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费马大定理-杜明全先生对费马定理的证明


The Proof of Fermat’s Last Theorem
Mingquan Du (Dept. of Math., Xinyang Normal University, 464000, Henan, China)
EMAIL:1456704562@QQ.COM

Abstract: This paper will prove Fermat’s Last Theorem by probabilistic methods and way of contradiction, i.e., when positive integer n > 30, inde?nite equation xn + y n = z n does not have positive integer solution.

Keywords: Fermat’s Last Theorem, probability space, random variable, independent speciality. 1. Introduction In 1637, Fermat a?rm: for n > 2, the equation xn + y n = z n does not have positive integer solution. The literatures [1,2,3] proved that the equation (1) does not have positive integer solution for 3 ≤ n ≤ 125000. By using probabilistic methods and way of contradiction, this paper will prove that the equation (1) does not have positive integer solution,i.e.,we have the following theorem. Theorem 1.1 Suppose n > 30 and n is integer, then the equation xn + y n = z n (2) (1)

does not have positive integer solution. To prove this theorem, we ?rst prove some lemmas in the next section. 2. Some lemmas Lemma 2.1 Suppose 0 < x < 1, then x , (1 ? x)2 m=1 ∞ x2 + x m2 xm = . (1 ? x)3 m=1


mxm =

Lemma 2.2 Let B(b) = 1 + b + (1 ? b)
1?b , 1+b

0 < b < 1,
1 4

then B(b) is a function of up concave of b, and exists is a descending function of b when 0 < b ≤ b0 . Proof. Because B (b) = 1 ? 3 1?b 1 1?b ? · 2 1+b 2 1+b 1 1?b 1?b = 1 ? (3 ? ) 2 1+b 1+b 1?b 2 + 4b = 1? 2(1 + b) 1 + b = 1? B ” (b) = 3 2 1 + 2b 1 ? b , 1+b 1+b 1 3 1+b · + 2 1 ? b (1 + b) 2 1?b 1+b

< b0 <

1 2

, such that B(b)

1 1?b · > 0, 1 + b (1 + b)2

hence the function B(b) is up concave. Because B (1) = 1 ? 4 9√ 15 < 0, 25 4√ B (1) = 1 ? 3 > 0, 2 9

then exists

1 4

< b0 <

1 2

such that B(b) is a descending function of 0 < b ≤ b0 .

Lemma 2.3 Let 0 < b < 1, 0 < a < 2, and A(a; b) = (1 + b)a2 ? 2(1 + b)2 a + 2b(3 + b2 ),
2

then A(a; b) > 0 when B(b) < a < 2. Proof. Easily know that B(b) is a bigger root of the quadratic tri-term form A(a; b), therefore A(a; b) > 0 when B(b) < a < 2. Lemma 2.4 If exists 0 < p < 1, 0 < q < 1, such that a = p + q, b = pq, A(a; b) > 0, then (1 ? p)2 (1 ? q)2 (1 + pq) > 1. (1 ? pq)3 Proof. Because a = p + q, b = pq, 0 < p < 1, 0 < q < 1 and A(a; b) = (1 + b)a2 ? 2(1 + b)2 a + 2b(3 + b2 ) > 0, therefore (1 ? b)3 < (1 + b)a2 ? 2(1 + b)2 a + 2b(3 + b2 ) + (1 ? b)3 = (1 + b)a2 ? 2(1 + b)2 a + 6(1 + b) ? 6 + 2(1 + b ? 1)3 ? (1 + b ? 2)3 = (1 + b)a2 ? 2(1 + b)2 a + 6(1 + b) ? 6 + 2(1 + b)3 ? 6(1 + b)2 + 6(1 + b) ? 2 ? (1 + b)3 + 6(1 + b)2 ? 12(1 + b) + 8 = (1 + b)[a2 ? 2a(1 + b) + (1 + b)2 ] = (1 + b)(a ? 1 ? b)2 , So (1 ? pq)3 < (1 + pq)(p + q ? 1 ? pq)2 = (1 + pq)(1 ? p)2 (1 ? q)2 Hence (1 ? p)2 (1 ? q)2 (1 + pq) > 1. (1 ? pq)3 Lemma 2.5 Suppose the equation xn + y n = z n
3

have a group positive integer solution (x0 , y0 , z0 ), x0 < y0 < z0 , (x0 , y0 , z0 ) = 1, when n > 30 is a positive integer , then 1
?

