Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.

I have managed to solve the base case, which gives 9, which is a multiple of 3.

From here on,

I have $(n+1)((n+1)^2 + 8)$

$n^3 + 3n^2 + 11n + 9$

$n(n^2 + 8) + 3n^2 + 3n + 9$

How can I show that $3n^2 + 3n + 9$ is a multiple of 3?

share|cite|improve this question
"How can I show that $3n^2 + 3n + 9 $ is a multiple of 3?". Come on. Think about it... – Ragib Zaman May 27 '12 at 17:08
@Ragib: I can prove that by induction. – TonyK May 27 '12 at 17:26
Owow, seven answers. Setting a new Guinness World Record. – Gigili May 27 '12 at 17:30
@Gigili haha for what? – AkshaiShah May 27 '12 at 17:32
@akshai5050: Seven answers for a not-so-difficult question. I assume they're all saying the same thing! – Gigili May 27 '12 at 17:35
up vote 4 down vote accepted

You've already solved the case $n=1$, so I'll not repeat that there.

Assuming as the induction hypothesis that $n$ has the property that $3|n(n^2+8)$, we can rewrite $(n+1)((n+1)^2+8)$ to obtain $$ \begin{split} (n+1)((n+1)^2+8) & = (n+1)(n^2+2n+1+8)\\ & =n(n^2+8)+n(2n+1)+((n+1)^2+8)\\ & = n(n^2+8)+3n^2+3n+9 \end{split} $$ In the latter, all the terms are divisible by $3$, hence it follows that $(n+1)((n+1)^2+8)$ is also divisible by $3$. This finishes the induction proof, so we may conclude that $n(n^2+8)$ is divisible by $3$ for all $n\geq 1$.

share|cite|improve this answer

Without induction:

Since $8=-1\pmod{3}$, $n(n^2+8)=n(n^2-1)=(n-1)n(n+1)\pmod{3}$. Since $n-1$, $n$ and $n+1$ are three consecutive integers, at least (and in fact, exactly) one of them is a multiple of $3$, hence their product is a multiple of $3$.

share|cite|improve this answer
I have to use induction – AkshaiShah May 27 '12 at 16:58
Any idea why? This only seems to make things more complicated... – Did May 27 '12 at 17:17
It's a past exam paper – AkshaiShah May 27 '12 at 17:22

Hint $\rm\ n\:\!(n^2\!-\!1+9) = (n\!-\!1)\:\!n\:\!(n\!+\!1) + 9\:\!n\:$ so it suffices to show $3$ divides one of $\rm\:\!n\!-\!1, n, n\!+\!1. $

The base case $\rm\:n=1\:$ is true since $3$ divides $\rm\:n\!-\!1 = 0.\:$ For the induction step notice that

$\quad 3$ divides one of $\rm\: n\!-\!1, n, n\!+\!1\:\Rightarrow\: 3$ divides one of $\rm \:n,n\!+\!1,n\!+\!2\:\ $ by $\rm\:\ n\!+\!2\:\! =\:\! n\!-\!1 + 3$

For more general methods see my many posts on telescopy.

share|cite|improve this answer

You start by supposing that $n(n^2+8)$ is a multiple of 3, and you need to show that $(n+1)((n+1)^2 + 8)$ is also a multiple of 3. I would start by simplifying the latter expression.

Then manipulate the expression until it is a sum of things that are multiples of 3. This includes expressions like $3a$, which is always a multiple of 3 for any $a$, and $n(n^2+8)$, by the induction hypothesis.

share|cite|improve this answer

If $n\equiv 0\pmod 3$ Ok. If $n\equiv 1\pmod 3$, we have \begin{equation} n^{2} + 8 \equiv 1^{2} + 2\equiv 0\pmod 3. \end{equation} If $n \equiv 2\pmod 3$ we have \begin{equation} n^{2} + 8 \equiv 2^{2} + 2\equiv 6 \equiv 0\pmod 3. \end{equation}

share|cite|improve this answer

Using induction it is obvious that the statement is true for $n=1$. Now suppose that it is true for $n=k$, then we have $k(k^2+8)=3m$, where m is an integer. Considering the case where $n=k+1$, we have; $$(k+1)[(k+1)^2+8]=k(k^2+2k+9)+k^2+2k+9$$ $$=k(k^2+8+2k+1)+k^2+2k+9=k(k^2+8)+k(2k+1)+k^2+2k+9$$ $$=k(k^2+8)+3k^2+3k+9=3m+3k^2+3k+9=3(m+k^2+k+3)$$

share|cite|improve this answer

Since you have proven that this formula is available for $n=1$ we suppose that formula is available for all natural numbers n=k $$k(k^2+8)$$ then according to axiomme of mathematical induction we need to prove that formula is valid for $n=k+1$ or

$$(k +1)((k+1)^2+8)$$ is multiple of 3

now we have

$$(k+1)((k+1)^2+8)=(k+1)(k^2+2k+1+8)=k^3+3k^2+11k+9=(k^3+8k)+(3k^2+3k+9)=k(k^2+8)+3(k^2+k+3)$$ first part of expression is factor of 3 by assumption an second part evidently is factor of 3

share|cite|improve this answer

If $(n+1)((n+1)^2+8)=(n+1)(n^2+2n+9)$ then if $(n+1)= 0\ mod\ 3$, we're done. If not, than $(n+1)=1\mod 3$ or $(n+1)= 2\mod 3$. So if $n+1=1\mod 3 $ then $n=0\mod 3$ so $3|(n^2+2n+9)$, and if $n+1=2\mod 3$ then $n^2=n=1\mod 3$ now let $n^2=3k+1$ and $n=3l+1$ so we have $(n^2+2n+9)=(3k+1+6l+2+9)=3(k+2l+1+3)$.

share|cite|improve this answer

Quote: How can I show that $3n^2 + 3n + 9$ is a multiple of $3$? End of quote

I'm surprised at the complexity of some of the answers.

$$ 3n^2 + 3n + 9 = 3(n^2 + n + 3) = (3\cdot\text{something}). $$

A "multiple of $3$" is anything that is $3$ times something (where "something" means of course an integer).

share|cite|improve this answer
Some of the answers were before I edited the question with that part. Thanks though – AkshaiShah May 27 '12 at 18:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.