Pell 方程递归解的幂型因子性质 及其在不定方程中的应用

Pell 方程递归解的幂型因子性质 及其在不定方程中的应用

华南师范大学附属中学: 赵玉博 指导教师: 郝保国

摘要

本文的目标是研究 Catlan 猜想(x^myⁿ 1除 x=3,m=2,y=2,n=3 之外无正整数解) 的一个推广，即：不定方程x^m2yⁿ 1都可能有哪些正整数解。为解决这一问题，

深入研究了pell方程的解的一些数论性质，即：当最小解确定时，方程的递归解 的因子性质。并将其结论应用于对这种不定方程的讨论中，在限定n为偶数的情 况下取得了较为成功的结果。此外，文中还包含了关于pell方程解的数论性质的 结论在一些其它问题中的应用。

关键词：pell方程、Catlan猜想

Some Arithmetic Properties about the Factor of the Solution to Pell Equation

and Its Applications in Diophantine Equations

The Affiliated High School of South China Normal University Zhao Yubo Directed by: Hao Baoguo

Abstract

This Paper is aimed to study a generalization of the Catlan Conjecture (The only nontrivial solution to

x^myⁿ 1

is x=3,m=2,y=2,n=3), that is, to determine the nontrivial solutions to

x^m2yⁿ 1

. To solve this problem, I study some arithmetic properties of the solutions to Pell equation, or more precisely, when the minimum solution is fixed, the properties about factors of the recursive solutions to the equation. Furthermore, I apply these properties into the study of

x^m2yⁿ 1 ,

and obtain some successful results when n is restricted to be even. What’s more, some applications of these arithmetic properties in other problems are contained in the paper as well.

Key Words and Phrases:

Pell equation, Catlan Conjecture

1 ． Preliminaries.

We begin with some elementary properties about the solutions of Pell equation.

（1）x²Dy² 1 （ D0 and has no divisor of form n²）is called positive Pell equation(or standard Pell equation). It always has a minimal positive solution( ,x y_{1 1})，and all solutions (x y_n, _n) are fixed by x_ny_n D(x₁y₁ D)ⁿ，

（or 2

n n

xn   ，

2

n n

yn

D

 

 ，where  x₁ y₁ D，  x₁ y₁ D）

Furthermore, the recursion formula ^{1 1 1 1}

1 1 1 1

( 2)

n n n

n n n

x x x Dy y y x y y x n

 

 

 

 

  



and ^{m n m n m n}( , ^*)

m n m n m n

x x x Dy y

m n N

y x y y x





 

 

  

 hold.

If we define x₀ 1，y₀ 0，x_n x_n，y_n  y_n，index in the recursion formulas above can be chosen from ℤ including negative integers.

（2）x²Dy²  1，is called negative Pell equation. If it has solution, it must have infinitely many solutions. Assume( , )a b_{1 1} is its minimal positive solution，then the general solutions (a b_n, _n) are fixed by a_nb_n D(a₁b₁ D)^2n1，（or

2 1 2 1

2

n n

an  ^  ^

 ，

2 1 2 1

2

n n

bn

D

 ^  ^

 ，where  a₁ b₁ D，  a₁ b₁ D）

What’s more, (a₁b₁ D)² x₁ y₁ D（As the symbols above，(x₁，y₁) is the minimal solution of the corresponding positive Pell equation）.

（3）x²Dy² K， K 1，is called quasi-Pell equation. If it has solution, it must have infinitely many solutions. In particular, when K 2，if we denote the minimal solution as ( , )u v_{1 1} , then the general solutions ( ,u v_{n n}) are fixed by

2 1

1 1

1

( )

2

n

n n n

u v D

u v D





   （or

2 1 2 1

2

n n

n n

u  ^  ^ ，

2 1 2 1

2

n n

n n

v

D

 ^  ^

 ，where

1 1

u v D

   ，  u₁ v₁ D）

（4）No matter what kind of Pell equation it is, the recursion formula always holds.

Where

( x y

_n

,

_n

)

and (x y^*_n, ^*_n) stand for the nth positive solution of

2 2

x Dy K and x²Dy² 1.

