Pell's equation
You are encouraged to solve this task according to the task description, using any language you may know.
Pell's equation (also called the Pell–Fermat equation) is a Diophantine equation of the form:
- x2 - ny2 = 1
with integer solutions for x and y, where n is a given non-square positive integer.
- Task requirements
-
- find the smallest solution in positive integers to Pell's equation for n = {61, 109, 181, 277}.
- See also
-
- Wikipedia entry: Pell's equation.
11l
F solvePell(n)
V x = Int(sqrt(n))
V (y, z, r) = (x, 1, x << 1)
BigInt e1 = 1
BigInt e2 = 0
BigInt f1 = 0
BigInt f2 = 1
L
y = r * z - y
z = (n - y * y) I/ z
r = (x + y) I/ z
(e1, e2) = (e2, e1 + e2 * r)
(f1, f2) = (f2, f1 + f2 * r)
V (a, b) = (f2 * x + e2, f2)
I a * a - n * b * b == 1
R (a, b)
L(n) [61, 109, 181, 277]
V (x, y) = solvePell(n)
print(‘x^2 - #3 * y^2 = 1 for x = #27 and y = #25’.format(n, x, y))- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Ada
with Ada.Text_Io;
with Ada.Numerics.Elementary_Functions;
with Ada.Numerics.Big_Numbers.Big_Integers;
procedure Pells_Equation is
use Ada.Numerics.Big_Numbers.Big_Integers;
type Pair is
record
V1, V2 : Big_Integer;
end record;
procedure Solve_Pell (N : Natural; X, Y : out Big_Integer) is
use Ada.Numerics.Elementary_Functions;
Big_N : constant Big_Integer := To_Big_Integer (N);
XX : constant Big_Integer := To_Big_Integer (Natural (Float'Floor (Sqrt (Float (N)))));
begin
if XX**2 = Big_N then
X := 1; Y := 0;
return;
end if;
declare
YY : Big_Integer := XX;
Z : Big_Integer := 1;
R : Big_Integer := 2 * XX;
E : Pair := Pair'(V1 => 1, V2 => 0);
F : Pair := Pair'(V1 => 0, V2 => 1);
begin
loop
YY := R * Z - YY;
Z := (Big_N - YY**2) / Z;
R := (XX + YY) / Z;
E := Pair'(V1 => E.V2, V2 => R * E.V2 + E.V1);
F := Pair'(V1 => F.V2, V2 => R * F.V2 + F.V1);
X := E.V2 + XX * F.V2;
Y := F.V2;
exit when X**2 - Big_N * Y**2 = 1;
end loop;
end;
end Solve_Pell;
procedure Test (N : Natural) is
package Natural_Io is new Ada.Text_Io.Integer_Io (Natural);
use Ada.Text_Io, Natural_Io;
X, Y : Big_Integer;
begin
Solve_Pell (N, X => X, Y => Y);
Put ("X**2 - ");
Put (N, Width => 3);
Put (" * Y**2 = 1 for X = ");
Put (To_String (X, Width => 22));
Put (" and Y = ");
Put (To_String (Y, Width => 20));
New_Line;
end Test;
begin
Test (61);
Test (109);
Test (181);
Test (277);
end Pells_Equation;
- Output:
X**2 - 61 * Y**2 = 1 for X = 1766319049 and Y = 226153980 X**2 - 109 * Y**2 = 1 for X = 158070671986249 and Y = 15140424455100 X**2 - 181 * Y**2 = 1 for X = 2469645423824185801 and Y = 183567298683461940 X**2 - 277 * Y**2 = 1 for X = 159150073798980475849 and Y = 9562401173878027020
ALGOL 68
Also tests for a trival solution only (if n is a perfect square only 1, 0 is solution).
BEGIN
# find solutions to Pell's eqauation: x^2 - ny^2 = 1 for integer x, y, n #
MODE BIGINT = LONG LONG INT;
MODE BIGPAIR = STRUCT( BIGINT v1, v2 );
PROC solve pell = ( INT n )BIGPAIR:
IF INT x = ENTIER( sqrt( n ) );
x * x = n
THEN
# n is a erfect square - no solution otheg than 1,0 #
BIGPAIR( 1, 0 )
ELSE
# there are non-trivial solutions #
INT y := x;
INT z := 1;
INT r := 2*x;
BIGPAIR e := BIGPAIR( 1, 0 );
BIGPAIR f := BIGPAIR( 0, 1 );
BIGINT a := 0;
BIGINT b := 0;
WHILE
y := (r*z - y);
z := ENTIER ((n - y*y) / z);
r := ENTIER ((x + y) / z);
e := BIGPAIR( v2 OF e, r * v2 OF e + v1 OF e );
f := BIGPAIR( v2 OF f, r * v2 OF f + v1 OF f );
a := (v2 OF e + x*v2 OF f);
b := v2 OF f;
a*a - n*b*b /= 1
DO SKIP OD;
BIGPAIR( a, b )
FI # solve pell # ;
# task test cases #
[]INT nv = (61, 109, 181, 277);
FOR i FROM LWB nv TO UPB nv DO
INT n = nv[ i ];
BIGPAIR r = solve pell(n);
print( ("x^2 - ", whole( n, -3 ), " * y^2 = 1 for x = ", whole( v1 OF r, -21), " and y = ", whole( v2 OF r, -21 ), newline ) )
OD
END- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Arturo
solvePell: function [n][
x: to :integer sqrt n
[y, z, r]: @[x, 1, shl x 1]
[e1, e2]: [1, 0]
[f1, f2]: [0, 1]
while [true][
y: (r * z) - y
z: (n - y * y) / z
r: (x + y) / z
[e1, e2]: @[e2, e1 + e2 * r]
[f1, f2]: @[f2, f1 + f2 * r]
[a, b]: @[e2 + f2 * x, f2]
if 1 = (a*a) - n*b*b ->
return @[a, b]
]
]
loop [61 109 181 277] 'n [
[x, y]: solvePell n
print ["x² -" n "* y² = 1 for (x,y) =" x "," y]
]
- Output:
x² - 61 * y² = 1 for (x,y) = 1766319049 , 226153980 x² - 109 * y² = 1 for (x,y) = 158070671986249 , 15140424455100 x² - 181 * y² = 1 for (x,y) = 2469645423824185801 , 183567298683461940 x² - 277 * y² = 1 for (x,y) = 159150073798980475849 , 9562401173878027020
C
For n = 277, the x value is not correct because 64-bits is not enough to represent the value.
#include <math.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
struct Pair {
uint64_t v1, v2;
};
struct Pair makePair(uint64_t a, uint64_t b) {
struct Pair r;
r.v1 = a;
r.v2 = b;
return r;
}
struct Pair solvePell(int n) {
int x = (int) sqrt(n);
if (x * x == n) {
// n is a perfect square - no solution other than 1,0
return makePair(1, 0);
} else {
// there are non-trivial solutions
int y = x;
int z = 1;
int r = 2 * x;
struct Pair e = makePair(1, 0);
struct Pair f = makePair(0, 1);
uint64_t a = 0;
uint64_t b = 0;
while (true) {
y = r * z - y;
z = (n - y * y) / z;
r = (x + y) / z;
e = makePair(e.v2, r * e.v2 + e.v1);
f = makePair(f.v2, r * f.v2 + f.v1);
a = e.v2 + x * f.v2;
b = f.v2;
if (a * a - n * b * b == 1) {
break;
}
}
return makePair(a, b);
}
}
void test(int n) {
struct Pair r = solvePell(n);
printf("x^2 - %3d * y^2 = 1 for x = %21llu and y = %21llu\n", n, r.v1, r.v2);
}
int main() {
test(61);
test(109);
test(181);
test(277);
return 0;
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 11576121209304062921 and y = 9562401173878027020
C++
As with the C solution, the output for n = 277 is not correct because 64-bits is not enough to represent the value.
