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Prove that $n(n^2 - 1)(n + 2)$ is divisible by $4$ for any integer $n$

I can not understand how to prove it. Please help me.

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7 Answers 7

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HINT :

$(n-1)n(n+1)(n+2)$

is "consective numbers".

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    $\begingroup$ Why the scare quotes? $\endgroup$
    – hmakholm left over Monica
    Dec 23, 2013 at 4:00
  • $\begingroup$ Well, nothing special:) $\endgroup$
    – mathlove
    Dec 23, 2013 at 4:55
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You can say more.... The number is divisible by 24. Why?

We have 4 consecutive numbers. So two of them have to be even and one of them should be divisible by 4. So 8 should divide the product.

Also, since we have 4 consecutive numbers, one of them is divisible by 3. Hence 24 divides the product.

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  • $\begingroup$ Also, for n=2, the product is exactly 24, so no larger number works. $\endgroup$
    – marty cohen
    Dec 23, 2013 at 4:39
  • $\begingroup$ Nice observation! $\endgroup$
    – user44197
    Dec 23, 2013 at 4:57
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$$\text{Here is a different way out; Recall that } \displaystyle \sum_{k=1}^m k = \dfrac{m(m+1)}2$$ $$\text{Hence, we have }1 + 2 + \cdots +(n-1) = \dfrac{n(n-1)}2 \text{ and } 1 + 2 + \cdots +(n+1) = \dfrac{(n+1)(n+2)}2$$ $$\text{Putting the above two together, we get that}$$ $$\color{red}{\left(1 + 2 + \cdots +(n-1) \right) \left(1 + 2 + \cdots + n + (n+1)\right)} = \color{blue}{\dfrac{n(n^2-1)(n+2)}4}$$ $$\text{ Clearly, the }\color{red}{\text{left side is an integer}}\text{ and hence }\color{blue}{\text{right side is also an integer}}.$$

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  • $\begingroup$ In the combinatorics spirit, it is also equal to $4! \binom{n+2}{4}$ $\endgroup$
    – chubakueno
    Jan 2, 2014 at 5:47
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Hint: Consider two cases: $n$ is even, or $n$ is odd. Can you find two parts of the product that are both even in either case? In the second, you may find it useful to write

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

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You have $n(n^2-1)(n+2)$ which can be seen as $(n-1)n(n+1)(n+2)$

Now, you should be able to see that this is divisible by $4$

Because... any four consecutive numbers will have "....."

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  • $\begingroup$ first thanks for all, $\endgroup$
    – user111554
    Dec 23, 2013 at 13:59
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Hint 1 $\ $ Partition the $\,4\,$ factors of the product into $\rm\color{#c00}{two}$ sets, each of which contains integers of opposite parity. Then each set contains an even number, so the product has $\rm\color{#c00}{two}$ factors of $\,2$.

Hint 2 $\ $ Consecutive integers have opposite parity.

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The first very obvious thing to note is that $n(n^2 - 1)(n+2)$ can be written as $n(n - 1)(n+1)(n+2)$ which means that they are four consecutive integers. Obviously one of these must be divisible by $4$.
The proof: Any number $n$ is either of the form $2m$ or $2m+1$.

Case I: $n = 2m$
$n(n^2 - 1)(n+2)$ = $2m(4m^2 - 1)(2m + 2) = 2m(4m^2 - 1)*2(m+1) = 4x$
Case II: $n = 2m+1$
$n(n^2 - 1)(n+2) = (2m+1)((2m+1)^2 - 1)(2m+1 + 2) = (2m+1)((4m^2 + 4m +1) - 1)(2m+3) = 4x$
Hence proved.

The link provided by @lab bhattacharjee is quite useful.

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