2

The full question is: Factorise $n^3 + 3n^2 + 2n$. Hence prove that when $n$ is a positive integer, $n^3 + 3n^2 + 2n$ is always divisible by $6$.

So i factorised and got $n(n+1)(n+2)$ which i think is right? I'm not sure how to actually prove this is divisible by $6$ though. Thanks for help and i apologise if someone has already asked this, i couldn't find it. Also i've not been told whether i have to do induction or not.

Will Orrick
  • 18,220
Callum Hemsley
  • 23

5 Answers5

7

$n^3+3n^2+2n=(n+1)(n+2)n$. One of the factors must be even and one must be a multiple of three. Hence the product is a multiple of both $2$ and $3$ and hence is divisible by the least common multiple of $2$ and $3$, which turns out to be $6$.

Asinomás
  • 105,651
2

The product of three consecutive integers is divisible by $2$ since one of the factors need to be even. It is also divisible by $3$. (why?) The the product of three consecutive integers is then divisible by the least common multiple of those two numbers $2\times 3 =6$.

Alternatively, note that $$n(n+1)(n+2) = 6 {n+2 \choose 3}$$

Zain Patel
  • 16,802
1

$$ n^3+3n^2+2n = 6\binom{n+2}{3} \in 6\mathbb{Z}.$$

Jack D'Aurizio
  • 353,855
  • Whoa, i don't even know what ∈ means, but i'm guessing can be divided by 6. – Callum Hemsley Jul 13 '15 at 14:33
  • $x \in A$ means "x is a member of the set A". $6 \mathbb{Z}$ is the set of integer multiples of 6. Anyway, this is more or less just stating the conclusion, since it requires you to know that $(n+2)(n+1)n$ is divisible by $6$ in order to prove that ${n+2 \choose 3}$ is an integer in the first place. But it does give some nice generality: in general $n(n+1)\dots(n+k)$ is divisible by $(k+1)!$. – Ian Jul 13 '15 at 14:34
  • @Ian Ah okay, thanks for clarifying. – Callum Hemsley Jul 13 '15 at 14:35
  • The benefit of this solution is that it helps in proving $(n+1)(n+2)(n+3)(n+4)$ is a multiple of $24$. – Asinomás Jul 13 '15 at 14:40
  • @dREaM The same basic argument you applied in your answer can be used to prove it. At least one of $n+1,n+2,n+3,n+4$ is a multiple of $4$, at least one is even but not a multiple of $4$, at least one is divisible by $3$. – user26486 Jul 13 '15 at 14:49
  • ok, but it gets harder and harder. How would you do it for $(n)(n+1)(n+2)(n+3)(n+4)(n+5)(n+6)(n+7)(n+8)(n+9)(n+10)$? Prove it is a multiple of $11!$ – Asinomás Jul 13 '15 at 14:50
0

Keep going with your idea. $n(n+1)$ is the product of two consecutive integer so one of them is even. If $n$ is even we are done and if not $n+1$ is and the product is therefore divisible by $2$.

Similarly $n(n+1)(n+2)$ is the product of three consecutive integers so one of them is divisible by $3$. Let's test the residues modulo $3$

$$\begin{array}{c | c c c} n & n+1 & n+2 & n(n+1)(n+2)\\ \hline 0 & 1 & 2 & 0\\ 1 & 2 & 0 & 0\\ 2 & 0 & 1 & 0 \end{array}$$

So the product is also divisible by $3$ and therefore the product $n(n+1)(n+2)\equiv 0\pmod 6$

One could also test directly the congruences modulo $6$

$$\begin{array}{c|c c c c} n & n^3 & 3n^2 & 2n & n^3+3n^2+2n\\ \hline 0 & 0 & 0 & 0 & 0\\ 1 & 1 & 3 & 2 & 0\\ 2 & 2 & 0 & 4 & 0\\ 3 & 3 & 3 & 0 & 0\\ 4 & 4 & 0 & 2 & 0\\ 5 & 5 & 3 & 4 & 0 \end{array}$$

And we have proven that $n^3+3n^2+2n\equiv 0\pmod 6$

marwalix
  • 16,773
  • I think i understand, so essentially $n(n+1)$ has to have an even number, and $n(n+1)(n+2)$ must be divisible by $3$.Then it's the lowest common multiple of these two combined? I just struggle to understand the concept of actually "proving" it. – Callum Hemsley Jul 13 '15 at 14:39
  • I edited the answer to incorporate more details – marwalix Jul 13 '15 at 16:29
0

Using only modular arithmetic, without factoring, you can see that with $p(n)=n^{3}+3 n^{2}+2 n$ we have if $n\cong0(mod 2)$ then $p(n)\cong0+0+0=0 (mod 2)$ if $n\cong1(mod 2)$ then $p(n)\cong1+3+2\cong0(mod 2)$ So $2|p(n)$. Similarly if $n\cong1(mod 3)$ then $p(n)\cong1+3+2\cong0(mod 3)$, if $n\cong2(mod 3)$ then $p(n)\cong8+12+4\cong0(mod 3)$ so $3|p(n)$. Since both 2 and 3 divide $p(n)$ then $p(n)$ is a multiple of 6.

John McGee
  • 1,223