2

?

x0 n y ) + ( 0 )n = 1, z0 z0 1 y x ( 0 )n < < ( 0 )n ; z0 2 z0 x0 n y0 n set b? = ( ) ( ) , then z0 z0 1 1 0 < b? < < b0 < , 4 2 ( B(b? ) > 2b0 ;
3 1 21? n > B( ); 4

3

?

4

?

there exists 0 < b? < b? , such that 2 > B(b? ) > 21? n ;
3

5 6

?

8(

B(b? ) n ) > 1; 2
z ln y0 0

?

let
B(b ) ln B(b? ) ln 8( 2 ? )n 0 < α < min{ z0 z0 , z z , z z } ln x0 y0 n ln x0 y0 3n ln x0 y0 0 0 0 0

then 0 < α < 1, exists 0 < β < 1 such that α + β = 1, α < β and ( 7
? 1 z0 z0 n z0 n B(b? ) n z0 3n · ) < min{( ) α , B α (b? ), ( ) 3α ( ) 3α } x0 y0 y0 2 x0

There exists constant c such that (
1 z0 z0 n z0 n B(b? ) n z0 3n · ) < c < min{( ) α , B α (b? ), ( ) 3α ( ) 3α } x0 y0 y0 2 x0

and 1 < cα < 2, z0 z0 ?2α · c n > 1, x0 y0 z0 ?α c n > 1; x0 x α 0 < 0 c n < 1, z0 y0 α 0 < c n < 1; z0
4

8

?

9

?

ln b1 γ
0

z z ln x0 y0 c 0 0
?

?2α n

<

2 ln B(b z ln x0 c 0

?) ?α n

ln c( z 0 )n

x

< 3, γ =

0

ln cα ( z 0 )n
0

x

< 1.

Proof. 1 Because (x0 , y0 , z0 ) is a group solution of the equation (2), and
n n x0 < y0 < z0 , hence xn + y0 = z0 . So 0

( and (
?

x0 n y ) + ( 0 )n = 1 z0 z0

x0 n 1 y ) < < ( 0 )n . z0 2 z0

2 Since ( z 0 )n + ( z0 )n = 1, hence
0 0

x

y

b? = ( From Lemma 2.2,
1 4

x0 n y0 n 1 ) ( ) < . z0 z0 4

< b0 < 1 , and B(b) is a descending function of b 2

when 0 < b ≤ b0 , so 1 B(b? ) > B( ) > B(b0 ) > 2b0 . 4 3 Because 1 1 1 B( ) = 1 + + (1 ? ) 4 4 4 so n > 30 > then 2.08 ln 8 ln 8 ln 8 > > 2 > 2 , 0.07 ln 1.08 ln 1.84 ln B( 1 )
4 ?

5 3 3 = + 5 4 4

3 < 1.84, 5

2 ln B( 1 ) 3 4 < n ln 2

therefore
3 1 21? n > B( ). 4

4 From 2 , 0 < b? = ( z 0 z0 )n <
0 0

?

?

x y

1 4

< b0 . From Lemma 2.2,we know B(b)is a

descending function
5

of b when 0 < b ≤ b0 , hence 1 B(b? ) > B( ). 4 From 3 ,if
3 1 2 > 21? n ≥ B(b? ) > B( ), 4 ?

then take 0 < b? < b? , such that
3 1 2 > B(b? ) > 21? n > B(b? ) > B( ), 4

if
3 1 2 > B(b? ) > 21? n > B( ), 4

then take 0 < b? < b? ,such that
3 1 2 > B(b? ) > B(b? ) > 21? n > B( ), 4

therefore there always exists 0 < b? < b? <

1 4

< b0 such that
3

2 > B(b? ) > 21? n . 5 From 4 ,then B(b? ) > 21? n , so 8( 6 Because 0<α< so β =1?α>1? and ( z0 z0 n z0 n · ) < ( )α . x0 y0 y0
6
z ln y0 0 z ln x0 · 0 z0 y0
? 3 ? ?

