Prove that $(n^2-1)\mid(n^3+1)$ iff $n=2$

Question

Seperating $n^2-1$ into $(n+1)(n-1)$.

I have noticed that $n^3+1=(n+1)(n^2-n+1)$, so we have $\forall n\geq 2$, $(n+1)\mid(n^3+1)$.

We now need to show that $(n-1)\mid(n^2-n+1)$ iff $n=2$

This breaks down to showing that $(n-1)\not\mid(n^2-n+1)$ for $n\geq 3$

If $n$ is odd then $(n-1)$ will be even. $n^2$ will be odd, n will be odd and since odd minus an odd is even, $n^2-n$ will be even, so $n^2-n+1$ will be odd and an even cannot divide an odd, so statement does not hold for any odd value for $n$.

Assume $n$ is even, then $n=2k$ for $k\in\mathbb{N}$

We must now show $(2k-1)\not\mid((2k)^2-2k+1)\Leftrightarrow(2k-1)\not\mid(4k^2-2k+1),\forall k\in\mathbb{N},k\geq 2.$

This is the part I am stuck on, how should I proceed? Am I going down the right lines?

Thanks in advance for the help.

Answer 1

We have $$ 4k^2 - 2k + 1 = 2k(2k-1) + 1 $$ This cannot be a multiple of $2k-1$ for any $k \geq 2$.

Answer 2

We know that $(n-1)|(n^2-n)$. If also $(n-1)|(n^2-n+1)$, then $n-1$ divides two consecutive integers, which is only possible if $n-1=1$.

Answer 3

Hint $\,\ n^2\!-\!1\mid \overbrace{n(n^2\!-\!1)\!+\!n\!+\!1}^{\Large n^3+1}\iff n^2\!-\!1\mid n\!+\!1\overset{{\rm cancel}\ n+1}\iff n\!-\!1\mid 1\ $ if $\,\ n\neq -1$

Remark $\ $ i.e. $\,{\rm mod}\ n^2\!-\!1\!:\,\ \color{#c00}{n^2\equiv 1}\,\Rightarrow\, n^3 \equiv n(\color{#c00}{n^2})\equiv n(\color{#c00}1)$

Answer 4

You can also proceed using polynomial long division: $$ n^2-n+1=(n-1)n+1$$ Dividing both sides by $n+1$ you obtain $$\frac{n^2-n+1}{n-1}=n+\frac{1}{n-1}$$

Since the left side of the equation is an integer, then so must be the left one. Therefore $\frac{1}{n-1}$ must be an integer, because $n$ is an integer as well. Hence the only possible value is $n-1=1 \implies n=2$