Diophantine equation : $N= \frac{x^2+y}{x+y^2}$ I am looking for information about the following diophantine equation : 
$N = \displaystyle\frac{x^2+y}{x+y^2}$  
Has it been studied ?
Is there any efficient algorithm to solve it?
Any links?
I have tried to solve it by myself this week-end, but haven't made any progress ... 
Thanks in advance
Philippe
P.S:
My first post. Sorry for being unclear.
Does this equation have solutions in integers x,y for all integer N > 0 ?

Thanks for @Gerry Myerson's solution.
As far as I'm concernd, the solution could be stated as follows.
Solution
The equation could be rewritten as $$u^2-nv^2=1-n^3 \tag1$$where $u=2nb-1,v=2a-n$.
$\textbf{Case 1:}$ $n$ is not a square number.
Take $x,y$ such that $$x^2-ny^2=1.\tag2$$. This is always possible, because $x^2-ny^2=1$ is a Pell equation. Then, \begin{align*}(x+ yn^2)^2-n(y+xn)^2&=x^2+2xyn^2+y^2n^4-ny^2-2xyn^2-x^2n^3\\&=(x^2-ny^2)+n^3(ny^2-x^2)\\&=1-n^3.\end{align*}This shows that $u=x+yn^2$ and $v=y+xn$ is a solution for $(1)$. That's to say, all $a,b$ such that $2nb-1=x+yn^2$ and $2a-n=y+xn$ could satisfy $(1)$. Hence,$$a=\frac{(x+1)n+y}{2},b=\frac{x+yn^2+1}{2n}.\tag3$$
But, the problem comes out here. We need that $a,b \in \mathbb{N+}$, but the $a,b$ obtained from $(3)$ are not always like thus for all $x,y$.
So, how to show that we can find $x,y$ such that $(2)$ hold and $a,b \in \mathbb{N+}$?