B(b? ) n ) > 1. 2
z ln y0 0 z ln x0 · 0 z0 y0

< 1,
z ln y0 0 z ln x0 · 0 z0 y0

=

z ln x0 0 z ln x0 · 0 z0 y0

>

>α>0

Because α< then ( Because ln 8( B(b? ) )n 2 α< z z , 3n ln x0 · y0 0 0 then ( z0 z0 n B(b? ) n z0 3n · ) <( ) 3α ( ) 3α , x0 y0 2 x0
1 2
1 z0 z0 n · ) < B α (b? ). x0 y0

ln B(b? ) z , z n ln x0 · y0 0 0

0 where using inequality ( x0 )n < z ? ?

of 1 .

0

7 From 6 , there exists constant c , such that ( hence 1 < c < B α (b? ), cα < B ( b? ) < 2.
0 But from 2 , ( x0 · z ? 1 1 z0 z0 n z0 n B(b? ) n z0 3n · ) < c < min{( ) α , B α (b? ), ( ) 3α ( ) 3α }, x0 y0 y0 2 x0

y0 n ) z0

< 1 , so 4 ( z0 z0 n · ) > 4 > c2α , x0 y0 z0 z0 ?2α · c n > 1. x0 y0

therefore

0 From 1 , ( x0 )n < 1 , so z 2

?

( then

z0 n ) > 2 > cα , x0 z0 ?α c n > 1. x0
7

0 8 From 1 , ( x0 )n < 1 , from 7 , z 2

?

0

0

( then

z0 n ) > 2 > cα , x0 x0 α c n < 1. z0

z From 7 , c < ( y0 ) α , so 0

?

n

cα < ( then

z0 n ) , y0

y0 α c n < 1. z0 9 From 4 , b? < b? . From 2 , B(b? ) > 2b0 . From Lemma 2.2, we know B(b? ) > B(b? ) > 2b0 , so 1 2 < . B(b? ) b0 From 7 , 1 < c( therefore (
? ? ? ? ?

x0 n y0 n x0 ) ( ) < c( )n , z0 z0 z0

2 ln c( x0 )n ( y0 )n 1 ln c( x0 )n ) z0 z0 < ( ) z0 . B(b? ) b0
x y x

From 6 , we know that α + β = 1, 0 < α < β < 1,therefore
2 ( B(b ) )
?

ln cα ( z 0 )n cβ ( z0 )n
0 0 y

< ( b10 )
x

ln c( z 0 )n
0

,
β

2 ( B(b ) )
? ?

ln cβ ( z0 )n
0

? < ( B(b0 ) ) 2b

ln cα ( z 0 )n
0

( b10 )ln c .

From 7 , c > 1, so cα < cβ . Hence ( 2 ln cα ( y0 )n B(b? ) ln cα ( x0 )n 1 ln cβ z0 z0 ) ) < ( ( ) B(b? ) 2b0 b0 x0 n B(b? ) ln cα ( z ) 0 = ( , d) 2b0 1 ln cα ( x0 )n z0 d=( ) , b0
8
ln cβ

where

Since d 1 =( ) b0 b0 where γ= therefore
? ( B(bγ ) ) 2b 0 ? ( B(bγ ) ) n 2b 0 1

1+

ln cβ x ln cα ( z 0 )n 0

=(
x

1 ) b0

x ln c( z 0 )n 0 x ln cα ( z 0 )n 0

=(

1 γ ) , b0

ln c( z 0 )n ln cα ( z 0 )n
0

x

0

< 1,

ln

1 z0 n ( ) cα x0

2 < ( B(b ) )
?

ln
1

1 z0 n ( ) cα y0

, ,

ln

z 1 +ln x0 cα 0 1

2 < ( B(b ) ) n
?

ln

z 1 +ln y0 cα 0

? ? ( B(bγ ) ) n ln cα ( B(bγ ) ) 2b 2 2b0 0

2

ln

z0 x0

2 < ( B(b ) )
?

ln

z0 y0

,
?

