Prove that $\dfrac{nm^2-n+1}{2mn-2n+1} \notin\mathbb{Z}$ when $m \geq 2$ Given: $n$ and $m$ are positive integers with
$m≥2$, show that:
$$\frac {nm²-n+1}{2mn - 2n+1}\notin \mathbb Z$$
My attempt:
$n, m \in \mathbb{Z}$
$nm^2 \in \mathbb{Z}$
$nm^2 - n \in \mathbb{Z}$
$2mn - 2n \in \mathbb{Z}$
So $\dfrac{nm^2 - n }{2mn - 2n+1}$ may or may not $\in \mathbb{Z}$
But $\dfrac{1}{2mn - 2n + 1} \in \mathbb{Z}$ only when
$$
\begin{aligned}
2mn - 2n +1&= 1 \\
2mn&= 2n \\
m&= 1
\end{aligned}
$$
Given $m \geq 2$, $\dfrac{1}{2mn - 2n + c} \notin \mathbb{Z}$
So $\dfrac{nm^2 - n + 1}{2mn - 2n+1} \notin \mathbb{Z}$
 A: If $\frac{2nm^2-2n+2}{2mn-2n+1}$ is not an integer, then the original fraction $\frac{nm^2-n+1}{2mn-2n+1}$ cannot be an integer either. So we show that $\frac{2nm^2-2n+2}{2mn-2n+1}$ is not an integer.
\begin{align*}
\frac{2nm^2-2n+2}{2mn-2n+1} &= m + \frac{2mn-2n-m+2}{2mn-2n+1} \\[6pt]
&=m + 1 + \frac{1-m}{2mn-2n+1} \\[6pt] &= m+1 - \frac{m-1}{2n(m-1)+1}.
\end{align*}
As $m+1$ is an integer, it is sufficient to show that the last term $\frac{m-1}{2n(m-1)+1}$ is not an integer.
Indeed, since $n>0$ and $m\geq 2$, we have
$$ 0<\frac{m-1}{2n(m-1)+1} \leq \frac{m-1}{2(m-1)+1}< \frac{m-1}{2(m-1)}=\frac{1}{2}$$
so $\frac{m-1}{2n(m-1)+1}\in\left(0,\frac{1}{2}\right)$ and in particular not an integer, as desired.
A: Suppose for contradiction that $$\frac{nm^2-n+1}{2nm-2n+1}=\frac{n(m-1)(m+1)+1}{2n(m-1)+1}$$ is some integer — say, $k+1$.  Then $$n(m-1)(m+1)+1=n(m-1)\cdot2(k+1)+k+1$$  Rearranging, $$n(m-1)(m-2k-1)=k$$  To make things pretty, I will let $l=m-1$; then $$nl(l-2k)=k\tag{1}$$
Set $a=l-2k$; then $k=nla$.  Rewriting (1) in terms of $a$, $$a=l-2k=l-2nla=l(1-2na)$$  Likewise, let $b=1-2na$; then $a=lb$, so that $$b=1-2na=1-2nlb$$  But now we can rearrange and notice that $$(1+2nl)b=1$$  Since this is an equation in integers, there are only two solutions: $$(b,1+2nl)\in\{(1,1),(-1,-1)\}$$  Since $nl>0$, neither case is possible, our desired contradiction.
