I'm studying a family of rational curves: (x^2-ax+b)/(S+x)-y=0, where a is 0 or 1, b and S are large integers for which factorization would not be feasible, the problem consist in finding the lattice points (x,y) on the curves.
I arrived to an expression in the form x1^2-y1^2=N,
x1 = x+a+2S
y1 = a+n-4 (b-n*S)
N = 4 (S^2+a*S+b)
Such expression can be solved in Z for x1, y1 easily
(x1-y1) (x1+y1) = 4N = p1 * p2
In fact it turns in the system of equations:
x1-y1=p1
x1+y1=p2
N= 4 (x^2+ax+b)
Example try p1=2 p2=N/2. x=S^2+a*S+b-S results in an integer solution for y.
BUT the other solution depend on the factorization of N that is too big to be factored.
So maybe the other lattice points could result in a linear combination of xn,yn? Still to be investigated.
I found that some more advanced algoritm for Pell equation would not work because the term in y1 is a perfect square but in "Solving the Pell equation" (HENDRIK W. LENSTRA, JR.), it is mentioned that the smooth numbers algoritm could be extended to any value of the coefficient of y1. Still to be verified.
I'm thinking to attack as follows
1 - By smooth numbers algoritms mentioned above
2 - By defining g(x)=(x^2-ax+b)/(S+x), Delta=g(x+1)-g(x), by Eulero Maclaurin the sum of Delta(x) from 0 to x returns y in terms of an equation in the form x-N/4 ln(f(x))+Eps. Being x integer, lattice solutions should be those for which ln(f(x))+Eps=0.
3 - by solving g(x)-Floor(g(x))=0 by Fourier analysis of Floor(g(x)).
Other things I noted:
For small x, y(x) tends to be constant over an interval of x (i.e. floor(g(x))= k over consecutive intervals of x) and thus a numeric approach could be to set y and solve for x. But this covers only a small portion of the possible values for x.
For x large enought y(x) tends to be linear and in particular A=S-x+a+y is constant over intervals of x. (i.e. Floor(g(x+1))=1+g(x)).
Hope you find this interesting and help me in solving this problem!
Regards, P.