

(2) Suppose for all integer $k \ge 1$, p(k) is true.
That is, $k(k^2+5)$ is divisible by 6

We must show that p(k+1) is true.
$(k+1)((k+1)^2+5)$=$k^3+3k^2+3k+1+5(k+1)$
=$k^3+3k^2+8k+6$
=$k(k^2+5)+3k^2+3k+6$

I'm stuck on this step. I feel I have to show $3k^2+3k+6$ is divisible by 6. But, how can I show $3k^2+3k+6$ is divisible by 6?

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    $\begingroup$ $k^3+3k^2+8k+6=k\left(k^2+5\right)+3\left(k^2+k+2\right)$. Notice $k^2+k+2=k(k+1)+2$ is always even for all $k\in\mathbb Z$. $\endgroup$
    – user236182
    Feb 16, 2016 at 10:25
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    $\begingroup$ Does it have to be by induction? It seems to be quicker to show that it is always congruent to 0 modulo 2 as well as modulo 3, and then appeal to Chinese remainders. $\endgroup$ Feb 16, 2016 at 10:36
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    $\begingroup$ @user236182 I edited it. $\endgroup$
    – buzzee
    Feb 16, 2016 at 11:37
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    $\begingroup$ @buzzee Now notice that $3k^2+3k+6=3(k(k+1)+2)$. This number is divisible by $6$ because $k(k+1)+2$ is divisible by $2$ (because $k(k+1)$ is divisible by $2$, because exactly one of $k,k+1$ is divisible by $2$). $\endgroup$
    – user236182
    Feb 16, 2016 at 11:41
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    $\begingroup$ $6|6$ so remains to show $6|3k^2 + 3k$. $3|3k^2 + 3k$ so remains to show $2|k^2 + k$. If $k$ is even $k^2 + k$ is even. If $k$ is odd$ then $k^2 + k$ is even. $\endgroup$
    – fleablood
    Feb 16, 2016 at 16:31

4 Answers 4

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If $f(n)=n(n^2+5)$

$f(k+1)-f(k)$ $=(k+1)\{(k+1)^2+5\}-k(k^2+5)=3k^2+3k+1+5=6\cdot\dfrac{k(k+1)}2+6$ which is divisible by $6$ as $k(k+1)$ is even

$\implies6\mid f(k)\iff6\mid f(k+1)$

If induction is not mandatory,

$$n(n^2+5)=\underbrace{(n-1)n(n+1)}_{\text{Product of Three consecutive integers }}+6n$$

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First, show that this is true for $n=1$:

$1(1^2+5)=6$

Second, assume that this is true for $n$:

$n(n^2+5)=6k$

Third, prove that this is true for $n+1$:

$(n+1)((n+1)^2+5)=$

$\color\red{n(n^2+5)}+3n^2+3n+6=$

$\color\red{6k}+3n^2+3n+6=$

$6\left(k+\frac{n(n+1)}{2}+1\right)$


Since either $n$ or $n+1$ is even, $\frac{n(n+1)}{2}$ is integer.

Please note that the assumption is used only in the part marked red.

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Hint: let $$f(n)=n(n^2+5),n\geq1$$ then $$f(n+1)=(n+1)(n^2+2n+6)=n^3+2n^2+6n+n^2+2n+6=n^3+3n^2+8n+6=$$ $$=n^3+5n+3n^2+3n+6=3(n^2+n)+6+n(n^2+5)=f(n)+6\left(\frac{n(n+1)}{2}+1\right)$$

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I'm utterly confused. Three hours before asking this, you asked (and had answered) this: Proof of for all integer $n \ge 2$, $n^3-n$ is divisible by 6 by mathematical induction.

In that one you asked, and got answer for, how to show $6| 3k^2 + 3k$. In this one you are asking for how to show $6| 3k^2 + 3k + 6$.

How can you know the answer to one but not the answer to the other?

Answer to both: $3|3*h$ for any integer h, so $3|3(k^2 + k)$. If $k$ is odd so is $k^2$ so $k^2 + k$ is the sum of two odd numbers and is even. If $k$ is even so is $k^2$ and so is $k^2 + k$. So $k^2 + k$ is even. So $2|(k^2 +k)$ so $6|3(k^2 + k)$ so $6|3(k^2 + k) + 6$.

BTW: $n(n^2 + 5) = n^3 + 5n = (n^3 -1) + 6n$ so one is divisible by 6 if and only if the other one is.

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