# Special Diophantine Equation, Special Pell's Equation

I'm dealing with special kind of "Diophantine Equation". I want to generate all the solutions for

$x^2 - aby^2 = a^2 - ab$

given $a,b \in \mathbb{N}$, $a > b$, $a$ and $b$ are co-primes

I can easily generate a solution for this problem by putting $x = a, y = 1$

Since, $gcd(a,1)=1$, this is also a "primitive solution"

Now, I want to generate all the solutions with $x,y \in \mathbb{N}$. How to generate other solutions?

It'll be helpful if someone can tell me via recurrence. something like

$X_{n+1} = p.X_n + q.Y_n + k$

$Y_{n+1} = r.X_n + s.Y_n + l$

Edit: Thanks to André Nicolas I know how to generate all solution from the primitive solution.

@MISC {228390, TITLE = {How to find solutions of $x^2-3y^2=-2$?}, AUTHOR = {André Nicolas (https://math.stackexchange.com/users/6312/andr%c3%a9-nicolas)}, HOWPUBLISHED = {Mathematics Stack Exchange}, NOTE = {URL:https://math.stackexchange.com/q/228390 (version: 2012-11-03)}, EPRINT = {https://math.stackexchange.com/q/228390}, URL = {https://math.stackexchange.com/q/228390} }

But, I'm still missing on the "primitive solutions". Apparently, sometimes there are more than 1 primitive solutions. But I don't know how to find other "primitive solutions"?

Edit2:

I actually found a way to solve these type of equations. http://www.jpr2718.org/FundSoln.pdf

But it requires finding $z^2 \equiv a.b \: (\mathbb{mod} \, a(a-b))$ (Quadratic residue). Which I don't know how to compute efficiently. (Tonelli and other algorithms only works for prime).

• math.stackexchange.com/questions/228356/… Commented Sep 6, 2017 at 12:44
• $$p^2-abs^2=1$$ $$x=a(p^2-2bps+abs^2)$$ $$y=p^2-2aps+abs^2$$ Commented Sep 6, 2017 at 15:09

My thoughts

Rearranging $x^2 - aby^2 = a^2 - ab$ gives $$x^2 - a^2 = ab \left( y^2 - 1 \right)$$

$a$ must be a divisor of either $x+a$ or $x-a$, therefore $a$ must be a divisor of $x$. Therefore let $x=na$, giving $$a(n^2-1)=b(y^2-1)$$

$a$ and $b$ only need compensate for prime factors not common to both $(n^2-1)$ and $(y^2-1)$, which means co-prime $a$ and $b$ can be chosen for any values of $n$ and $y$ with $y>n$.

• since, $a$ and $b$ are co-primes we can conclude $n^2-1 = kb$ and $y^2-1=ka$ for some $k \in \mathbb{N}$ . I tried over iterating n for finding k. But i'm getting timeout on large a,b. Is there any other method? Commented Sep 7, 2017 at 1:29
• Choose $y$. There will be a solution for every positive $n$, $(n< y)$. For a particular $n$, $k=gcd(y^2-1,n^2-1)$. Once $k$ is determined $a$, $b$ and $x$ can be found. Commented Sep 7, 2017 at 15:22
• $a$ and $b$ is given, we need to find $n$ and $y$. I understood the gcd part. Can you please explain how to generate the solutions? Commented Sep 7, 2017 at 17:52
• Sorry I missed the fact that $a$ and $b$ are given. If you calculate $\frac{y^2-n^2}{a-b}-\frac{y^2+n^2-2}{a+b}$ on a spreadsheet by varying $y$ and $n$ (e.g. with $a=12$ $b=5$) you get a surface that passes though the zero plane. I don't think there is a quick way to locate integer co-ordinates that give zero, along the line of intersection with the zero plane (that is at $y=17$, $n=11$) Commented Sep 7, 2017 at 20:30
• I actually found a way to solve these type of equations, but it requires finding $z^2 \equiv a.b \: (\mathbb{mod} \, a(a-b))$ (Quadratic residue). Which I don't know how to compute efficiently. jpr2718.org/FundSoln.pdf Commented Sep 7, 2017 at 21:44