#include <iomanip>
#include <iostream>
#include <tuple>
std::tuple<uint64_t, uint64_t> solvePell(int n) {
int x = (int)sqrt(n);
if (x * x == n) {
// n is a perfect square - no solution other than 1,0
return std::make_pair(1, 0);
}
// there are non-trivial solutions
int y = x;
int z = 1;
int r = 2 * x;
std::tuple<uint64_t, uint64_t> e = std::make_pair(1, 0);
std::tuple<uint64_t, uint64_t> f = std::make_pair(0, 1);
uint64_t a = 0;
uint64_t b = 0;
while (true) {
y = r * z - y;
z = (n - y * y) / z;
r = (x + y) / z;
e = std::make_pair(std::get<1>(e), r * std::get<1>(e) + std::get<0>(e));
f = std::make_pair(std::get<1>(f), r * std::get<1>(f) + std::get<0>(f));
a = std::get<1>(e) + x * std::get<1>(f);
b = std::get<1>(f);
if (a * a - n * b * b == 1) {
break;
}
}
return std::make_pair(a, b);
}
void test(int n) {
auto r = solvePell(n);
std::cout << "x^2 - " << std::setw(3) << n << " * y^2 = 1 for x = " << std::setw(21) << std::get<0>(r) << " and y = " << std::setw(21) << std::get<1>(r) << '\n';
}
int main() {
test(61);
test(109);
test(181);
test(277);
return 0;
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 11576121209304062921 and y = 9562401173878027020
C#
using System;
using System.Numerics;
static class Program
{
static void Fun(ref BigInteger a, ref BigInteger b, int c)
{
BigInteger t = a; a = b; b = b * c + t;
}
static void SolvePell(int n, ref BigInteger a, ref BigInteger b)
{
int x = (int)Math.Sqrt(n), y = x, z = 1, r = x << 1;
BigInteger e1 = 1, e2 = 0, f1 = 0, f2 = 1;
while (true)
{
y = r * z - y; z = (n - y * y) / z; r = (x + y) / z;
Fun(ref e1, ref e2, r); Fun(ref f1, ref f2, r); a = f2; b = e2; Fun(ref b, ref a, x);
if (a * a - n * b * b == 1) return;
}
}
static void Main()
{
BigInteger x, y; foreach (int n in new[] { 61, 109, 181, 277 })
{
SolvePell(n, ref x, ref y);
Console.WriteLine("x^2 - {0,3} * y^2 = 1 for x = {1,27:n0} and y = {2,25:n0}", n, x, y);
}
}
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1,766,319,049 and y = 226,153,980 x^2 - 109 * y^2 = 1 for x = 158,070,671,986,249 and y = 15,140,424,455,100 x^2 - 181 * y^2 = 1 for x = 2,469,645,423,824,185,801 and y = 183,567,298,683,461,940 x^2 - 277 * y^2 = 1 for x = 159,150,073,798,980,475,849 and y = 9,562,401,173,878,027,020
D
import std.bigint;
import std.math;
import std.stdio;
void fun(ref BigInt a, ref BigInt b, int c) {
auto t = a;
a = b;
b = b * c + t;
}
void solvePell(int n, ref BigInt a, ref BigInt b) {
int x = cast(int) sqrt(cast(real) n);
int y = x;
int z = 1;
int r = x << 1;
BigInt e1 = 1;
BigInt e2 = 0;
BigInt f1 = 0;
BigInt f2 = 1;
while (true) {
y = r * z - y;
z = (n - y * y) / z;
r = (x + y) / z;
fun(e1, e2, r);
fun(f1, f2, r);
a = f2;
b = e2;
fun(b, a, x);
if (a * a - n * b * b == 1) {
return;
}
}
}
void main() {
BigInt x, y;
foreach(n; [61, 109, 181, 277]) {
solvePell(n, x, y);
writefln("x^2 - %3d * y^2 = 1 for x = %27d and y = %25d", n, x, y);
}
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Delphi
program Pells_equation;
{$APPTYPE CONSOLE}
uses
System.SysUtils,
Velthuis.BigIntegers;
type
TPellResult = record
x, y: BigInteger;
end;
function SolvePell(nn: UInt64): TPellResult;
var
n, x, y, z, r, e1, e2, f1, t, u, a, b: BigInteger;
begin
n := nn;
x := nn;
x := BigInteger.Sqrt(x);
y := BigInteger(x);
z := BigInteger.One;
r := x shl 1;
e1 := BigInteger.One;
e2 := BigInteger.Zero;
f1 := BigInteger.Zero;
b := BigInteger.One;
while True do
begin
y := (r * z) - y;
z := (n - (y * y)) div z;
r := (x + y) div z;
u := BigInteger(e1);
e1 := BigInteger(e2);
e2 := (r * e2) + u;
u := BigInteger(f1);
f1 := BigInteger(b);
b := r * b + u;
a := e2 + x * b;
t := (a * a) - (n * b * b);
if t = 1 then
begin
with Result do
begin
x := BigInteger(a);
y := BigInteger(b);
end;
Break;
end;
end;
end;
const
ns: TArray<UInt64> = [61, 109, 181, 277];
fmt = 'x^2 - %3d*y^2 = 1 for x = %-21s and y = %s';
begin
for var n in ns do
with SolvePell(n) do
writeln(format(fmt, [n, x.ToString, y.ToString]));
{$IFNDEF UNIX} readln; {$ENDIF}
end.
Factor
USING: formatting kernel locals math math.functions sequences ;
:: solve-pell ( n -- a b )
n sqrt >integer :> x!
x :> y!
1 :> z!
2 x * :> r!
1 0 :> ( e1! e2! )
0 1 :> ( f1! f2! )
0 0 :> ( a! b! )
[ a sq b sq n * - 1 = ] [
r z * y - y!
n y sq - z / floor z!
x y + z / floor r!
e2 r e2 * e1 + e2! e1!
f2 r f2 * f1 + f2! f1!
e2 x f2 * + a!
f2 b!
] until
a b ;
{ 61 109 181 277 } [
dup solve-pell
"x^2 - %3d*y^2 = 1 for x = %-21d and y = %d\n" printf
] each
- Output:
x^2 - 61*y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109*y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181*y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277*y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
FreeBASIC
for n = 277 the result is wrong, I do not know if you can represent such large numbers in FreeBasic!