2 n 2 n

z z 2 ? ? ln c1 ln B(bγ ) + ln x0 ln B(bγ ) < ln y0 ln B(b ) , α 2b 0 0
2 2b0 0

z z 2 ? ln c1 ln B(bγ ) + ln x0 ln b1 ? ln x0 ln B(b γ α 0 0
2 2b0 0

?)

z 2 < ln y0 ln B(b ) , 0
?

2 n

z z z 2 ? ln c1 ln B(bγ ) + ln x0 ln b1 < ln x0 y0 ln B(b ) , γ α 0 0 0
2 2b0 0 ?

ln

z0 1 ln γ x 0 b0

< ln

z0 z0 2 B(b? ) 2 + ln cα ln ln γ x0 y0 B(b? ) n 2b 2
0
γ

2 z0 z0 2 2 2b0 = ln ln ? ln cα ln x0 y0 B(b? ) n B(b? ) γ z0 z0 2 2 2 2 2 = ln ln ? ln cα ln ? ln cα ln b0 x0 y0 B(b? ) n B(b? ) n z0 z0 2 2 1 1 2 = ln ln ? ln cα ln + ln cα ln γ , x0 y0 B(b? ) n B(b? ) n b0

z (ln x0 ? 0

1 n

z z ln cα ) ln b1 < (ln x0 y0 ? γ 0 0
0

2 n

2 ln cα ) ln B(b ) ,
?

z ln x0 c 0

?α n

z z ln b1 < ln x0 y0 c γ 0 0
0

? 2α n

2 ln B(b ) .
?

From 7 we have ln b1 γ
0

?

z z ln x0 y0 c? n 0 0



<

2 ln B(b z ln x0 c 0

?) ?α n

,γ =

ln c( z 0 )n ln cα ( z 0 )n
0

x

x

0

Then we have proved the ?rst inequality is true.
9

In the following we will prove the second inequality is true also. In fact, from 7 we have
z c < ( B(b? ) ) 3α ( x0 ) 3α , 2 0
n 3n 0

therefore
z c3α < ( B(b? ) )n ( x0 )3n , 2 0 z 2 ( B(b ) )n < ( x0 )3n c?3α , 0
?

2 B(b? ) 2 ln B(b

z < ( x0 )3 c? n , 0 z < 3 ln x0 c? n . 0
α



?)

Then from 7 we have
ln ln
2 B(b? ) z0 ? α c n x0

0

< 3.

Lemma 2.6 Suppose when n > 30 is a positive integer, the equation xn +y n =z n have a group positive integer solution (x0 , y0 , z0 ) , x0 < y0 < z0 , (x0 , y0 , z0 ) = 1 , then there exists 0 < p < 1, 0 < q < 1 , such that 2 > p + q > B(b? ) and b? < pq < bγ 0 Proof. From 9 of Lemma 2.5, we have ln b1 γ
0 0

ln

z0 z0 ? 2α c n x0 y0

<

2 ln B(b

ln

z0 c x0

?) ?α n

,γ =

ln c( z 0 )n ln cα ( z 0 )n
0

x

x

0

Now we take k such that ln b1 γ
0

ln
0

z0 z0 ? 2α c n x0 y 0

<k<

2 ln B(b

ln

?) z0 ? α c n x0

,

From 7 of Lemma 2.5, we have 1 z0 z0 ? 2α k c n) γ < ( b0 x0 y0

10

and 2 z0 α > ( c? n )k , B(b? ) x0 then we have
0 0 bγ > ( x0 c n )k ( y0 c n )k 0 z z α α

