Positive Integer points of $f(x)=\frac{1}{c-\frac{1}{x}}$, where c is fixed So I am looking for the integer solutions of $f(x)=\frac{1}{c-\frac{1}{x}}$  for fixed $c\in \mathbb{Q}$ i.e. points $(x,f(x))\in \mathbb{N}\times \mathbb{N}$. (The c equals $\frac{4}{n}-\frac{1}{k}$ where $n,k\in \mathbb{N}$).
1)Are there any methods for computing that. Or at least getting some upper and lower bounds? Or at least references for methods for any curve and then I can look more deeply?
2)I know I can probably find that using algebraic number theory but I am really curious of any geometric/analysis methods out there.
3)Does that equation relate to Diophantine equations? Are there any ways of relating general curves to Diophantine? Like how elliptic curves are analyzed with Diophantine equations.
Thank you so much
 A: I'm not sure about the deeper questions you ask, but here is an elementary observation.
You are looking for positive integer solutions solutions $(x,y)$ for the curve
$$y = \frac{1}{c-1/x}$$
given a rational number $c$.
Just rewrite this as:
$$\frac{1}{x} + \frac{1}{y} = c$$
and the question becomes much more transparent.  You will have one or two solutions whenever $c$ can be written as $c = 1/x + 1/y = (x+y)/xy$; otherwise there will be no solutions.
More interesting is the question as whether $c$ of the form $c = 4/n - 1/k$ for positive integers $k$ and $n$ can satisfy the above condition.  We can rewrite this as integer solutions to
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{k} = \frac{4}{n}.$$
A few obvious solutions come to mind: $x = y = k = 3$ and $n = 1$, or $x = y = 2$, $k = n = 1$ and its obvious permutations.  This is a homogenous equation so integer multiples of those solutions work as well.  In fact, we can choose any positive rational numbers $x,y,k$ and then solve for $n$; then we can clear denominators to get integer solutions.  So there are lots of solutions.
To attempt to address some of your other questions, we can rewrite this equation as a homogenous cubic
$$kxn + kyn + xyn - 4xyk = 0$$
and (aside from your restictions on the sign of the solutions), it suffices to look for rational solutions $[x,y,z,k] \in \mathbb{P}^3(\mathbb{Q})$ because we can always clear denominator to get integral solutions.  We can let $n = -4z$ and get a more symmetric equation,
$$kxz + kyz + xyz + xyk = 0$$
and now we are looking for rational solutions to this elementary symmetric cubic polynomial.  This famous cubic surface is known as Cayley's nodal cubic surface.  It has four singular points: $[1,0,0,0], [0,1,0,0], [0,0,1,0], [0,0,0,1]$.