Sub Fun(Byref a As LongInt, Byref b As LongInt, c As Integer)
Dim As LongInt t
t = a : a = b : b = b * c + t
End Sub
Sub SolvePell(n As Integer, Byref a As LongInt, Byref b As LongInt)
Dim As Integer z, r
Dim As LongInt x, y, e1, e2, f1, f2
x = Sqr(n) : y = x : z = 1 : r = 2 * x
e1 = 1 : e2 = 0 : f1 = 0 : f2 = 1
While True
y = r * z - y : z = (n - y * y) / z : r = (x + y) / z
Fun(e1, e2, r) : Fun(f1, f2, r) : a = f2 : b = e2 : Fun(b, a, x)
If a * a - n * b * b = 1 Then Exit Sub
Wend
End Sub
Dim As Integer i
Dim As LongInt x, y
Dim As Integer n(0 To 3) = {61, 109, 181, 277}
For i = 0 To 3 ''n In {61, 109, 181, 277}
SolvePell(n(i), x, y)
Print Using "x^2 - ### * y^2 = 1 for x = ##################### and y = #####################"; n(i); x; y
Next i
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = -6870622864405488695 and y = -8884342899831524596
Go
package main
import (
"fmt"
"math/big"
)
var big1 = new(big.Int).SetUint64(1)
func solvePell(nn uint64) (*big.Int, *big.Int) {
n := new(big.Int).SetUint64(nn)
x := new(big.Int).Set(n)
x.Sqrt(x)
y := new(big.Int).Set(x)
z := new(big.Int).SetUint64(1)
r := new(big.Int).Lsh(x, 1)
e1 := new(big.Int).SetUint64(1)
e2 := new(big.Int)
f1 := new(big.Int)
f2 := new(big.Int).SetUint64(1)
t := new(big.Int)
u := new(big.Int)
a := new(big.Int)
b := new(big.Int)
for {
t.Mul(r, z)
y.Sub(t, y)
t.Mul(y, y)
t.Sub(n, t)
z.Quo(t, z)
t.Add(x, y)
r.Quo(t, z)
u.Set(e1)
e1.Set(e2)
t.Mul(r, e2)
e2.Add(t, u)
u.Set(f1)
f1.Set(f2)
t.Mul(r, f2)
f2.Add(t, u)
t.Mul(x, f2)
a.Add(e2, t)
b.Set(f2)
t.Mul(a, a)
u.Mul(n, b)
u.Mul(u, b)
t.Sub(t, u)
if t.Cmp(big1) == 0 {
return a, b
}
}
}
func main() {
ns := []uint64{61, 109, 181, 277}
for _, n := range ns {
x, y := solvePell(n)
fmt.Printf("x^2 - %3d*y^2 = 1 for x = %-21s and y = %s\n", n, x, y)
}
}
- Output:
x^2 - 61*y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109*y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181*y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277*y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Haskell
pell :: Integer -> (Integer, Integer)
pell n = go (x, 1, x * 2, 1, 0, 0, 1)
where
x = floor $ sqrt $ fromIntegral n
go (y, z, r, e1, e2, f1, f2) =
let y' = r * z - y
z' = (n - y' * y') `div` z
r' = (x + y') `div` z'
(e1', e2') = (e2, e2 * r' + e1)
(f1', f2') = (f2, f2 * r' + f1)
(a, b) = (f2', e2')
(b', a') = (a, a * x + b)
in if a' * a' - n * b' * b' == 1
then (a', b')
else go (y', z', r', e1', e2', f1', f2')
λ> mapM_ print $ pell <$> [61,109,181,277] (1766319049,226153980) (158070671986249,15140424455100) (2469645423824185801,183567298683461940) (159150073798980475849,9562401173878027020)
J
NB. sqrt representation for continued fraction
sqrt_cf =: 3 : 0
rep=. '' [ 'm d'=. 0 1 [ a =. a0=. <. %: y
while. a ~: +: a0 do.
rep=. rep , a=. <. (a0+m) % d=. d %~ y - *: m=. m -~ a*d
end. a0;rep
)
NB. find x,y such that x^2 - n*y^2 = 1 using continued fractions
pell =: 3 : 0
n =. 1 [ 'a0 as' =. x: &.> sqrt_cf y
while. 1 do. cs =. 2 x: (+%)/\ a0, n$as NB. convergents
if. # sols =. I. 1 = (*: cs) +/ . * 1 , -y do. cs {~ {. sols return. end.
n =. +: n
end.
)
Check that task is actually solved
verify =: 3 : 0
assert. 1 = (*: xy) +/ . * 1 _61 [ echo 61 ; xy =. pell 61
assert. 1 = (*: xy) +/ . * 1 _109 [ echo 109 ; xy =. pell 109
assert. 1 = (*: xy) +/ . * 1 _181 [ echo 181 ; xy =. pell 181
assert. 1 = (*: xy) +/ . * 1 _277 [ echo 277 ; xy =. pell 277
)
- Output:
verify '' ┌──┬────────────────────┐ │61│1766319049 226153980│ └──┴────────────────────┘ ┌───┬──────────────────────────────┐ │109│158070671986249 15140424455100│ └───┴──────────────────────────────┘ ┌───┬──────────────────────────────────────┐ │181│2469645423824185801 183567298683461940│ └───┴──────────────────────────────────────┘ ┌───┬─────────────────────────────────────────┐ │277│159150073798980475849 9562401173878027020│ └───┴─────────────────────────────────────────┘
Java
import java.math.BigInteger;
import java.text.NumberFormat;
import java.util.ArrayList;
import java.util.List;
public class PellsEquation {
public static void main(String[] args) {
NumberFormat format = NumberFormat.getInstance();
for ( int n : new int[] {61, 109, 181, 277, 8941} ) {
BigInteger[] pell = pellsEquation(n);
System.out.printf("x^2 - %3d * y^2 = 1 for:%n x = %s%n y = %s%n%n", n, format.format(pell[0]), format.format(pell[1]));
}
}
private static final BigInteger[] pellsEquation(int n) {
int a0 = (int) Math.sqrt(n);
if ( a0*a0 == n ) {
throw new IllegalArgumentException("ERROR 102: Invalid n = " + n);
}
List<Integer> continuedFrac = continuedFraction(n);
int count = 0;
BigInteger ajm2 = BigInteger.ONE;
BigInteger ajm1 = new BigInteger(a0 + "");
BigInteger bjm2 = BigInteger.ZERO;
BigInteger bjm1 = BigInteger.ONE;
boolean stop = (continuedFrac.size() % 2 == 1);
if ( continuedFrac.size() == 2 ) {
stop = true;
}
while ( true ) {
count++;
BigInteger bn = new BigInteger(continuedFrac.get(count) + "");
BigInteger aj = bn.multiply(ajm1).add(ajm2);
BigInteger bj = bn.multiply(bjm1).add(bjm2);
if ( stop && (count == continuedFrac.size()-2 || continuedFrac.size() == 2) ) {
return new BigInteger[] {aj, bj};
}
else if (continuedFrac.size() % 2 == 0 && count == continuedFrac.size()-2 ) {
stop = true;
}
if ( count == continuedFrac.size()-1 ) {
count = 0;
}
ajm2 = ajm1;
ajm1 = aj;
bjm2 = bjm1;
bjm1 = bj;
}
}
private static final List<Integer> continuedFraction(int n) {
List<Integer> answer = new ArrayList<Integer>();
int a0 = (int) Math.sqrt(n);
answer.add(a0);
int a = -a0;
int aStart = a;
int b = 1;
int bStart = b;
while ( true ) {
//count++;
int[] values = iterateFrac(n, a, b);
answer.add(values[0]);
a = values[1];
b = values[2];
if (a == aStart && b == bStart) break;
}
return answer;
}
// array[0] = new part of cont frac
// array[1] = new a
// array[2] = new b
private static final int[] iterateFrac(int n, int a, int b) {
int x = (int) Math.floor((b * Math.sqrt(n) - b * a)/(n - a * a));
int[] answer = new int[3];
answer[0] = x;
answer[1] = -(b * a + x *(n - a * a)) / b;
answer[2] = (n - a * a) / b;
return answer;
}
}
- Output:
x^2 - 61 * y^2 = 1 for:
x = 1,766,319,049
y = 226,153,980
x^2 - 109 * y^2 = 1 for:
x = 158,070,671,986,249
y = 15,140,424,455,100
x^2 - 181 * y^2 = 1 for:
x = 2,469,645,423,824,185,801
y = 183,567,298,683,461,940
x^2 - 277 * y^2 = 1 for:
x = 159,150,073,798,980,475,849
y = 9,562,401,173,878,027,020
x^2 - 8941 * y^2 = 1 for:
x = 2,565,007,112,872,132,129,669,406,439,503,954,211,359,492,684,749,762,901,360,167,370,740,763,715,001,557,789,090,674,216,330,243,703,833,040,774,221,628,256,858,633,287,876,949,448,689,668,281,446,637,464,359,482,677,366,420,261,407,112,316,649,010,675,881,349,744,201
y = 27,126,610,172,119,035,540,864,542,981,075,550,089,190,381,938,849,116,323,732,855,930,990,771,728,447,597,698,969,628,164,719,475,714,805,646,913,222,890,277,024,408,337,458,564,351,161,990,641,948,210,581,361,708,373,955,113,191,451,102,494,265,278,824,127,994,180
jq
Works with gojq, the Go implementation of jq
Preliminaries
# If $j is 0, then an error condition is raised;
# otherwise, assuming infinite-precision integer arithmetic,
# if the input and $j are integers, then the result will be an integer.
def idivide($i; $j):
($i % $j) as $mod
| ($i - $mod) / $j ;
def idivide($j):
idivide(.; $j);
# input should be a non-negative integer for accuracy
# but may be any non-negative finite number
def isqrt:
def irt:
. as $x
| 1 | until(. > $x; . * 4) as $q
| {$q, $x, r: 0}
| until( .q <= 1;
.q |= idivide(4)
| .t = .x - .r - .q
| .r |= idivide(2)
| if .t >= 0
then .x = .t
| .r += .q
else .
