Given the expression
$$ x = \frac{n^2 -bn+c}{(2n+k)-b}$$
For $b$ and $c$ integer coefficients, $k$ is any integer constant, and index $-\infty <n<\infty$,
1. What is the relationship that must exist between the coefficients $b$ and $c$ so that, at any constant $k$, $x$ will be a sequence of integers generated by the index $n$?
2. As a consequence, why there is no solution for $b=c≠0$?
I suppose it is better to divide the solutions between $k=\text{even}$ and $k=\text{odd}$.
There is no possible choice of $b,c,k$ at all such that $$ x_n = \frac{n^2-bn+c}{2n+k-b} $$ is an integer for every integer $n$.
To see this, temporarily switch variable to $u=n+k/2-b/2$. Our expression now has the form $$ x(u) = \frac{u^2+pu+q}{2u} $$ for some real constants $p$ and $q$ that we don't need to find explicitly. If we allow $u$ to be real (i.e. not necessarily an integer), differentiating this gives $$ \frac{dx}{du} = \frac12 - \frac{q}{2u^2} $$ which converges to $\frac12$ when $u\to\infty$. So for $u$ large enough we always have, say, $\frac13 < \frac{dx}{du} < \frac23 $, and therefore by the mean value theorem, $\frac13 < x(u+1)-x(u) < \frac23$.
Switching back to the original variable $n$, we see that $x_{n+1}-x_n$ is exactly something of the form $x(u+1)-x(u)$. But if $x_{n+1}$ and $x_n$ are both integers, their difference cannot lie between $\frac13$ and $\frac23$.