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I'm struggling with this problem:

For any natural number $n$, prove that $n^3 + 3n^2 + 2n$ is a multiple of $3$.

That $n^3 + 3n^2 + 2n$ is a multiple of $3$ means that: $n^3 + 3n^2 + 2n = 3 \times k$ where $k \in \mathbb N$ So I tried to find a the number $k$.

The best result I found was: $n^3 + 3n^2 + 2n = n(n+1)(n+2)$ But I'm lagging at the last step, to prove that it is a multiple of $3$.

(However, I got the intuition, If you see the multiples of $3$: $\{0, 3, 6, 9,\dotsc\}$ there is a difference of $3$ between them.

So $n(n+1)(n+2)$ incorporates that difference. This means that if you take a number $n$ then $n$ or $n+1$ or $n+2$ could be a multiple of $3$ and so their multiplication is a multiple of $3$) But I couldn't extend that idea into a consistent mathematical proof.

Also this problem doesn't help either: Proof that $n^3+2n$ is divisible by $3$

Community
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  • In $\mathbb{Z}_3$, n is either $0,1$ or $2$. From Euclidian algorithm $n=3q$ or $n=3q$ +1 or $n=3q +2$, with q $\in \mathbb{Z}$. Replace $n$ in the equation by each case and see the result. – ALM Oct 06 '13 at 18:38
  • http://mathforum.org/library/drmath/view/61347.html – hjpotter92 Oct 06 '13 at 18:42

4 Answers4

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Among three consecutive integers, one must be a multiple of three. Reason: if $n=3k$, we're done. If $n=3k+1$ then $n+2=3j$ is a multiple of three. If $n=3k+2$, then $n+1=3m$ is a multiple of three. In any case, $3\mid n(n+1)(n+2)$.

Pedro
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  • That's what I found but couldn't express as a consistent mathematical proof. –  Oct 06 '13 at 18:37
  • @Adobe Added something. – Pedro Oct 06 '13 at 18:37
  • I don't get your reasoning. Not every natural number can be expressed as $3k+1$ or $3k$ –  Oct 06 '13 at 18:43
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    @Adobe True. But any natural number can be expressed either as $3k,3k+1$ or $3k+2$. So we look into the three possible cases. All of them yield a positive result, so we're done. – Pedro Oct 06 '13 at 18:43
  • That was right, but I have to prove then that every natural number can be expressed either as $3k$, $3k+1$ or $3k+2$. If you have any clue that would be very helpful. Thanks. –  Oct 06 '13 at 18:47
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    @Adobe Do you know what the Euclidean algorithm is? – Pedro Oct 06 '13 at 18:52
  • Sorry I'm not aware of it. –  Oct 06 '13 at 19:02
  • @Adobe You may pick any book on elementary number theory and read about it, then. You'll find it useful. Burton's "Elementary Number Theory" comes to mind. It's essentially what you learned in high-school: for any pair of integers $a,b$ with $b\neq 0$ we can find a unique pair $c,d$ such that $$a=bc+d$$ and either $d=0$ or $0<d<|b|$. Usually $c$ is called the quotient of $a$ over $b$ and $d$ the remainder. – Pedro Oct 06 '13 at 19:08
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    Anyway, I found a little proof (not really one) without using what I haven't already learnt. So thank you so much @PedroTamaroff for your effort, that was really helpful! ;-) –  Oct 06 '13 at 19:54
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HINT: $$n^3+3n^2+2n=n^3-n+3n^3+3n\equiv \underbrace{(n-1)n(n+1)}_{\text{ product of three consecutive integers}}\pmod3$$

Among three consecutive integers, on must be divisible by $3$

lab bhattacharjee
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    $n(n+1)(n+2)$ that I found is also the product of three consecutive integers but I couldn't express the "one must be divisible by 3" mathematically. –  Oct 06 '13 at 18:38
  • @Adobe, though over-killing, we can use induction as follows: If $F(n)=(n+1)n(n-1),F(n+1)-F(n)=(n+2)(n+1)n-(n+1)n(n-1)=3(n+1)n$ Now, what is $F(0)$ or $F(1)?$ – lab bhattacharjee Oct 06 '13 at 18:42
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Yes, you're on the right track and have realized the important points. Now consider using the pigeonhole principle: It's clear that none of the numbers $n$, $n + 1$, $n + 2$ leave the same remainder when divided by $3$, and the remainders lie in $\{0, 1, 2\}$. Three remainders, three distinct numbers: One of them has to be zero.

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In response to your inductive proof link to another example, I thought you might find a similar proof useful to prove that $n^3 + 3n^2 + 2n = n(n+1)(n+2)$ is a multiple of $3$ (I will say divisible by 3, you can modify it yourself if you care).

$\textbf{Proof:}$ (by induction on $n$):

For the basis $n=1$, the result is trivial.

Assume the divisibility holds for $n=k$, that is, $$3 \, | \, k(k+1)(k+2).$$

Let $n=k+1$. Then

$$(k+1)(k+2)(k+3)=k(k+1)(k+2) + 3(k+1)(k+2).$$

Now clearly $$3 \, | \, 3(k+1)(k+2),$$ and from our inductive hypothesis, we have that $$3 \, | \, k(k+1)(k+2).$$ Hence $$3 \, | \, (k(k+1)(k+2) + 3(k+1)(k+2)).$$ Therefore $3 \, | \, n(n+1)(n+2)\, \forall \, n \in \mathbb{N}$.

J. W. Perry
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  • Eloquently explained, thanks! :-) –  Oct 06 '13 at 19:56
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    You could have made that algebra much easier in the middle by just splitting the factor of k+3 and leaving the other two factors alone. – Chris Oct 06 '13 at 22:49
  • Its not a bad way. I just got to the end and thought "oooh, that looks like a nice result. Oooh, look it factorises nicely to a really nice result! oh wait, that's what we started with..." – Chris Oct 06 '13 at 23:01
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    @Chris : On afterthought, I removed the entire middle of that algebra as it is so painfully obvious. Maybe I should not do algebra while eating omelettes. Nice call there, thanks for the read. – J. W. Perry Oct 06 '13 at 23:18
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    Don't you need to prove that 3| (+1)(+2)? – Peter Schorn Oct 29 '20 at 13:46