end)
| .r ;
if type == "number" and (isinfinite|not) and (isnan|not) and . >= 0
then irt
else "isqrt requires a non-negative integer for accuracy" | error
end ;The Task
def solvePell:
. as $n
| ($n|isqrt) as $x
| { $x,
y : $x,
z : 1,
r : ($x * 2),
v1 : 1,
v2 : 0,
f1 : 0,
f2 : 1 }
| until(.emit;
.y = .r*.z - .y
| .z = idivide($n - .y*.y; .z)
| .r = idivide(.x + .y; .z)
| .v1 as $t
| .v1 = .v2
| .v2 = .r*.v2 + $t
| .f1 as $t
| .f1 = .f2
| .f2 = .r*.f2 + $t
| (.v2 + .x*.f2) as $a
| .f2 as $b
| if ($a*$a - $n*$b*$b == 1) then .emit = [$a, $b] else . end
).emit ;
(61, 109, 181, 277)
| solvePell as $res
| "x² - \(.)y² = 1 for x = \($res[0]) and y = \($res[1])"- Output:
x² - 61y² = 1 for x = 1766319049 and y = 226153980 x² - 109y² = 1 for x = 158070671986249 and y = 15140424455100 x² - 181y² = 1 for x = 2469645423824185801 and y = 183567298683461940 x² - 277y² = 1 for x = 159150073798980475849 and y = 9562401173878027020
Julia
function pell(n)
x = BigInt(floor(sqrt(n)))
y, z, r = x, BigInt(1), x << 1
e1, e2, f1, f2 = BigInt(1), BigInt(0), BigInt(0), BigInt(1)
while true
y = r * z - y
z = div(n - y * y, z)
r = div(x + y, z)
e1, e2 = e2, e2 * r + e1
f1, f2 = f2, f2 * r + f1
a, b = f2, e2
b, a = a, a * x + b
if a * a - n * b * b == 1
return a, b
end
end
end
for target in BigInt[61, 109, 181, 277]
x, y = pell(target)
println("x\u00b2 - $target", "y\u00b2 = 1 for x = $x and y = $y")
end
- Output:
x² - 61y² = 1 for x = 1766319049 and y = 226153980 x² - 109y² = 1 for x = 158070671986249 and y = 15140424455100 x² - 181y² = 1 for x = 2469645423824185801 and y = 183567298683461940 x² - 277y² = 1 for x = 159150073798980475849 and y = 9562401173878027020
Kotlin
import java.math.BigInteger
import kotlin.math.sqrt
class BIRef(var value: BigInteger) {
operator fun minus(b: BIRef): BIRef {
return BIRef(value - b.value)
}
operator fun times(b: BIRef): BIRef {
return BIRef(value * b.value)
}
override fun equals(other: Any?): Boolean {
if (this === other) return true
if (javaClass != other?.javaClass) return false
other as BIRef
if (value != other.value) return false
return true
}
override fun hashCode(): Int {
return value.hashCode()
}
override fun toString(): String {
return value.toString()
}
}
fun f(a: BIRef, b: BIRef, c: Int) {
val t = a.value
a.value = b.value
b.value = b.value * BigInteger.valueOf(c.toLong()) + t
}
fun solvePell(n: Int, a: BIRef, b: BIRef) {
val x = sqrt(n.toDouble()).toInt()
var y = x
var z = 1
var r = x shl 1
val e1 = BIRef(BigInteger.ONE)
val e2 = BIRef(BigInteger.ZERO)
val f1 = BIRef(BigInteger.ZERO)
val f2 = BIRef(BigInteger.ONE)
while (true) {
y = r * z - y
z = (n - y * y) / z
r = (x + y) / z
f(e1, e2, r)
f(f1, f2, r)
a.value = f2.value
b.value = e2.value
f(b, a, x)
if (a * a - BIRef(n.toBigInteger()) * b * b == BIRef(BigInteger.ONE)) {
return
}
}
}
fun main() {
val x = BIRef(BigInteger.ZERO)
val y = BIRef(BigInteger.ZERO)
intArrayOf(61, 109, 181, 277).forEach {
solvePell(it, x, y)
println("x^2 - %3d * y^2 = 1 for x = %,27d and y = %,25d".format(it, x.value, y.value))
}
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1,766,319,049 and y = 226,153,980 x^2 - 109 * y^2 = 1 for x = 158,070,671,986,249 and y = 15,140,424,455,100 x^2 - 181 * y^2 = 1 for x = 2,469,645,423,824,185,801 and y = 183,567,298,683,461,940 x^2 - 277 * y^2 = 1 for x = 159,150,073,798,980,475,849 and y = 9,562,401,173,878,027,020
Lambdatalk
Computing big numbers requires the lib_BN library.
{def pell
{lambda {:n}
{let { {:n :n}
{:x {BN.intPart {BN.sqrt :n}}} // x=int(sqrt(n))
} {pell.r :n :x :x 1 {* 2 :x} 1 0 0 1}
}}}
-> pell
{def pell.r
{lambda {:n :x :y :z :r :e1 :e2 :f1 :f2}
{let { {:n :n} {:x :x} {:z :z} {:r :r} // no closure ->
{:e1 :e1} {:e2 :e2} {:f1 :f1} {:f2 :f2} // must reassign :(
{:y {BN.- {BN.* :r :z} :y}} // y=rz-y
} {let { {:n :n} {:x :x} {:y :y} {:r :r}
{:e1 :e1} {:e2 :e2} {:f1 :f1} {:f2 :f2}
{:z {BN.intPart
{BN./ {BN.- :n {BN.* :y :y}} :z}}} // z=(n-y*y)//z
} {let { {:n :n} {:x :x} {:y :y} {:z :z}
{:e1 :e1} {:e2 :e2} {:f1 :f1} {:f2 :f2}
{:r {BN.intPart {BN./ {BN.+ :x :y} :z}}} // r= (x+y)//z
} {let { {:n :n} {:x :x} {:y :y} {:z :z} {:r :r}
{:e1 :e2} // e1=e2
{:e2 {BN.+ {BN.* :r :e2} :e1}} // e2=r*e2+e1
{:f1 :f2} // f1=f2
{:f2 {BN.+ {BN.* :r :f2} :f1}} // f2=r*f2+f1
} {let { {:n :n} {:x :x} {:y :y} {:z :z} {:r :r}
{:e1 :e1} {:e2 :e2} {:f1 :f1} {:f2 :f2}
{:a {BN.+ :e2 {BN.* :x :f2}}} // a=e2+x*f2
{:b :f2} // b=f2
} {if {= {BN.compare {BN.- {BN.* :a :a}
{BN.* :n {BN.* :b :b}}}
1}
0} // a*a-n*b*b == 1
then {div}x{sup 2} - n*y{sup 2} = 1 for n=:n, x=:a, y=:b
else {pell.r :n :x :y :z :r :e1 :e2 :f1 :f2} // do it again
}}}}}}}}
-> pell.r
{S.map pell 61 109 181 277}
->
x^2 - n*y^2 = 1 for n=61, x=1766319049, y=226153980
x^2 - n*y^2 = 1 for n=109, x=158070671986249, y=15140424455100
x^2 - n*y^2 = 1 for n=181, x=2469645423824185801, y=183567298683461940
x^2 - n*y^2 = 1 for n=277, x=159150073798980475849, y=9562401173878027020
langur
val solvePell = fn(n) {
val x = trunc(n ^/ 2)
var y, z, r = x, 1, x * 2
var e1, e2, f1, f2 = 1, 0, 0, 1
for {
y = r * z - y
z = (n - y * y) \ z
r = (x + y) \ z
e1, e2 = fun(e1, e2, r)
f1, f2 = fun(f1, f2, r)
val b, a = fun(e2, f2, x)
if a^2 - n * b^2 == 1: return [a, b]
}
}
val C = fn(x) {
# format number string with commas
var neg, s = "", x -> string
if s[1] == '-' {
neg, s = "-", less(s, of=1)
}
neg ~ join(split(s, by=-3), by=",")
}
for n in [61, 109, 181, 277, 8941] {
val x, y = solvePell(n)
writeln "x² - {{n}}y² = 1 for:\n\tx = {{x:fn C}}\n\ty = {{y:fn C}}\n"
}- Output:
