Divisibility Problem: Show That n=3

[geoffrey159]

Homework Statement

Show that the only $n \in \mathbb{N}-\{0,1\}$ such that $2n-1|(3n^2-3n+1)(3n^2-3n+2)$ is 3.

Homework Equations

$P_n = (3n^2-3n+1)(3n^2-3n+2)$
Addition and multiplication in $\mathbb{Z}/(2n-1)\mathbb{Z}$

The Attempt at a Solution

Hello, I'm not 100% sure my proof is correct for this problem, can you check it please ? Do you see a simpler way of doing it ?

I want to show that in $\mathbb{Z}/(2n-1)\mathbb{Z}$, $\hat {P_n} = \hat 0 \Rightarrow n = 3$ .

First, I rewrite $P_n = ( (2n-1)^2 + n(1-n) ) ( (2n-1)^2 + n(1-n) + 1 )$.

Using the fact that $\hat 0 = \widehat {2n-1}$ and $\hat n = \hat 1 - \hat n$, I get that $\hat {P_n} = \hat {n^2} ( \hat {n^2} + \hat 1)$

Now, assuming that $\hat {P_n} = \hat 0$, I get that
$\hat 0 = \hat{4} \hat {P_n} = \widehat{4n^4} + \widehat{4n^2} = \widehat{ (2n^2)^2 } + \widehat { (2n)^2 } = \widehat{ (2n)^2 } .\widehat{n^2} + \hat 1 . \hat 1 = \hat{n^2} + \hat 1$

Multiplying by $\hat{2}$ left and right :
$\hat 0 = \widehat{2n}.\hat{n} + \hat 2 = \hat n + \hat 2 = \hat 1 - \hat n + \hat 2 = \widehat{3 -n}$

So $2n-1 | 3 - n$. It is always true that $2n-1 | 0$ so $n = 3$ is a solution. It is the only possibility because $2n-1\ge |3 - n|$ so it does not divide $3 - n$ for any other $n\in \mathbb{N}-\{0,1\}$

 

[mfb]

Looks right.

 

[geoffrey159]

Thank you !

 

[geoffrey159]

Do you know a simpler way of solving this problem ?

 

[mfb]

I don't see one.