Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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've been asked to show that:

$n(n+1)(2n+1) \equiv 0 \pmod 6$

I found in a previous question that:

$n(n+1)$ was divisible by $2$ and resulted in an even number e.g

$n(n+1) \equiv 0 \pmod 2$

so I figured I needed to find:

$(2n+1) \equiv 0 \pmod 3$ in order to complete

$n(n+1)(2n+1) \equiv 0 \pmod 6$

but I am unsure on how to find $(2n+1) \equiv 0 \pmod 3$

Is this the right way to find the mod 6 and if so could you tell me how I could find

$(2n+1) \equiv 0 \pmod 3?$

share|cite|improve this question

A rather unconventional way to solve this is by using the identity $$\sum^n_{k=0}k^2=\frac{n(n+1)(2n+1)}{6}$$ Since $\dfrac{n(n+1)(2n+1)}{6}$ is a sum of integers, it must be an integer as well. Therefore $n(n+1)(2n+1)$ is divisible by $6$.

share|cite|improve this answer
It's not unconventional but natural if one knows telescopy - see my answer. – Bill Dubuque Jul 27 '14 at 4:15
If needed, here is link to a post about the sum of squares formula used in your post:… – Martin Sleziak Nov 23 '14 at 11:08

${\rm mod}\ 6\!:\ f(n) = n(n\!+\!1)(2n\!+\!1)\,$ is constant by $\,f(n)-f(n\!-\!1) = 6n^2\equiv 0,\,$ so $\,f(n)\equiv f(0)\equiv 0$.

Remark $\ $ Summing the above difference, using telescopy, we obtain $\,f(n) = 6\sum_{k=1}^n k^2\equiv 0$

share|cite|improve this answer

The numbers $2n$, $2n+1$, and $2n+2$ are three consecutive numbers, so one of them is divisible by $3$. If $2n+1$ is not, then one of $2n$ or $2n+2$ is. But if $2n$ is divisible by $3$, so is $n$. And if $2n+2$ is divisible by $3$, so is $n+1$. Thus one of $n$, $2n+1$, and $n+1$ is divisible by $3$.

share|cite|improve this answer

$$n(n+1)(2n+1)=n(n+1)(2n-2+3)=2\underbrace{(n-1)n(n+1)}_{3\text{ consecutive integers}}+3\underbrace{n(n+1)}_{2\text{ consecutive integers}}$$

Reference :

  1. The product of n consecutive integers is divisible by n factorial

  2. The product of n consecutive integers is divisible by n! (without using the properties of binomial coefficients)

share|cite|improve this answer

It is not necessary to have $\displaystyle 2n+1\equiv0\pmod3\iff2n\equiv-1\equiv2$

$\displaystyle\iff n\equiv1\pmod3$ as $(2,3)=1$

As for $\displaystyle n\not\equiv1,n\equiv0$ or $-1\pmod3$

In either case, $\displaystyle3|n(n+1)\implies 3|n(n+1)(2n+1)$

Alternatively, for any integer $\displaystyle n, n\equiv-1,0$ or $1\pmod3$

For the first two cases, $\displaystyle3|n(n+1)\implies 3|n(n+1)(2n+1)$

If $n\equiv1\pmod3,2n+1\equiv0$

share|cite|improve this answer

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.