x² - 61y² = 1 for: x = 1,766,319,049 y = 226,153,980 x² - 109y² = 1 for: x = 158,070,671,986,249 y = 15,140,424,455,100 x² - 181y² = 1 for: x = 2,469,645,423,824,185,801 y = 183,567,298,683,461,940 x² - 277y² = 1 for: x = 159,150,073,798,980,475,849 y = 9,562,401,173,878,027,020 x² - 8941y² = 1 for: x = 2,565,007,112,872,132,129,669,406,439,503,954,211,359,492,684,749,762,901,360,167,370,740,763,715,001,557,789,090,674,216,330,243,703,833,040,774,221,628,256,858,633,287,876,949,448,689,668,281,446,637,464,359,482,677,366,420,261,407,112,316,649,010,675,881,349,744,201 y = 27,126,610,172,119,035,540,864,542,981,075,550,089,190,381,938,849,116,323,732,855,930,990,771,728,447,597,698,969,628,164,719,475,714,805,646,913,222,890,277,024,408,337,458,564,351,161,990,641,948,210,581,361,708,373,955,113,191,451,102,494,265,278,824,127,994,180
Mathematica /Wolfram Language
FindInstance[x^2 - 61 y^2 == 1, {x, y}, PositiveIntegers]
FindInstance[x^2 - 109 y^2 == 1, {x, y}, PositiveIntegers]
FindInstance[x^2 - 181 y^2 == 1, {x, y}, PositiveIntegers]
FindInstance[x^2 - 277 y^2 == 1, {x, y}, PositiveIntegers]
- Output:
{{x -> 1766319049, y -> 226153980}}
{{x -> 158070671986249, y -> 15140424455100}}
{{x -> 2469645423824185801, y -> 183567298683461940}}
{{x -> 159150073798980475849, y -> 9562401173878027020}}
Nim
import math, strformat
import bignum
func solvePell(n: int): (Int, Int) =
let x = newInt(sqrt(n.toFloat).int)
var (y, z, r) = (x, newInt(1), x shl 1)
var (e1, e2) = (newInt(1), newInt(0))
var (f1, f2) = (newInt(0), newInt(1))
while true:
y = r * z - y
z = (n - y * y) div z
r = (x + y) div z
(e1, e2) = (e2, e1 + e2 * r)
(f1, f2) = (f2, f1 + f2 * r)
let (a, b) = (f2 * x + e2, f2)
if a * a - n * b * b == 1:
return (a, b)
for n in [61, 109, 181, 277]:
let (x, y) = solvePell(n)
echo &"x² - {n:3} * y² = 1 for (x, y) = ({x:>21}, {y:>19})"
- Output:
x² - 61 * y² = 1 for (x, y) = ( 1766319049, 226153980) x² - 109 * y² = 1 for (x, y) = ( 158070671986249, 15140424455100) x² - 181 * y² = 1 for (x, y) = ( 2469645423824185801, 183567298683461940) x² - 277 * y² = 1 for (x, y) = (159150073798980475849, 9562401173878027020)
Pascal
A console application in Free Pascal, created with the Lazarus IDE.
Pascal has no built-in support for arbitrarily large integers. The program below requires integers larger than 64 bits, and therefore uses an external library. The code could easily be adapted to solve the negative Pell equation x^2 - n*y^2 = -1, or show that no solution exists.
program Pell_console;
uses SysUtils,
uIntX; // uIntX is a unit in the library IntXLib4Pascal.
// uIntX.TIntX is an arbitrarily large integer.
// For the given n: if there are non-trivial solutions of x^2 - n*y^2 = 1
// in non-negative integers (x,y), return the smallest.
// Else return the trivial solution (x,y) = (1,0).
procedure SolvePell( n : integer; out x, y : uIntX.TIntX);
var
m, a, c, d : integer;
p, q, p_next, q_next, p_prev, q_prev : uIntX.TIntX;
evenNrSteps : boolean;
begin
if (n >= 0) then m := Trunc( Sqrt( 1.0*n + 0.5)) // or use Rosetta Code Isqrt
else m := 0;
if n <= m*m then begin // if n is not a positive non-square
x := 1; y := 0; exit; // return a trivial solution
end;
c := m; d := 1;
p := 1; q := 0;
p_prev := 0; q_prev := 1;
a := m;
evenNrSteps := true;
repeat
// Get the next convergent p/q in the continued fraction for sqrt(n)
p_next := a*p + p_prev;
q_next := a*q + q_prev;
p_prev := p; p := p_next;
q_prev := q; q := q_next;
// Get the next term a in the continued fraction for sqrt(n)
Assert((n - c*c) mod d = 0); // optional sanity check
d := (n - c*c) div d;
a := (m + c) div d;
c := a*d - c;
evenNrSteps := not evenNrSteps;
until (c = m) and (d = 1);
{
If the first return to (c,d) = (m,1) occurs after an even number of steps,
then p^2 - n*q^2 = 1, and there is no solution to x^2 - n*y^2 = -1.
Else p^2 - n*q^2 = -1, and to get a solution to x^2 - n*y^2 = 1 we can
either continue until we return to (c,d) = (m,1) for the second time,
or use the short cut below.
}
if evenNrSteps then begin
x := p; y := q;
end
else begin
x := 2*p*p + 1; y := 2*p*q
end;
end;
// For the given n: show the Pell solution on the console.
procedure ShowPellSolution( n : integer);
var
x, y : uIntX.TIntX;
lineOut : string;
begin
SolvePell( n, x, y);
lineOut := SysUtils.Format( 'n = %d --> (', [n]);
lineOut := lineOut + x.ToString + ', ' + y.ToString + ')';
WriteLn( lineOut);
end;
// Main routine
begin
ShowPellSolution( 61);
ShowPellSolution( 109);
ShowPellSolution( 181);
ShowPellSolution( 277);
end.
- Output:
n = 61 --> (1766319049, 226153980) n = 109 --> (158070671986249, 15140424455100) n = 181 --> (2469645423824185801, 183567298683461940) n = 277 --> (159150073798980475849, 9562401173878027020)
Perl
sub solve_pell {
my ($n) = @_;
use bigint try => 'GMP';
my $x = int(sqrt($n));
my $y = $x;
my $z = 1;
my $r = 2 * $x;
my ($e1, $e2) = (1, 0);
my ($f1, $f2) = (0, 1);
for (; ;) {
$y = $r * $z - $y;
$z = int(($n - $y * $y) / $z);
$r = int(($x + $y) / $z);
($e1, $e2) = ($e2, $r * $e2 + $e1);
($f1, $f2) = ($f2, $r * $f2 + $f1);
my $A = $e2 + $x * $f2;
my $B = $f2;
if ($A**2 - $n * $B**2 == 1) {
return ($A, $B);
}
}
}
foreach my $n (61, 109, 181, 277) {
my ($x, $y) = solve_pell($n);
printf("x^2 - %3d*y^2 = 1 for x = %-21s and y = %s\n", $n, $x, $y);
}
- Output:
x^2 - 61*y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109*y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181*y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277*y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Phix
This now ignores the nonsquare part of the task spec, returning {1,0}.
