2n^3+3n^2+n\equiv 0 (mod 6)

**dwsmith** · July 8th 2010, 03:35 AM

$2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

$6|(2n^3+3n^2+n)\rightarrow 6k=2n^3+3n^2+n$

???

**aman_cc** · July 8th 2010, 04:45 AM

Hint: $2n^3+3n^2+n = n(n+1)(2n+1)$

This is a rather easy exercise - using either induction or direct proof (unless I am missing something in the question)

**Soroban** · July 8th 2010, 07:30 AM

Hello, dwsmith!

I have a back-door approach to this one . . .

$2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

As aman-cc pointed out: . $2n^3 +3n^2 + n \:=\:n(n+1)(2n+1)$

I recognized this is as part of the formula for the sum of consecutive squares:

. . $1^2 + 2^2 + 3^2 + \hdots + n^2 \:=\:\dfrac{n(n+1)(2n+1)}{6}$

Since the left side is an integer, the right side must be an integer.
. . That is, $n(n+1)(2n+1)$ is a multiple of 6.

Therefore: . $2n^3 + 3n^2 + n \:\equiv\:0\text{ (mod 6)}$ .for all integer values of $n.$

**simplependulum** · July 8th 2010, 07:14 PM

Consider $2n^3 + 3n^2 + n = 2(n^3 - n ) + 3n(n+1) = 2(n-1)(n)(n+1) + 3n(n+1)$ , use the property of product of few consecutive integers .

**melese** · July 8th 2010, 11:44 PM

$2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

This a solution I saw once:
$2n^3+3n^2+n=n(2n^2+3n+1)=n(n+1)(2n+1)$ . One of $n$ and $n+1$ is divisible by $2$ .

$2\equiv -1(mod\ 3)$ ;
$n(n+1)(2n+1)\equiv -(n-1)n(n+1)(mod\ 3)$ . Then at least one of $n-1$ , $n$ , and $n+1$ is divisible by $3$ .

Since $2$ and $3$ are relatively prime $2n^3+3n^2+n$ is divisible by $6$ .

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