$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement.

Question

"Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"

Resolution: $$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$ Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.

score 5 · Accepted Answer · edited Apr 13 '17 at 12:20

5

We know from here or here, the product $n$ consecutive integers is divisible by $n!$ for integer $n>0$

As $24=4!$and we already have $n(n^2-1)=(n-1)n(n+1)$ the product of $3$ consecutive integers

So, if we can arrange the next or previous term as multiplier, we shall have the product of $4$ consecutive integers, hence divisible by $4!=24$

If we consider the previous term $(n-2),$

$3n+2=3(n-2)+4$

$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n-2)+4\}$$ $$=3\underbrace{(n+1)n(n-1)(n-2)}_{\text{ the product of } 4 \text{ consecutive integers }} +4 \underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$

If we consider the next term $(n+2),$

$3n+2=3(n+2)-4$

$$\implies n(n^2-1)(3n+2)=n(n^2-1)\{3(n+2)-4\}$$ $$=3\underbrace{(n+2)(n+1)n(n-1)}_{\text{ the product of } 4 \text{ consecutive integers }}-4\underbrace{(n+1)n(n-1)}_{\text{ the product of } 3\text{ consecutive integers }}$$

edited Apr 13 '17 at 12:20

Community

1

answered Jul 29 '13 at 18:37

lab bhattacharjee

274,582

Right, got it. How do these factorizations so "miraculous"? – benjamin_ee Jul 29 '13 at 18:39
@marcelolpjunior, does it look so now? – lab bhattacharjee Jul 29 '13 at 18:42
Crystal clear... Tank YOu... – benjamin_ee Jul 29 '13 at 18:59
1

@marcelolpjunior, my pleasure. Using induction let $f(n)=n(n^2-1)(3n+2)=(3n+2)(n+1)n(n-1)$
and $f(m)$ be divisible by $24$

$\implies f(m+1)=(3m+5)(m+2)(m+1)m$

$\implies f(m+1)-f(m)=(3m+5)(m+2)(m+1)m-(3m+2)(m+1)m(m-1)$ $=m(m+1){(3m+5)(m+2)-(3m+2)(m-1)}=12m(m+1)(m+1)$ which is clearly divisible by $24$ as one of $m,m+1$ must be even

Now, $f(1)=0$ is divisible by $24$ ...
– lab bhattacharjee Jul 29 '13 at 19:02
Heck, pretty cool, here in Brazil induction work differently, but the reasoning is very similar, here it is as if they were steps ... But at the same, I liked ... Thank you. – benjamin_ee Jul 29 '13 at 19:10
@marcelolpjunior, my pleasure. Would you please your induction method how it's different? – lab bhattacharjee Jul 30 '13 at 16:38
+1 Now I see where OP's fascination to products of consecutive integers came from (he asked for it in another question). A perfect match to that. Well done. – Jyrki Lahtonen Aug 05 '13 at 12:25
@JyrkiLahtonen, I have used products of consecutive integers here too: http://math.stackexchange.com/questions/459334/prove-that-360-divides-a-2a-1a-a-a1a2 – lab bhattacharjee Aug 05 '13 at 13:33
Good. May be use another trick next time :-) – Jyrki Lahtonen Aug 05 '13 at 15:00

score 1 · Answer 2 · answered Jul 29 '13 at 18:39

1

Hint:

$(n-1)n(n+1)$ is product of three consecutive numbers. So they are always divisible by $8$ and $3$ if $n$ is odd. You don't even have to bother about $3n+2$. Now think, what happens to $3n+2$ if $n$ is even.

answered Jul 29 '13 at 18:39

Kunnysan

2,050

score 1 · Answer 3 · answered Jul 29 '13 at 18:45

To prove $8|n(n^2-1)(3n+2)$ you can simply break the problem in three cases:

Case 1 $n$ odd.

Then $n^2-1=(n-1)(n+1)$ is the product of two consecutive even numbers. Thus one must be divisible by $4$ and the other by $2$.

Case 2 $n$ multiple of 4.

Then $4|n$ and $2|3n+2$.

Case 3 $n$ even but not divisible by 4.

Then $2|n$ and $n-2$ is divisible by $4$. Hence

$$4|3n+2=4n-(n-2) \,.$$

score 1 · Answer 4 · answered Jul 29 '13 at 18:49

Another approach, perhaps more elementary (but lengthier):

$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)$$

(1) Clearly, exactly one of $\,n-1\,,\,n\,,\,n+1\;$ is divisible by $\,3\,$

(2) If $\,n\,$ is odd, then $\,n-1\,,\,n+1\;$ are even with exactly one of them divisible by $\,4\,\implies (n-1)n(n+1)\,$ is divisible by $\,2\cdot 4=8\;$

(3) If $\,n\,$ is even, then also $\;3n+2\;$ is even, and exactly one of them is divisible by $\,4\,$ , since

$$n=4k+2\implies 3n+2=12k+8=4(3k+2) $$

and again $\;n(3n+2)\;$ is divisible by $\;4\cdot 2=8\;$.

The above cover all the possible cases and in each one the given expression is divisible by $\;3\;$ and by $\;8\;$ , so also by $\,3\cdot 8 =24\;$

Reasoning interesting, thank you. – benjamin_ee Jul 29 '13 at 18:52 — benjamin_ee, Jul 29 '13 at 18:52

score 0 · Answer 5 · answered Feb 18 '24 at 01:02

0

Newton's interpolation formula gives $$ n(n^2-1)(3n+2) = 48 \binom{n}{2} + 120 \binom{n}{3} + 72 \binom{n}{4} $$ which is clearly a multiple of $24$.

answered Feb 18 '24 at 01:02

lhf

216,483

$24\mid n(n^{2}-1)(3n+2)$ for all $n$ natural problems in the statement.

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