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

• Jul 8th 2010, 03:35 AM
dwsmith
2n^3+3n^2+n\equiv 0 (mod 6)
$\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

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

???
• Jul 8th 2010, 04:45 AM
aman_cc
Hint: $\displaystyle 2n^3+3n^2+n = n(n+1)(2n+1)$

This is a rather easy exercise - using either induction or direct proof (unless I am missing something in the question)
• Jul 8th 2010, 07:30 AM
Soroban
Hello, dwsmith!

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

Quote:

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

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

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

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

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

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

• Jul 8th 2010, 07:14 PM
simplependulum
Consider $\displaystyle 2n^3 + 3n^2 + n = 2(n^3 - n ) + 3n(n+1) = 2(n-1)(n)(n+1) + 3n(n+1)$ , use the property of product of few consecutive integers .
• Jul 8th 2010, 11:44 PM
melese
$\displaystyle 2n^3+3n^2+n\equiv 0 \ \mbox{(mod 6)}$

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

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

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