Pell’s equation

Pell’s equation

Pell’s equation

Pell’s equation

refers generally to the equation

x

2

-

A

⁢

y

2

=

1

where

A

is a

positive integer

not a square. This equation is now known to have an

infinite

number of solutions in positive integers

x

and

y

.

Fermat, in 1657, proposed to English mathematicians that they find a method of solving this equation in general, for integral

x

and

y

. Fermat apparently already possessed a method, for doing so. According to

[

1

]

, the Greeks and Indians had detailed knowledge of this problem; the solutions

577

2

-

2

⋅

408

2

=

1

1151

2

-

92

⋅

120

2

=

1

were both known by Indian mathematicians B.C. In fact, the solution given above for

A

=

92

is the smallest solution for this

A

, and was given by Brahmagupta (b. 598 A.D.) together with a method for its

derivation

. The cyclic method, described in detail below, was given by Bháscara Achárya (b. 1114 A.D.).

The English, notably Wallis, developed a different but provably almost identical method, presumably in response to Fermat’s challenge. Both methods produce, in time, all integral solutions to the equation.

As can be seen from the entry on Pell’s equation and

simple continued fractions

, solutions of Pell’s equation can also be found by developing the

continued fraction

for

A

.

Here is a table of the minimum value of

y

for some values of

A

; a

complete

table for

A

<

150

can be found in

[

1

]

:

A

y

2

2

22

42

37

12

61

226153980

109

15140424455100

136

3

143

1

According to

[

1

]

, Pell had nothing to do with the development of this problem or its solution; its attribution to Pell is due to an error on the part of Euler.

The Cyclic Method

The cyclic method relies on successive approximation and on a hope that the approximation process will

converge

. Of course, it can be proven that the process does converge and that it results

eventually

in all solutions of the

Pell equation

. The method relies on the well-known

identity

(

x

2

-

A

⁢

y

2

)

⁢

(

w

2

-

A

⁢

z

2

)

=

(

x

⁢

w

+

A

⁢

y

⁢

z

)

2

-

A

⁢

(

x

⁢

z

+

y

⁢

w

)

2

(1)

Given an equation

c

2

-

A

⁢

d

2

=

k

this identity can be used to produce another solution.

So, for example, let’s take the case

A

=

22

. Start by choosing

x

so that

x

2

-

22

⋅

1

2

is as small in

absolute value

as possible, i.e.

x

=

5

and

5

2

-

22

⋅

1

2

=

3

. If we multiply that by the equation

r

2

-

22

⋅

1

2

=

s

, we get, using (

1

),

(

5

⁢

r

+

22

)

2

-

22

⋅

(

r

+

5

)

2

=

3

⁢

s

Remember that we are trying to find values so that the right-hand is

1

. However, simply choosing a value of

r

that makes

s

as small as possible will not achieve any progress, since we will then have

r

=

5

and

s

=

3

, so we will get a right-hand side of

9

. Instead, we will choose

r

so that

r

+

5

is

divisible

by

3

; then

5

⁢

r

+

22

must be divisible by

3

as well, so

s

is divisible by

3

and we may divide the resulting equation through by

9

. To make

r

+

5

≡

0

(

mod

3

)

we need

r

≡

1

(

mod

3

)

. We want to make

s

as small as possible in magnitude subject to this condition, so we choose

r

=

4

and then

s

=

-

6

. The resulting equation, after dividing through by

9

, is

14

2

-

22

⋅

3

2

=

-

2

We are not done, so let’s repeat the

operation

. Multiply again by

r

2

-

22

⋅

1

2

=

s

to get

(

14

⁢

r

+

66

)

2

-

22

⋅

(

3

⁢

r

+

14

)

2

=

-

2

⁢

s

and we must choose

r

such that

3

⁢

r

+

14

≡

0

(

mod

2

)

(i.e.

r

≡

0

(

mod

2

)

) to make

s

as small as possible in magnitude. This happens for

r

=

4

,

s

=

-

6

; substituting and dividing through by

2

2

=

4

we get

61

2

-

22

⋅

13

2

=

3

Repeating, we get

(

61

⁢

r

+

22

⋅

13

)

2

-

22

⋅

(

13

⁢

r

+

61

)

2

=

3

⁢

s

so

13

⁢

r

+

61

≡

0

(

mod

3

)

, or

r

≡

3

(

mod

3

)

. Minimizing

s

leads to

r

=

5

,

s

=

3

, whence, dividing the result by

3

2

=

9

we get

197

2

-

22

⋅

42

2

=

1

and we are done. The

sequence

of equations was

1

2

-

22

⋅

0

2

=

1

5

2

-

22

⋅

1

2

=

3

14

2

-

22

⋅

3

2

=

-

2

61

2

-

22

⋅

13

2

=

3

197

2

-

22

⋅

42

2

=

1

The fact that the sequence of

differences

is palindromic is not an accident; it can be proven that this is always the case.

The English Method

The English method uses the same basic algorithm as the cyclic method, except that when choosing

r

, it the largest

r

such that

r

2

<

A

, which

s

<

0

. Thus if two successive equations in this method are

x

2

-

A

⁢

y

2

=

k

X

2

-

A

⁢

Y

2

=

K

it is clear that

K

=

r

2

-

A

k

and thus that

K

and

k

have different signs. So in the English method, the values of

k

alternate in sign.

For the case

A

=

22

, above, the sequence of equations is

1

2

-

22

⋅

0

2

=

1

4

2

-

22

⋅

1

2

=

-

6

5

2

-

22

⋅

1

2

=

3

14

2

-

22

⋅

3

2

=

-

2

61

2

-

22

⋅

13

2

=

3

136

2

-

22

⋅

29

2

=

-

6

197

2

-

22

⋅

42

2

=

1

In general, the English method results in the same solutions as the cyclic method, but it gets there more slowly - in the English method, an additional entry is inserted in between any two entries that have the same sign.

References

1

H.M. Edwards,

Fermat’s Last

Theorem

: A Genetic

Introduction

to

Algebraic Number Theory

, Springer-Verlag 1977.

2

B.L. van der Waerden, ”Pell’s Equation in Greek and Hindu Mathematics”, Russ. Math Surveys, 1976, 31 (5), 210-225.

3

C.O. Selenius, ”Rationale of the chakravala process of Jayadeva and Bhaskara II”, Historia Mathematica, 2 (1975), 167-184.

Title

Pell’s equation

Canonical name

PellsEquation

Date of creation

2013-03-22 16:28:07

Last modified on

2013-03-22 16:28:07

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

8

Author

rm50 (10146)

Entry type

Topic

Classification

msc 11D09

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