2. Main Theorems

In this paper, we will obtain the following two main theorems

Theorem 1 (1) For standard Pell equation x²Dy² 1，assume d is a divisor of D, d is greater than 3, and ( , ) 1y d₁  ，then d^k y_n d^k n

(

nN^*，

kN

.

The symbol

“ ”

means: b^a a  b^a|a，b^a1|a) (2) For（I）negative Pell equation x²Dy²  1，

（II）quasi-Pell equation x²Dy²   2, 4， （III）quasi-Pell equation x²Dy² 4. Similar conclusion holds, that is：

In equation of type （I）， if d D| and ( ,d y₁) =1， d3 ，then

2 1

k k

d yn d n .

In equation of type （II），if d D| , ( ,d y₁)=1，d3，and d is odd integer，

then d^k y_n d^k 2n1.

In equation of type （III），if d D| , ( ,d y₁)=1，d3，and d is odd integer，

then d^k y_n d^k n.

(3) In particular， for standard Pell equation x²Dy² 1，we have：assume the minimal solution is ( ,x y_{1 1})，d D| ，d3, and d^ y₁，then if ( , y¹) 1

d d^  ，we have

k k

d^{ } yn d n.

Remark: In (2), among the equations of type （I），the situation that 2 |d and 0

k will never occurat all，since this will yield 2 |y_n 4 |Dy_n²x_n²  1(mod 4)， which leads to a contradiction. So in this case d must be odd.

Theorem 2 (1) Denote the general solutions of standard Pell equation

2 2

1

x Dy  as (x y_n, _n)，then  t N^*among the following propositions either (i) or (ii) will hold; and either (iii) or (iv) will hold：

(i)n t x, | _n.

(ii) f t( )N^* which is uniquely determined by t，such that |

n ( ) t x n

 f t is a positive odd integer.

(iii)n t, | y_n.

（iv） g t( )N^* which is uniquely determined by t，such that |t y_n g t( ) |n.

(2) x²Dy²  2，denote its general solutions as (x y_n, _n)，for _ odd integer t，among the following propositions either (i) or (ii) will hold; and either (iii) or (iv) will hold：

(i) n N t x^*, | _n.

(ii)  f t( )N^* which is uniquely determined by t ， such that

| _n ( ) | 2 1 t x  f t n .

(iii) n N t^*, | y_n.

（ iv）  g t( )N^* which is uniquely determined by t ， such that

| _n ( ) | 2 1 t y  g t n .

(3)More generally, we can obtain：

For fixed tN^*，we denote the following statement as proposition （I）：either we cannot find an x_n such that t x| _n ，or f t( )N^* such that t x| _n

(2 1) ( )

n k f t

   ，kN^*；either we cannot find an y_n such that |t y_n，or ( ) *

g t N

  such that |t y_n  n kg t( )，kN^*.

In parallel, we denote the following statement as proposition （II）：either we cannot find an x_n such that |t x_n，or f t( )N^* such that |t x_n 2n 1 kf t( )，

kN*；either we cannot find an y_n such that t y| _n，or g t( )N^* such that

| _n

t y 2n 1 kg t( )，kN^*. Then we have：

（i）For x²Dy² 1，proposition （I） holds.

（ii）For x²Dy²  1，proposition（II） holds.

（iii）For x²Dy²  2，proposition（II） holds if we require in which t is odd.

（iv）对x²Dy²  4，proposition（I）holds if we require in which t is odd.

This paper will be organized as follows：in section 3 we will prove theorem 1，

and several applications of theorem 1 will be given in section 4；in section 5 we will prove theorem 2，and several applications of theorem 2 will be given in section 6. In the applications in section 4 and 6，we will obtain parts of the results of the generalized Catlan’s Conjecture.

3. Proof of Theorem 1

Obviously, theorem 1.(1) is the corollary of 1.(3)，so we prove 1.(3) first.

The proof of 1.(3) is divided into three parts:

In standard Pell equation x²Dy² 1，assume d is divisor of D, d is greater than 3, and d^ y₁，( , y¹) 1

d d^  (N ).