with javascript_semantics include mpfr.e procedure fun(mpz a,b,t, integer c) -- {a,b} = {b,c*b+a} (and t gets trashed) mpz_set(t,a) mpz_set(a,b) mpz_mul_si(b,b,c) mpz_add(b,b,t) end procedure function SolvePell(integer n) integer x = floor(sqrt(n)), y = x, z = 1, r = x*2 mpz e1 = mpz_init(1), e2 = mpz_init(), f1 = mpz_init(), f2 = mpz_init(1), t = mpz_init(0), u = mpz_init(), a = mpz_init(1), b = mpz_init(0) if x*x!=n then while mpz_cmp_si(t,1)!=0 do y = r*z - y z = floor((n-y*y)/z) r = floor((x+y)/z) fun(e1,e2,t,r) -- {e1,e2} = {e2,r*e2+e1} fun(f1,f2,t,r) -- {f1,f2} = {f2,r*r2+f1} mpz_set(a,f2) mpz_set(b,e2) fun(b,a,t,x) -- {b,a} = {f2,x*f2+e2} mpz_mul(t,a,a) mpz_mul_si(u,b,n) mpz_mul(u,u,b) mpz_sub(t,t,u) -- t = a^2-n*b^2 end while end if return {a, b} end function function split_into_chunks(string x, integer one, rest) sequence res = {x[1..one]} x = x[one+1..$] integer l = length(x) while l do integer k = min(l,rest) res = append(res,x[1..k]) x = x[k+1..$] l -= k end while return join(res,"\n"&repeat(' ',29))&"\n"&repeat(' ',17) end function sequence ns = {4, 61, 109, 181, 277, 8941} for i=1 to length(ns) do integer n = ns[i] mpz {x, y} = SolvePell(n) string xs = mpz_get_str(x,comma_fill:=true), ys = mpz_get_str(y,comma_fill:=true) if length(xs)>97 then xs = split_into_chunks(xs,98,96) ys = split_into_chunks(ys,99,96) end if printf(1,"x^2 - %3d*y^2 = 1 for x = %27s and y = %25s\n", {n, xs, ys}) end for
- Output:
x^2 - 4*y^2 = 1 for x = 1 and y = 0
x^2 - 61*y^2 = 1 for x = 1,766,319,049 and y = 226,153,980
x^2 - 109*y^2 = 1 for x = 158,070,671,986,249 and y = 15,140,424,455,100
x^2 - 181*y^2 = 1 for x = 2,469,645,423,824,185,801 and y = 183,567,298,683,461,940
x^2 - 277*y^2 = 1 for x = 159,150,073,798,980,475,849 and y = 9,562,401,173,878,027,020
x^2 - 8941*y^2 = 1 for x = 2,565,007,112,872,132,129,669,406,439,503,954,211,359,492,684,749,762,
901,360,167,370,740,763,715,001,557,789,090,674,216,330,243,703,833,
040,774,221,628,256,858,633,287,876,949,448,689,668,281,446,637,464,
359,482,677,366,420,261,407,112,316,649,010,675,881,349,744,201
and y = 27,126,610,172,119,035,540,864,542,981,075,550,089,190,381,938,849,116,
323,732,855,930,990,771,728,447,597,698,969,628,164,719,475,714,805,
646,913,222,890,277,024,408,337,458,564,351,161,990,641,948,210,581,
361,708,373,955,113,191,451,102,494,265,278,824,127,994,180
Prolog
pell(A, X, Y) finds all solutions X, Y s.t. X^2 - A*Y^2 = 1. Therefore the once() predicate can be used to only select the first one.
% Find the square root as a continued fraction
cf_sqrt(N, Sz, [A0, Frac]) :-
A0 is floor(sqrt(N)),
(A0*A0 =:= N ->
Sz = 0, Frac = []
;
cf_sqrt(N, A0, A0, 0, 1, 0, [], Sz, Frac)).
cf_sqrt(N, A, A0, M0, D0, Sz0, L, Sz, R) :-
M1 is D0*A0 - M0,
D1 is (N - M1*M1) div D0,
A1 is (A + M1) div D1,
(A1 =:= 2*A ->
succ(Sz0, Sz), revtl([A1|L], R, R)
;
succ(Sz0, Sz1), cf_sqrt(N, A, A1, M1, D1, Sz1, [A1|L], Sz, R)).
revtl([], Z, Z).
revtl([A|As], Bs, Z) :- revtl(As, [A|Bs], Z).
% evaluate an infinite continued fraction as a lazy list of convergents.
%
convergents([A0, As], Lz) :-
lazy_list(next_convergent, eval_state(1, 0, A0, 1, As), Lz).
next_convergent(eval_state(P0, Q0, P1, Q1, [Term|Ts]), eval_state(P1, Q1, P2, Q2, Ts), R) :-
P2 is Term*P1 + P0,
Q2 is Term*Q1 + Q0,
R is P1 rdiv Q1.
% solve Pell's equation
%
pell(N, X, Y) :-
cf_sqrt(N, _, D), convergents(D, Rs),
once((member(R, Rs), ratio(R, P, Q), P*P - N*Q*Q =:= 1)),
pell_seq(N, P, Q, X, Y).
ratio(N, N, 1) :- integer(N).
ratio(P rdiv Q, P, Q).
pell_seq(_, X, Y, X, Y).
pell_seq(N, X0, Y0, X2, Y2) :-
pell_seq(N, X0, Y0, X1, Y1),
X2 is X0*X1 + N*Y0*Y1,
Y2 is X0*Y1 + Y0*X1.
- Output:
% show how we can keep generating solutions for x^2 - 3y^2 = 1 ?- pell(3,X,Y). X = 2, Y = 1 ; X = 7, Y = 4 ; X = 26, Y = 15 ; X = 97, Y = 56 ; X = 362, Y = 209 ; X = 1351, Y = 780 ; X = 5042, Y = 2911 . % solve the task ?- forall((member(A, [61, 109, 181, 277]), once(pell(A, X, Y))), (write(X**2-A*Y**2=1), nl)). 1766319049**2-61*226153980**2=1 158070671986249**2-109*15140424455100**2=1 2469645423824185801**2-181*183567298683461940**2=1 159150073798980475849**2-277*9562401173878027020**2=1 true.
Python
import math
def solvePell(n):
x = int(math.sqrt(n))
y, z, r = x, 1, x << 1
e1, e2 = 1, 0
f1, f2 = 0, 1
while True:
y = r * z - y
z = (n - y * y) // z
r = (x + y) // z
e1, e2 = e2, e1 + e2 * r
f1, f2 = f2, f1 + f2 * r
a, b = f2 * x + e2, f2
if a * a - n * b * b == 1:
return a, b
for n in [61, 109, 181, 277]:
x, y = solvePell(n)
print("x^2 - %3d * y^2 = 1 for x = %27d and y = %25d" % (n, x, y))
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Raku
(formerly Perl 6)
use Lingua::EN::Numbers;
sub pell (Int $n) {
my $y = my $x = Int(sqrt $n);
my $z = 1;
my $r = 2 * $x;
my ($e1, $e2) = (1, 0);
my ($f1, $f2) = (0, 1);
loop {
$y = $r * $z - $y;
$z = Int(($n - $y²) / $z);
$r = Int(($x + $y) / $z);
($e1, $e2) = ($e2, $r * $e2 + $e1);
($f1, $f2) = ($f2, $r * $f2 + $f1);
my $A = $e2 + $x * $f2;
my $B = $f2;
if ($A² - $n * $B² == 1) {
return ($A, $B);
}
}
}
for 61, 109, 181, 277, 8941 -> $n {
next if $n.sqrt.narrow ~~ Int;
my ($x, $y) = pell($n);
printf "x² - %sy² = 1 for:\n\tx = %s\n\ty = %s\n\n", $n, |($x, $y)».,
}
- Output:
x² - 61y² = 1 for: x = 1,766,319,049 y = 226,153,980 x² - 109y² = 1 for: x = 158,070,671,986,249 y = 15,140,424,455,100 x² - 181y² = 1 for: x = 2,469,645,423,824,185,801 y = 183,567,298,683,461,940 x² - 277y² = 1 for: x = 159,150,073,798,980,475,849 y = 9,562,401,173,878,027,020 x² - 8941y² = 1 for: x = 2,565,007,112,872,132,129,669,406,439,503,954,211,359,492,684,749,762,901,360,167,370,740,763,715,001,557,789,090,674,216,330,243,703,833,040,774,221,628,256,858,633,287,876,949,448,689,668,281,446,637,464,359,482,677,366,420,261,407,112,316,649,010,675,881,349,744,201 y = 27,126,610,172,119,035,540,864,542,981,075,550,089,190,381,938,849,116,323,732,855,930,990,771,728,447,597,698,969,628,164,719,475,714,805,646,913,222,890,277,024,408,337,458,564,351,161,990,641,948,210,581,361,708,373,955,113,191,451,102,494,265,278,824,127,994,180
REXX
A little extra code was added to align and commatize the gihugeic numbers for readability.
