# Finding the number of Integer Solutions

I was hoping someone could give me some insight on a question I had for personal research. I have an equation: $C = 6xy + x + y$ where $x,y$ are positive integer variables, and where $C$ is a predefined integer constant (ex: $24 = 6xy + x + y$). Because $x\text{ and }y$ must be integers, this can be classified as a Diophantine Equation, however doesn't quite fit as a linear type equation or any other general type format that I can find for Diophantine Equations.

However, I am not truly interested in the actual solutions to the equation, just how many exist. For instance I know $x = 0, y = C\text{ and }x = C, y = 0$ is a solution for all values of $C$, but for some values of $C$, such as 17, there are no other solutions and other values of $C$, such as 15, $x = 1, y = 2$ is also a solution, giving me 3 total solutions.

So what is the quickest way in determining how many solutions said equation has, or at least verify that the equation has more than just $x = 0, y = C\text{ and }x = C, y = 0$ as solutions?

Thanks, DevenJ

-
$(6x+1)(6y+1)=(6C+1)$ –  lab bhattacharjee Nov 29 '12 at 17:56
@user50237: For truly huge numbers, the computations could be difficult, for factoring can be hard to do. –  André Nicolas Nov 29 '12 at 18:00
you don't have to factor $6C+1$ just try $6x+1$ and $6y+1$ between $\pm \sqrt{6C+1}$. That's $O(n)$.. can we beat that? –  user50336 Nov 29 '12 at 22:06
@cassandrao, what's $n$? I don't see any $n$ in the question. And isn't trying every $6x+1$ up to the square root just a version of factoring by trial division? –  Gerry Myerson Nov 29 '12 at 23:08