（i）d^{ 1}|y_n d n|

since ¹

1 1 3 3 3

1 1 1 1

1 1 1

1 1

1 1

( ) ( )

...

2

(mod )

n n

n n

n n n

n

x D y x D y

y C x y C x y D

D nx y d^

 

 

  

      



and by the original equation we have ( , ) 1x d₁  ，then by ( ,d y₁)d^ the conclusion follows.

（ii）d^{ 1} y_d If not，assume

1

2 1 3 3 3

1 1 1

| _d ^d _d ^d ...

d^ y dx ^ y C x ^ y D Then

1

2 1 3 3

1 1 1

( 1)( 2)

| ...

6

d d d d d

d^{ } dx ^ y   x ^ y D

 

1

1 1 3 3

1 1 1

( 1)( 2)

| ...

6

d D d d d

d^{ } x ^ y   x ^ y

 

If 3 |d，then 3 | (d1)(d2) ，2 | (d1)(d2) so 6 | (d1)(d2)

then

1

1 3 3 1 1

1 1 1

( 1)( 2)

| |

6

d d

d d

d^ D   x ^ y d^ x ^ y ，which leads to a contradiction.

If 3 |d，then assume d 3d₁，then d₁1

so 1

1 1 3 3 1 1

1 1 1 1 1 1 1

( 1)( 2)

| ... |

3 2

d D d d d d

d^ x ^ y    x ^ y  d^ x ^ y however (x y d1 1, ^)d^ and d₁1，which leads to a contradiction. So （ii） holds.

（iii）Now we prove the theorem directly

By the preliminary proposition （4），

2 3 2

2 2 2 2 3

2

2 2 (mod )

(mod )

d d d d d

d d d d

y x y x y d

x x Dy x d









  

   

 . Generally，

if we have proven

1 2 3

2 3

(mod )

(mod )

k

kd d d

k

kd d

y kx y d

x x d





 



 

 

 then

(k 1)d kd d kd d

y _  y x x y kx_d^k1y_d x_d x y_d^k _d (k1)x y_d^k _d (modd^23)

(k 1)d kd d kd d

x _ x x Dy y x_d^kx_d  D kx_d^k1y_dy_d x_d^k1 (modd^23) so

2 3

1 2 3

(mod )

(mod )

n

nd d

n

nd d d

x x d

y nx y d







 

 

 



so whe d |n we have d^{ 2} | y_nd，otherwise d^{ 2}|nx_d^n1y_d d nx| _d^n1.

by ( ,d x_n) 1 d n| ，this is a contradiction. This deduction also proves when d n| ,

2| _nd.

d^{ } y 2

2

d^ yd .

so d^{ 2}|y_n d²|n， ² ₂ d^ yd .

Generally, assume that we have proven d^{ s} |y_n d^s |n，and s

s

d

^

y

d ， sN^.

then

2 2 2 2 2 1

2

2 2 1

2

(mod )

2 2 (mod )

s s s s

s s s s s

s

d d d d

s

d d d d d

x x Dy x d

y x y x y d





 

 

   

   



^.

If we have proven

2 2 1

1 2 2 1

(mod )

(mod )

s s

s s s

k s

k d d

k s

k d d d

x x d

y kx y d





 

   



 

  



^.

then

2

( 1) ^{s s s s s s s s}

k s k

k d kd d kd d d d d d

x

_

 x  x   D y  y  x  x  Dx y

1 2 2 1

(mod )

s

k s

x

d^

d

^{ }



.

(k 1)d^s kd^s d^s kd^s d^s

y

_

 x y  y x

^k_{s s} ^k_s¹ _s

d d d d

x y kx

^

x

 

2 2 2

( 1)

s s

(mod )

k s

d d

k x y d

^{ }

 

so

2 2 2

1 2 2 2

(mod )

, (mod )

s s

s s s

n s

nd d

n s

nd d d

x x d

n y nx y d





 

  

 

    

^.

So naturally we have d^{  s 1}|y_n s^s1|n ， 1

1

s

s

d

^

y

d

  .

