# Recurrence relation between solutions of a quadratic Diophantine equation

I have the fundamental solution of the following Diophantine equation: $$\frac{x(x-1)}{y(y-1)}=\frac{m}{n} \hspace{5 mm}, \hspace{5 mm} m \le n$$ $$nx^2-my^2-nx+my=0$$ Is it possible to derive a recurrence relation using which we can get the complete set of solution?

For example, if $m=1$ and $n=2$, then initial solution is $(x,y)=(3,4)$ and we can get the next solution by the linear transformation $(x,y) \rightarrow (3x+2y-2,4x+3y-3)$.

Can it be generalised for all $(m,n)$?

My approach:

Let the transform be $(x,y) \rightarrow (ax+by+c,px+qy+r)$, then substituting these in original equation and after some simplification,

$$(na^2-mp^2)x^2-(mq^2-nb^2)y^2-(an-pm+2prm-2acn)x+(2bcn-bn-2qrm+qm)y+(2abn-2pqm)xy+(nc^2-nc-mr^2-mr)=0$$ Comparing with original equation we have to solve following set of equations: $$na^2-mp^2=n$$ $$mq^2-nb^2=m$$ $$an-pm+2prm-2acn=n$$ $$2bcn-bn-2qrm+qm=m$$ $$2abn-2pqm=0$$ $$nc^2-nc-mr^2-mr=0$$ Is there any method using I can solve these equations efficiently to get $(a,b,c)$ and $(p,q,r)$?

$UPDATE:$

Using the method http://www.alpertron.com.ar/METHODS.HTM, I have to solve $r^2-mns^2=1$ and find the fundamental solutions.

Let $X_n$ and $Y_n$ be the initial solution, then next solutions will be:

$$X_{n+1}=(r^2+mns^2)X_n+2mrsY_n-(mns^2+mrs)$$ $$Y_{n+1}=2nrsX_n+(r^2+mns^2)Y_n-(mns^2+nrs)$$

For $m=1$ and $n=2$, $r=3$ and $s=2$, $$X_{n+1}=17X_n+12Y_n-14$$ $$Y_{n+1}=24X_n+17Y_n-20$$

Which gives the solution $(x,y):(3,4) \rightarrow (85,120)$ instead of $(x,y):(3,4) \rightarrow (15,21)$. This recurrence does not give complete set of solutions.

• I think your best bet is to punch this system into maple or sage or some other piece of software and see if you can get an answer – Jeff Strom Sep 1 '15 at 4:57
• The answer is positive. You can see the method here or you can find the solution here – GAVD Sep 1 '15 at 5:01
• I am trying to solve hackerrank.com/contests/projecteuler/challenges/euler100 I think it should be solvable using standard programming languages like C++ / JAVA / Python – guest123456 Sep 1 '15 at 5:01
• Thanks @GAVD for the link :) . The method is so much complicated. Anyway I should be able to implement the idea. It will require quite some time. – guest123456 Sep 1 '15 at 5:07
• It is not clear. What else could be the issue? You also clearly shown that such an equation always can be associated with the Pell equation. math.stackexchange.com/questions/1414779/… Solutions of Pell equations standard procedure requiring the continued fraction decomposition. To abandon Pell equations fail. – individ Sep 1 '15 at 5:35

The condition can also be written as $$\iff n(x-\tfrac12)^2-m(y-\tfrac12)^2=\frac{m+n}4$$ $$\iff (2nx-n)^2-mn(2y-1)^2=n(m+n)$$ So with the substitution $u=n(2x-1), v = 2y-1$, we are looking for solutions to the Pell type equation $u^2-(mn)v^2=n(m+n)$, which allows recursive solutions (if you have any particular solution) depending on the value of $mn$.