Diophantine equation : $N= \frac{x^2+y}{x+y^2}$

I am seeking details concerning the adhering to diophantine formula:

$N = \displaystyle\frac{x^2+y}{x+y^2}$

Has it been researched? Exists any kind of reliable algorithm to address it?¢ Any web links?

I have actually attempted to address it by myself today - end, yet have not made any kind of progression

Thanks in advance¢ Philippe

P.S: ¢ My first blog post. Sorry for being unclear.¢ Does this formula have remedies in integers x, y for all integer N > 0?

2022-06-07 14:31:41
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Answers: 1

Follow user9325 is pointer concerning finishing the squares, and afterwards (seek out and also) use the concept of Pell formulas.

Edit: OK, I presume you really did not get anything out of user9325 is pointer, 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 remedy $U=-1$, $V=\pm N$. The remedy $U=-1$, $V=N$ represents $x=N$, $y=0$, which currently reveals that there is a remedy for each and every $N$, yet possibly that is also unimportant. After that take any kind of remedy to $a^2-Nb^2=1$ and also you get an additional remedy, $U=-a\pm bN^2$, $V=-b\pm aN$. Currently $a^2-Nb^2=1$ has great deals of remedies - that is the Pellian I mentioned. For $y$ to be an integer, you require $a\equiv 1\pmod N$, so you need to research sufficient of the concept to see if that can be made to take place.

If $N$ is a square, claim, $N=m^2$, after that the Pellian does not use, yet you have something less complex ; $(U+mV)(U-mV)=1-m^6$. Currently you'll access the majority of finitely several remedies, given that there are just finitely several means to variable $1-m^6$. Below is one instance ; take $N=4=2^2$ so $m=2$ and also $1-m^6=-63$ ; take $U+2V=63$, $U-2V=-1$ to get $U=31$, $V=16$ ; after that $x=10$, $y=4$.

2022-06-07 14:53:14