By induction we have  k N^*， d^{ k} y_n d^k n.

Proof of 1．(2)：

Firstly，the minimal solution of x²Dy² h is ( ,x y_{1 1})，then the general solution (x y_n, _n) is given by ^{( 1 1 )}

an

n n

n

x y D

x y D

b

   .

Where when h 1，a_n 2n1，b_n 1； when h4 ，a_n n，b_n 2^n1；

when h 4，a_n 2n1，b_n 2^2n2； when h 2，a_n 2n1，_{bn 2}^n1.

The proof will still be divided into three parts：

（i）d y| _n d a| _n

since

1

1 3

1 3 3

1 1 1

1 1 1 1

1

1 1

( ) ( ) ...

2 (mod )

n n

n n

n n

n

a a

a a

a a

n

n n a

n

C x y C x y D

x D y x D y

y b D b

a x y d

 



    

  

 



By the original equation we have ( , ) 1x d₁  ，then by ( ,d y₁) 1 the conclusion holds.

（ii）

cd

d y

Where when h   1, 4, 2， ¹

d 2 c d

 ，when h4，c_d d. Then we have

cd

a d.

If not，assume ¹

1 3 3 3

1 1 1

2 ...

|

cd cd

d cd

d

d

a a

c a

c

c

a x y C x y D d y

b

 

 



Then

1

2 1 3 3

1 1 1

( 1)( 2)

| ...

6

d d d d d

d dx ^ y   x ^ y D

  

1

1 3 3

1 1 1

( 1)( 2)

| ...

6

d D d d d

d x ^ y   x ^ y

  

If 3 |d，then 3 | (d1)(d2) ，2 | (d1)(d2), so 6 | (d1)(d2)

and | ^{( 1)( 2)} | ₁ ¹ ₁ 6

d d d

d D   d x ^ y

  ，which is a contradiction.

If d²|nu_d^n1v_d d nu| _d^n1.3 |d，then assume d 3d₁，d₁1 so

1

1 3 3 1

1 1 1 1 1 1 1

( 1)( 2)

| ... |

3 2

d D d d d d

d x ^ y     x ^ y  d x ^ y

but (x y d_{1 1}, ) 1 and d₁ 1，which is a contradiction. So （ii）holds.

（iii）Now we prove the theorem directly

Let u_n Dv_n(x₁ D y₁)ⁿ，then actually we have n ^an n

x u

 b

_{， n} ^an

n

y v

 b _un，

vnsatisfy ^{m n m n m n}

m n m n n m

u u u Dv v v u v u v





 

  

 Thus what we have got in the previous two steps is exactly d v| _n d n| ，d v_d.

Then we have

3 2

2 2 2 3

2

2 2 (mod )

(mod )

d d d d d

d d d d

v u v u v d

u u Dv u d

  

   



Generally, assume we have proven

1 3

3

(mod ) (mod )

k

kd d d

k

kd d

v ku v d

u u d

  

 



Then v_(k1)d v u_{kd d} u v_{kd d} ku_d^k1v_du_d u v_d^k _d (k1)u v_d^k _d (modd³) u_(k1)d u u_{kd d} Dv v_{kd d} u_d^ku_d  D ku_d^k1v_dv_d u_d^k1(modd³) So we have

3

1 3

(mod ) (mod )

n

nd d

n

nd d d

u u d

v nu ^v d

 

 



So when d |n we have d² |v_nd，otherwise d²|nu_d^n1v_d d nu| _d^n1.