/*REXX program to solve Pell's equation for the smallest solution of positive integers. */
numeric digits 2200 /*ensure enough decimal digs for answer*/
parse arg $ /*obtain optional arguments from the CL*/
if $=='' | $=="," then $= 61 109 181 277 /*Not specified? Then use the defaults*/
d= 28 /*used for aligning the output numbers.*/
do j=1 for words($); #= word($, j) /*process all the numbers in the list. */
parse value pells(#) with x y /*extract the two values of X and Y.*/
cx= comma(x); Lcx= length(cx); cy= comma(y); Lcy= length(cy)
say 'x^2 -'right(#, max(4, length(#))) "* y^2 == 1" ,
' when x='right(cx, max(d, Lcx)) " and y="right(cy, max(d, Lcy))
end /*j*/
exit 0 /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
comma: parse arg ?; do jc=length(?)-3 to 1 by -3; ?= insert(',', ?, jc); end; return ?
floor: procedure; parse arg x; _= x % 1; return _ - (x < 0) * (x \= _)
/*──────────────────────────────────────────────────────────────────────────────────────*/
iSqrt: procedure; parse arg x; r= 0; q= 1; do while q<=x; q= q * 4; end
do while q>1; q= q%4; _= x-r-q; r= r%2; if _>=0 then do; x= _; r= r+q; end; end
return r /*R: is the integer square root of X. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
pells: procedure; parse arg n; x= iSqrt(n); y=x /*obtain arg; obtain integer sqrt of N*/
parse value 1 0 with e1 e2 1 f2 f1 /*assign values for: E1, E2, and F2, F1*/
z= 1; r= x + x
do until ( (e2 + x*f2)**2 - n*f2*f2) == 1
y= r*z - y; z= floor( (n - y*y) / z)
r= floor( (x + y ) / z)
parse value e2 r*e2 + e1 with e1 e2
parse value f2 r*f2 + f1 with f1 f2
end /*until*/
return e2 + x * f2 f2
- output when using the default inputs:
x^2 - 61 * y^2 == 1 when x= 1,766,319,049 and y= 226,153,980 x^2 - 109 * y^2 == 1 when x= 158,070,671,986,249 and y= 15,140,424,455,100 x^2 - 181 * y^2 == 1 when x= 2,469,645,423,824,185,801 and y= 183,567,298,683,461,940 x^2 - 277 * y^2 == 1 when x= 159,150,073,798,980,475,849 and y= 9,562,401,173,878,027,020
Ruby
def solve_pell(n)
x = Integer.sqrt(n)
y = x
z = 1
r = 2*x
e1, e2 = 1, 0
f1, f2 = 0, 1
loop do
y = r*z - y
z = (n - y*y) / z
r = (x + y) / z
e1, e2 = e2, r*e2 + e1
f1, f2 = f2, r*f2 + f1
a, b = e2 + x*f2, f2
break a, b if a*a - n*b*b == 1
end
end
[61, 109, 181, 277].each {|n| puts "x*x - %3s*y*y = 1 for x = %-21s and y = %s" % [n, *solve_pell(n)]}
- Output:
x*x - 61*y*y = 1 for x = 1766319049 and y = 226153980 x*x - 109*y*y = 1 for x = 158070671986249 and y = 15140424455100 x*x - 181*y*y = 1 for x = 2469645423824185801 and y = 183567298683461940 x*x - 277*y*y = 1 for x = 159150073798980475849 and y = 9562401173878027020
Rust
use num_bigint::{ToBigInt, BigInt};
use num_traits::{Zero, One};
//use std::mem::replace in the loop if you want this to be more efficient
fn main() {
test(61u64);
test(109u64);
test(181u64);
test(277u64);
}
struct Pair {
v1: BigInt,
v2: BigInt,
}
impl Pair {
pub fn make_pair(a: &BigInt, b: &BigInt) -> Pair {
Pair {
v1: a.clone(),
v2: b.clone(),
}
}
}
fn solve_pell(n: u64) -> Pair{
let x: BigInt = ((n as f64).sqrt()).to_bigint().unwrap();
if x.clone() * x.clone() == n.to_bigint().unwrap() {
Pair::make_pair(&One::one(), &Zero::zero())
} else {
let mut y: BigInt = x.clone();
let mut z: BigInt = One::one();
let mut r: BigInt = ( &z + &z) * x.clone();
let mut e: Pair = Pair::make_pair(&One::one(), &Zero::zero());
let mut f: Pair = Pair::make_pair(&Zero::zero() ,&One::one());
let mut a: BigInt = Zero::zero();
let mut b: BigInt = Zero::zero();
while &a * &a - n * &b * &b != One::one() {
//println!("{} {} {}", y, z, r);
y = &r * &z - &y;
z = (n - &y * &y) / &z;
r = (&x + &y) / &z;
e = Pair::make_pair(&e.v2, &(&r * &e.v2 + &e.v1));
f = Pair::make_pair(&f.v2, &(&r * &f.v2 + &f.v1));
a = &e.v2 + &x * &f.v2;
b = f.v2.clone();
}
let pa = &a;
let pb = &b;
Pair::make_pair(&pa.clone(), &pb.clone())
}
}
fn test(n: u64) {
let r: Pair = solve_pell(n);
println!("x^2 - {} * y^2 = 1 for x = {} and y = {}", n, r.v1, r.v2);
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Scala
def pellFermat(n: Int): (BigInt,BigInt) = {
import scala.math.{sqrt, floor}
val x = BigInt(floor(sqrt(n)).toInt)
var i = 0
// Use the Continued Fractions method
def converge(y:BigInt, z:BigInt, r:BigInt, e1:BigInt, e2:BigInt, f1:BigInt, f2:BigInt ) : (BigInt,BigInt) = {
val a = f2 * x + e2
val b = f2
if (a * a - n * b * b == 1) {
return (a, b)
}
val yh = r * z - y
val zh = (n - yh * yh) / z
val rh = (x + yh) / zh
converge(yh,zh,rh,e2,e1 + e2 * rh,f2,f1 + f2 * rh)
}
converge(x,BigInt("1"),x << 1,BigInt("1"),BigInt("0"),BigInt("0"),BigInt("1"))
}
val nums = List(61,109,181,277)
val solutions = nums.map{pellFermat(_)}
{
println("For Pell's Equation, x\u00b2 - ny\u00b2 = 1\n")
(nums zip solutions).foreach{ case (n, (x,y)) => println(s"n = $n, x = $x, y = $y")}
}
- Output:
For Pell's Equation, x² - ny² = 1 n = 61, x = 1766319049, y = 226153980 n = 109, x = 158070671986249, y = 15140424455100 n = 181, x = 2469645423824185801, y = 183567298683461940 n = 277, x = 159150073798980475849, y = 9562401173878027020
Sidef
func solve_pell(n) {
var x = n.isqrt
var y = x
var z = 1
var r = 2*x
var (e1, e2) = (1, 0)
var (f1, f2) = (0, 1)
loop {
y = (r*z - y)
z = floor((n - y*y) / z)
r = floor((x + y) / z)
(e1, e2) = (e2, r*e2 + e1)
(f1, f2) = (f2, r*f2 + f1)
var A = (e2 + x*f2)
var B = f2
if (A**2 - n*B**2 == 1) {
return (A, B)
}
}
}
for n in [61, 109, 181, 277] {
var (x, y) = solve_pell(n)
printf("x^2 - %3d*y^2 = 1 for x = %-21s and y = %s\n", n, x, y)
}
- Output:
x^2 - 61*y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109*y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181*y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277*y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
