A natural number equation For what values of $n$ the equation $x^2 - (2n+1)xy + y^2 + x = 0$ has no solution in natural numbers ? (for $n=1$ it has a trivial solution). 
 A: Letting $m=2n+1$ just for brevity, we see that:
$$(x-y)^2 = x((m-2)y-1)\\
(x+y)^2 = x((m+2)y-1)$$
Now, note that if $p|x$, then $p|x+y$ so $p|y$ and therefore $p\not\mid (m\pm2)y-1$.
Therefore, $x,(m-2)y-1$ are relatively prime, so must be perfect squares. Similarly, $(m+2)y-1$ must be a perfect square.
In the original equation, we can substitute $x=x_0^2$ and $y=x_0y_0$ (necessarily, $x_0|y$) and dividing out the common factors to get:
Substituting, this means we need to solve:
$$x_0^2 -mx_0y_0 + y_0^2 + 1 = 0$$
This has an integer solution if and only if $D = (my_0)^2 - 4(y_0^2+1) = (m^2-4)y_0^2 - 4$ is a perfect square, which means if and only if we have  solution to:
$$U^2 - (m^2-4)V^2 = -4$$
Mike Bennett below in comments asserts this does not have a solution except when $m=\pm 3$, which coincides with $n=-2,1$.
This is not hard to reduce this to the equation: $$W^2-(m^2-4)Z^2=-1$$
The continued fraction expansion of $\sqrt{m^2-4}$, when $m>3$ is odd, can be written as:
$$\sqrt{m^2-4} = \left[m-1,\overline{1,\frac{m-3}{2},2,\frac{m-3}{2},1,2(m-1)},\dots\right]$$
Since the length of the repetition is even, there is no solution to $W^2-(m^2-4)Z^2=-1$.
