Continued fractions and Pell's equation - numerical issues - math

Question

Mathematical background
Continued fractions are a way to represent numbers (rational or not), with a basic recursion formula to calculate it. Given a number r, we define r[0]=r and have:
for n in range(0..N):
a[n] = floor(r[n])
if r[n] == [an]: break
r[n+1] = 1 / (r[n]-a[n])
where a is the final representation. We can also define a series of convergents by
h[-2,-1] = [0, 1]
k[-2, -1] = [1, 0]
h[n] = a[n]*h[n-1]+h[n-2]
k[n] = a[n]*k[n-1]+k[n-2]
where h[n]/k[n] converge to r.
Pell's equation is a problem of the form x^2-D*y^2=1 where all numbers are integers and D is not a perfect square in our case. A solution for a given D that minimizes x is given by continued fractions. Basically, for the above equation, it is guaranteed that this (fundamental) solution is x=h[n] and y=k[n] for the lowest n found which solves the equation in the continued fraction expansion of sqrt(D).
Problem
I am failing to get this simple algorithm work for D=61. I first noticed it did not solve Pell's equation for 100 coefficients, so I compared it against Wolfram Alpha's convergents and continued fraction representation and noticed the 20th elements fail - the representation is 3 compared to 4 that I get, yielding different convergents - h[20]=335159612 on Wolfram compared to 425680601 for me.
I tested the code below, two languages (though to be fair, Python is C under the hood I guess), on two systems and get the same result - a diff on loop 20. I'll note that the convergents are still accurate and converge! Why am I getting different results compared to Wolfram Alpha, and is it possible to fix it?
For testing, here's a Python program to solve Pell's equation for D=61, printing first 20 convergents and the continued fraction representation cf (and some extra unneeded fluff):
from math import floor, sqrt # Can use mpmath here as well.
def continued_fraction(D, count=100, thresh=1E-12, verbose=False):
cf = []
h = (0, 1)
k = (1, 0)
r = start = sqrt(D)
initial_count = count
x = (1+thresh+start)*start
y = start
while abs(x/y - start) > thresh and count:
i = int(floor(r))
cf.append(i)
f = r - i
x, y = i*h[-1] + h[-2], i*k[-1] + k[-2]
if verbose is True or verbose == initial_count-count:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
if x**2 - D*y**2 == 1:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
print(cf)
return
count -= 1
r = 1/f
h = (h[1], x)
k = (k[1], y)
print(cf)
raise OverflowError(f"Converged on {x} {y} with count {count} and diff {abs(start-x/y)}!")
continued_fraction(61, count=20, verbose=True, thresh=-1) # We don't want to stop on account of thresh in this example
A c program doing the same:
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int main() {
long D = 61;
double start = sqrt(D);
long h[] = {0, 1};
long k[] = {1, 0};
int count = 20;
float thresh = 1E-12;
double r = start;
long x = (1+thresh+start)*start;
long y = start;
while(abs(x/(double)y-start) > -1 && count) {
long i = floor(r);
double f = r - i;
x = i * h[1] + h[0];
y = i * k[1] + k[0];
printf("%ld\u00B2-%ldx%ld\u00B2 = %lf\n", x, D, y, x*x-D*y*y);
r = 1/f;
--count;
h[0] = h[1];
h[1] = x;
k[0] = k[1];
k[1] = y;
}
return 0;
}

Answer 1

mpmath, python's multi-precision library can be used. Just be careful that all the important numbers are in mp format.
In the code below, x, y and i are standard multi-precision integers. r and f are multi-precision real numbers. Note that the initial count is set higher than 20.
from mpmath import mp, mpf
mp.dps = 50 # precision in number of decimal digits
def continued_fraction(D, count=22, thresh=mpf(1E-12), verbose=False):
cf = []
h = (0, 1)
k = (1, 0)
r = start = mp.sqrt(D)
initial_count = count
x = 0 # some dummy starting values, they will be overwritten early in the while loop
y = 1
while abs(x/y - start) > thresh and count > 0:
i = int(mp.floor(r))
cf.append(i)
x, y = i*h[-1] + h[-2], i*k[-1] + k[-2]
if verbose or initial_count == count:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
if x**2 - D*y**2 == 1:
print(f'{x}\u00B2-{D}x{y}\u00B2 = {x**2-D*y**2}')
print(cf)
return
count -= 1
f = r - i
r = 1/f
h = (h[1], x)
k = (k[1], y)
print(cf)
raise OverflowError(f"Converged on {x} {y} with count {count} and diff {abs(start-x/y)}!")
continued_fraction(61, count=22, verbose=True, thresh=mpf(1e-100))
Output is similar to wolfram's:
...
335159612²-61x42912791² = 3
1431159437²-61x183241189² = -12
1766319049²-61x226153980² = 1
[7, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1, 14, 1, 4, 3, 1, 2, 2, 1, 3, 4, 1]