Swift
func solvePell<T: BinaryInteger>(n: T, _ a: inout T, _ b: inout T) {
func swap(_ a: inout T, _ b: inout T, mul by: T) {
(a, b) = (b, b * by + a)
}
let x = T(Double(n).squareRoot())
var y = x
var z = T(1)
var r = x << 1
var e1 = T(1)
var e2 = T(0)
var f1 = T(0)
var f2 = T(1)
while true {
y = r * z - y
z = (n - y * y) / z
r = (x + y) / z
swap(&e1, &e2, mul: r)
swap(&f1, &f2, mul: r)
(a, b) = (f2, e2)
swap(&b, &a, mul: x)
if a * a - n * b * b == 1 {
return
}
}
}
var x = BigInt(0)
var y = BigInt(0)
for n in [61, 109, 181, 277] {
solvePell(n: BigInt(n), &x, &y)
print("x\u{00b2} - \(n)y\u{00b2} = 1 for x = \(x) and y = \(y)")
}
- Output:
x² - 61y² = 1 for x = 1766319049 and y = 226153980 x² - 109y² = 1 for x = 158070671986249 and y = 15140424455100 x² - 181y² = 1 for x = 2469645423824185801 and y = 183567298683461940 x² - 277y² = 1 for x = 159150073798980475849 and y = 9562401173878027020
Visual Basic .NET
Imports System.Numerics
Module Module1
Sub Fun(ByRef a As BigInteger, ByRef b As BigInteger, c As Integer)
Dim t As BigInteger = a : a = b : b = b * c + t
End Sub
Sub SolvePell(n As Integer, ByRef a As BigInteger, ByRef b As BigInteger)
Dim x As Integer = Math.Sqrt(n), y As Integer = x, z As Integer = 1, r As Integer = x << 1,
e1 As BigInteger = 1, e2 As BigInteger = 0, f1 As BigInteger = 0, f2 As BigInteger = 1
While True
y = r * z - y : z = (n - y * y) / z : r = (x + y) / z
Fun(e1, e2, r) : Fun(f1, f2, r) : a = f2 : b = e2 : Fun(b, a, x)
If a * a - n * b * b = 1 Then Exit Sub
End While
End Sub
Sub Main()
Dim x As BigInteger, y As BigInteger
For Each n As Integer In {61, 109, 181, 277}
SolvePell(n, x, y)
Console.WriteLine("x^2 - {0,3} * y^2 = 1 for x = {1,27:n0} and y = {2,25:n0}", n, x, y)
Next
End Sub
End Module
- Output:
x^2 - 61 * y^2 = 1 for x = 1,766,319,049 and y = 226,153,980 x^2 - 109 * y^2 = 1 for x = 158,070,671,986,249 and y = 15,140,424,455,100 x^2 - 181 * y^2 = 1 for x = 2,469,645,423,824,185,801 and y = 183,567,298,683,461,940 x^2 - 277 * y^2 = 1 for x = 159,150,073,798,980,475,849 and y = 9,562,401,173,878,027,020
Wren
import "./big" for BigInt
import "./fmt" for Fmt
var solvePell = Fn.new { |n|
n = BigInt.new(n)
var x = n.isqrt
var y = x.copy()
var z = BigInt.one
var r = x * 2
var e1 = BigInt.one
var e2 = BigInt.zero
var f1 = BigInt.zero
var f2 = BigInt.one
while (true) {
y = r*z - y
z = (n - y*y) / z
r = (x + y) / z
var t = e1.copy()
e1 = e2.copy()
e2 = r*e2 + t
t = f1.copy()
f1 = f2.copy()
f2 = r*f2 + t
var a = e2 + x*f2
var b = f2.copy()
if (a*a - n*b*b == BigInt.one) return [a, b]
}
}
for (n in [61, 109, 181, 277]) {
var res = solvePell.call(n)
Fmt.print("x² - $3dy² = 1 for x = $-21i and y = $i", n, res[0], res[1])
}
- Output:
x² - 61y² = 1 for x = 1766319049 and y = 226153980 x² - 109y² = 1 for x = 158070671986249 and y = 15140424455100 x² - 181y² = 1 for x = 2469645423824185801 and y = 183567298683461940 x² - 277y² = 1 for x = 159150073798980475849 and y = 9562401173878027020
Zig
Neither C nor C++ gave the correct answer for 277 because of integer overflow which was performed silently as undefined behaviour.
Zig gives the correct answer for 277.
- Zig does not allow integer overflow at runtime on operators such as + and -.
- Zig provides integer operators which can detect overflow, e.g.
@subWithOverflow()
- Zig supports arbitrary bit-width integers, e.g.
u256
const std = @import("std");
pub fn main() !void {
const writer = std.io.getStdOut().writer();
try printSolvedPell(61, writer);
try printSolvedPell(109, writer);
try printSolvedPell(181, writer);
try printSolvedPell(277, writer);
}
const Pair = struct {
v1: u256,
v2: u256,
fn init(a: u256, b: u256) Pair {
return Pair{
.v1 = a,
.v2 = b,
};
}
};
fn solvePell(n: u256) Pair {
const x: u256 = std.math.sqrt(n);
// n is a perfect square - no solution other than 1,0
if (x * x == n)
return Pair.init(1, 0);
// there are non-trivial solutions
var y = x;
var z: u256 = 1;
var r = 2 * x;
var e = Pair.init(1, 0);
var f = Pair.init(0, 1);
var a: u256 = 0;
var b: u256 = 0;
while (true) {
y = r * z - y;
z = (n - y * y) / z;
r = (x + y) / z;
e = Pair.init(e.v2, r * e.v2 + e.v1);
f = Pair.init(f.v2, r * f.v2 + f.v1);
a = e.v2 + x * f.v2;
b = f.v2;
const ov = @subWithOverflow(a * a, n * b * b);
if (ov[1] != 0)
continue;
if (ov[0] == 1) // a * a, n * b * b == 1
break;
}
return Pair.init(a, b);
}
fn printSolvedPell(n: u256, writer: anytype) !void {
const r = solvePell(n);
try writer.print("x^2 - {d:3} * y^2 = 1 for x = {d:21} and y = {d:19}\n", .{ n, r.v1, r.v2 });
}
- Output:
x^2 - 61 * y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109 * y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181 * y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277 * y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
zkl
GNU Multiple Precision Arithmetic Library
var [const] BI=Import("zklBigNum"); // libGMP
fcn solve_pell(n){
x,y,z,r := BI(n).root(2), x.copy(), BI(1), x*2;
e1,e2, f1,f2 := BI(1), BI(0), BI(0), BI(1);
reg t; // a,b = c,d is a=c; b=d
do(30_000){ // throttle this in case of screw up
y,z,r = (r*z - y), (n - y*y)/z, (x + y)/z;
t,e2,e1 = e2, r*e2 + e1, t;
t,f2,f1 = f2, r*f2 + f1, t;
A,B := e2 + x*f2, f2;
if (A*A - B*B*n == 1) return(A,B);
}
}foreach n in (T(61, 109, 181, 277)){
x,y:=solve_pell(n);
println("x^2 - %3d*y^2 = 1 for x = %-21d and y = %d".fmt(n,x,y));
}- Output:
x^2 - 61*y^2 = 1 for x = 1766319049 and y = 226153980 x^2 - 109*y^2 = 1 for x = 158070671986249 and y = 15140424455100 x^2 - 181*y^2 = 1 for x = 2469645423824185801 and y = 183567298683461940 x^2 - 277*y^2 = 1 for x = 159150073798980475849 and y = 9562401173878027020
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