By ( ,d u_n) 1 d n| ，which is a contradiction. This also proves when d n| we have d²|v_nd. 2

2

d vd

So d²|v_n d²|n， ² ₂. d vd

Generally, assume we have proven d^s|v_n d^s|n，and s

s

d vd ， then

2 2 2 2

2

2 2

(mod )

2 2 (mod ).

s s s s

s s s s s

s

d d d d

s

d d d d d

u u Dv u d

v u v u v d





   

  



If we have proven

2

1 2

(mod ) (mod ).

s s

s s s

k s

k d d

k s

k d d d

u u d

v ku v d





 



 

  



Then

2

( 1) ^{s s s s s s s s}

k s k

k d kd d kd d d d d d

u

_

 u  u   D v  v  u  u  Du v

1 2

(mod ).

s

k s

u

d^

d

^



(k 1)d^s kd^s d^s kd^s d^s

v

_

 u v  v u

s s s¹ s

k k

d d d d

u v ku

^

u

 

( 1)

s s

(mod

2

)

k s

d d

k u v d

^

 

so

2

1 2

(mod )

, (mod ).

s s

s s s

n s

nd d

n s

nd d d

u u d

n v nu v d



 

 

    

so naturally we have

d

^s1

| v

_n

 d

^s1

| n

， 1

1

s

s

d



v

d 

Then by induction we have  k N^*， d^k v_n d^k n. That is exactly d^k y_n d^k a_n

4. Applications of Theorem 1

Example 1 x²24^2n11

x

^，

n

N.

This equation has an obvious solution x5，n0. Assume ( , )x n is a non-negative solution of x²24^2n11 such that n0，then in the Pell equation

2 2

24 1

x  y  ，assume y_m 24 ,ⁿ n0，then 24ⁿ y_m 24ⁿ m. So ym y₂₄n. Notice that

y

₂₄n

 C

¹₂₄n

  5 C

₂₄³n

  5 24 ...

³



 5 24ⁿ 24ⁿ  y_m This is a contradiction.

So this equation has no non-negative integral solution other than ( , )x n (5, 0). This example can be directly generalized into the following result：

2 2 2 1

( 1) ⁿ 1

x  a  ^  （a2 is fixed，

x

^，

n

are variables）has no non-negative solutions.

A more general question is: Besides n0, is there any non-negative integral solution of x²y^2n11? We can only obtain parts of answer to this question here.

Example 2 x²y^2n1 1(n0)when 2n 1 p is prime，we must have 2 | y，

| p x.

Proof：First we prove2 |y. Otherwise 2 | y，then 2 |x. So x1，x1 are both odd.

By (x1)(x 1) y^2n1，and (x1)(x 1) y^2n1 So

2 1

2 1 2 1

2 1

1 2

1

n

n n

n

x u

u v

x v

  



    

  . So uv|u^2n1v^2n1 2.

Since u，v are either both odd or both even, we must have u v 2. So

2 1 2 1 2 1

(v2) ^n v ^n 2 ^n 2. This is a contradiction.

So 2 |y.

By the original equation, we have x² (y1)(y^p1y^p2 ... 1).

1 2

... 1 1 ( 1) ... 1 (mod 1)

p p

y ^ y ^         p y .

If p y| 1 then p x| ²  p x| , and the conclusion follows.

Next we assume p| y1 so (y1,y^p1y^p2  ... 1) 1.

So

2

1 2 2

1 ... 1

p p

y a

y ^ y ^ b

  

    

 since y2 and 2 | y so y8，a2. Now the original equation turns into x²(a²1)^2n1 1 (a2)

By the generalization of example 1 we know this equation has no non-negative integral solutions such that n0.

In fact x²y^2n1 1 is a special case of Catalan’s Conjecture：the Diophantine equation x^a y^b 1 has only one solutionx3，y2，a2，b3 if we require the positive integers x，_y，a，b are greater than 1. This conjecture has been solved by Preda Mihăilescu in 2002.

Now we deal with the generalized Catlan’s conjecture： the Diophantine euqation

2 1

m n

x  y  .

When m=n=2, it is just the standard Pell equationx²2y² 1，we have known it has infinitely many solutions with minimal solution (3,2)，general solutions can be given by x_n 2y_n (3 2 2)ⁿ.

When m=p or 2p，p is odd prime，the following several examples show some necessary conditions of the solutions of this equation.

Example 3 x^p2y² 1(x3) (p is odd prime), its any positive integral solution must satisfy p y| .

Proof：(x^p1)(x^p1  ... 1) 2y² x^p1  ... 1 p(mod(x1))

Ifp x| 1，then obviously p y| .

