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 ?
 A: Follow user9325's suggestion about completing the squares, and then (look up and) apply the theory of Pell equations. 
Edit: OK, I guess you didn't get anything out of user9325's suggestion, so I'll take it up for you. 
$N=(x^2+y)/(x+y^2)$, $Ny^2-y=x^2-Nx$, $U^2-NV^2=1-N^3$ where $U=2Ny-1$, $V=2x-N$. This has the solution $U=-1$, $V=\pm N$. The solution $U=-1$, $V=N$ corresponds to $x=N$, $y=0$, which already shows that there's a solution for each $N$, but maybe that's too trivial. Then take any solution to $a^2-Nb^2=1$ and you get another solution, $U=-a\pm bN^2$, $V=-b\pm aN$. Now $a^2-Nb^2=1$ has lots of solutions - that's the Pellian I alluded to. For $y$ to be an integer, you need $a\equiv 1\pmod N$, so you have to study enough of the theory to see if that can be made to happen. 
If $N$ is a square, say, $N=m^2$, then the Pellian doesn't apply, but you have something simpler; $(U+mV)(U-mV)=1-m^6$. Now you'll get at most finitely many solutions, since there are only finitely many ways to factor $1-m^6$. Here's one example; take $N=4=2^2$ so $m=2$ and $1-m^6=-63$; take $U+2V=63$, $U-2V=-1$ to get $U=31$, $V=16$; then $x=10$, $y=4$. 
A: 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 $a,b \in \mathbb{N+}$?
