Suppose you have an equation of the form
$$ a(n^2 - m^2) + b(n-m) + c = 0 $$
With given integers $a$, $b$ and $c$.
Is there a necessary and sufficient condition that the equation has no solutions $n$ and $m$ in $\mathbb{N}$?
Are there simple sufficient conditions that it has no solutions in $\mathbb{N}$ which are helpful to show this in practice (when you have concrete numbers for $a$,$b$ and $c$?)
$$(n-m)(a(n+m)+b)+c=0$$
If $c=0$, $n=m=1$ is always a solution. Otherwise consider $c \not =0$.
If $a=0$, then the equation becomes $b(n-m)+c=0$, so there is no solution iff $b \nmid c$.
Otherwise consider $a \not =0$ as well. Let $n-m=d \mid c$, where $d$ can be negative. There are only finitely many possibilities for $d$, depending on $c$. Then $a(n+m)+b+\frac{c}{d}=0$, so $n+m=-\frac{b+\frac{c}{d}}{a}$. This gives
$$(n, m)=(\frac{-\frac{b+\frac{c}{d}}{a}+d}{2}, \frac{-\frac{b+\frac{c}{d}}{a}-d}{2})=(\frac{ad^2-bd-c}{2ad}, \frac{-ad^2-bd-c}{2ad})$$
Thus there is no solution iff the above expression for $m,n$ does not give a pair of natural numbers for all positive and negative divisors $d$ of $c$.