If p x| 1，then (x1,x^p1  ... 1) 1. Since x is odd，x^p1 ... 1 is odd, too.

so x^p1  ... 1 a²，_{x 1 2b}².

The equation turns to (2b²1)^p2y² 1 ，so

1

2 2

(2 , (2 y b  1) )

^p is solution of quasi-Pell equation X²2(2b²1)Y²  2.

The minimal solution of this equation is (X Y₁, )₁ (2 ,1)b ，assume

1

2 2

(2 , (2 1) ) ( , )

p

n n

y b X Y







，

Then by theorem 1.(2) we have

1

2 2

(2 1)

p

b n

  , thus

1

2 2

(2 1)

p

n b

 



Now 2



²



^{2 1}

1

2 2(2 1)

2(2 1)

2

n

n n n

b b

X b Y





 

   ，

So ²¹

2 2

2 1

(2 1)(2 )

2 1 (2 1)

2

n p

n n n

n b

Y n n b Y

 



        ，this is a contradiction.

So p x| 1，p y| .

Completely in parallel, we can prove any positive integral solution of

2 2 1

xp y － (x3) (p is odd prime) must satisfyp y| .

Example 4 x^2p 2y² 1 (x3) (pis odd prime), its any positive integral solution must satisfy p y| .

Proof：x^2py⁴ (y²1)²，and x is oddx^2p 1(mod8)，so y is even.

By results from Pythagoras equation，a b, with one odd and the other even，ab， such that x^p a²b²，y² 2ab，y²1a²b².

so a²b²  1 2ab a b 1.

so x^p 2b1，y² 2 (b b1).

When b is odd，(2(b1), )b 1，so 2(b 1) u²，bv²，yuv， 2 2 1

xp v  ，by Example 3，p v| so p y| .

When b is even，_{(2 , (b b1))1}，so 2bu²，(b 1) v²，_yuv，

x^p2v²  1，by the remark after Example 3，p v| so p y| .

Example 5 The Diophantine equation 3^2n12y² 1 has no positive integral solution when 2n1 is a composite number.

Lemma：a，m，nN^*，则(a^m1,aⁿ 1) a^{( , )m n} 1. Proof of lemma：For m n , by Euclidean algorithm we have，

1 1

mq n r nq r_{2 1}r₂ …

r_k q_k2r_k1r_k2 …

r_s q_s2r_s1 thus r_s1 ( , )m n .

Then we have (a^m1,aⁿ1)

((a^m 1) (aⁿ 1),aⁿ1) (a^ma aⁿ, ⁿ 1)

(a aⁿ( ^{m n} 1),aⁿ 1) (a^{m n} 1,aⁿ1)

…

( a

^{m q n 1}

 1, a

ⁿ

  1) ( a

ⁿ

 1, a

^r1

 1)

…

( a

^r1

 1, a

^r2

 1)

…(a^rs 1,a^rs1 1) 

a

^rs1

 1

a^{( , )m n} 1. The lemma is proven.

Now return to Example 5. Denote prime divisors of 2n1 as p₁，p₂，…，p_k. since 2n1 is a composite number，

So for   1 i k，

2 1

pi n . Now we have

2 1

3 2 2 1

i i

n p

p y

  

 

 

 

  ^{, since}

2 1

3

ⁱ

3

n p





，by conclusion of

Example 3 we know for   1 i k，p y_i| . So 3^2n1 1(modp_i)，  1 i k.

Without losing generality, we can assume p₁ is the minimal prime divisor of 2n1，then p₁ is an odd integer that is greater than 1，and we have

2 1

1| 3 ⁿ 1

p ^  ， p₁| 3^p111 By the lemma，p₁| 3^{(2n1,p11)}1.

Notice that all the prime divisors of p₁1 is smaller than p₁，by our assumption ofp₁, 2n1 cannot divided by them. Sop₁1 is relative prime to 2n1.

Thus (2n1,p₁ 1) 1， p₁| 3 1 2  ， this is a contradiction.

Thus we have proven when 2n1 is a composite number, 3^2n12y² 1 has no positive integral solutions.