# Integer solutions of a cubic equation

With $\mathrm {gcd}(x,y)=1$ I have the following equation: $$x^3-xy^2+1=N$$
I want to find the integer solutions, given an N, of the variables $x$ and $y$. I have tried factoring the equation into $x(x^2-y^2)=N-1$ and then tried to link it to Pell's Equation but so far I've got nothing, I don't even understand how Pell's Equation is solved (even though I know the method, I cannot grasp the intuition or reasoning behind it), I would appreciate any help.

• there is no link to Pell or much of anything else. The reasonable procedure is to find all the divisors of $N-1,$ also the negatives of these. Then all three of $x,x-y,x+y$ must be in the list. Where did you get this problem??????? – Will Jagy Jul 20 '15 at 0:43
• @WillJagy, is there a method that does not require brute force? I was trying to find the link to Pell because the Pell Equation does not require much brute force after finding the approximation via continued fractions. I was hoping this expression had similar solutions. Is the origin of the problem important? If so I can edit the post to bring in more detail regarding its genesis. But generally I thought such details were superfluous :) – Guacho Perez Jul 20 '15 at 1:11
• I think you had better tell the whole story of this, and give some information on your own background as relates to mathematics. The one you gave has a wonderful algorithmic answer, just not closed form. Oh, the whole point of Pell is indefinite quadratic forms that cannot be factored and the solution set is infinite, you have $x(x+y)(x-y)$ and finitely many solutions. – Will Jagy Jul 20 '15 at 1:34
• @willjagy alright, but does it really have no closed form solution? – Guacho Perez Jul 21 '15 at 13:04
• Compare $xy+1=N$, which you would take to $xy=N-1$ and factor $N-1$. Is that a closed form? In this case you have three factors and need an arithmetic progression between them. – Ross Millikan Jul 24 '15 at 15:06

You could always solve for $y$:$$y=\pm\sqrt{\frac{N-1-x^3}{-x}}$$ From here, you could use brute force to find some integer $x$ value that returns integer $y$ value.
You could always do it backwards as well, solving for $x$.
Sadly, you would still have to use brute force methods. But the cube roots and such will make finding a $y$ value that will work much narrower.