and
0 0 0 B(b? ) < 2( x0 c n )k < ( x0 c n )k + ( y0 c n )k . z z z α α α

From 8 of Lemma 2.5 we know
0 0 0 < ( x0 c n )k < ( y0 c n )k < 1 . z z α α

0

Set
0 0 p = ( x0 c n )k , q = ( y0 c n )k , z z α α

then 0 < p < 1, 0 < q < 1 ,and bγ > pq , 0 B(b? ) < p + q < 2. Finally we will prove pq > b? . In fact, from 7 of Lemma 2.5 we know c > 1. From the supposition of this lemma and 9 of Lemma 2.5 we know n > 30 > 3 > k. From 4 of Lemma 2.5 , we know b? < b? . Therefore pq = ( x0 α k y0 α k cn ) ( cn ) z0 z0 x0 y0 k 2αk ) c n = ( z0 z0 x0 y0 k > ( ) z0 z0 x0 y0 n > ( ) = b? > b ? . z0 z0
0 0 0

3. The proof of Theorem 1.1 Suppose the equation (2) has a group positive integer solution (x0 , y0 , z0 ). Without loss of generality , we may suppose x0 < y0 < z0 , (x0 , y0 , z0 ) = 1 . Now in the same probability space we de?ne two independent random variable
11

ξ and η . The random variable ξ obeys the following distribution: P (ξ = (xx0 )n ) = where
0 p = ( x0 c n )k , x = 1, 2, · · · z α

(1?p)2 xpx , p

From 8 of Lemma 2.5 ,0 < p < 1 and from Lemma 2.1 we have
∞ x=1

0

P (ξ = (xx0 )n ) = 1.

The random variable η obeys the following distribution: P (η = (yy0 )n ) = where
0 q = ( y0 c n )k , y = 1, 2, · · · z α

(1?q)2 yq y , q

From 8 of Lemma 2.5 ,0 < q < 1 and from Lemma 2.1 we have
∞ y=1

0

P (η = (yy0 )n ) = 1.

Now we will prove P(


{ξ + η = (zz0 )n }) > 1

z=1

and therefore lead contradiction to complete the proof. In fact, because events {ξ + η = (zz0 )n }, z = 1, 2, · · ·also disjoint each other and
∞ z=1 ∞ x=1

{ξ + η = (zz0 )n } ?

{ξ = (xx0 )n } ,

where events {ξ = (xx0 )n }, x = 1, 2, · · ·also disjoint each other , so


P( = P (( = P(


{ξ + η = (zz0 )n })


z=1 ∞

{ξ + η = (zz0 ) })

n

(
x=1

{ξ = (xx0 )n })) ({ξ = (xx0 )n }))

z=1 ∞ ∞

({ξ + η = (zz0 )n })

z=1 x=1 ∞

=
z=1 x=1

P (ξ = (xx0 )n )P (η = (zz0 )n ? (xx0 )n )

12



z

=
z=1 x=1 ∞

P (ξ = (xx0 )n )P (η = (zz0 )n ? (xx0 )n ) P (ξ = (zz0 )n )P (η = (zz0 )n ? (zx0 )n )
z=1 ∞

≥ ≥
z=1 ∞

P (ξ = (zz0 )n )P (η = (zy0 )n ) (1 ? p)2 z (1 ? q)2 z zp zq p q


=
z=1

(1 ? p)2 (1 ? q)2 = pq

z 2 (pq)z
z=1

(1 ? p)2 (1 ? q)2 (pq)2 + pq = · pq (1 ? pq)3 (1 ? p)2 (1 ? q)2 (1 + pq) = , (1 ? pq)3 where using that (x0 , y0 , z0 ) is a group solution of the equation (2), Lemma 2.1 and that ξ and η are two independent random variables. Set a = p + q, b = pq, from Lemma 2.6 we have 2 > a > B(b? ) and b? < b < b γ . 0 From Lemma 2.2 ,B(b) is a descending function of b when 0 < b ≤ b0 ,therefore 2 > a > B(b? ) > B(b ) . From Lemma 2.3 we have A(a ; b ) > 0, where a = p + q, b = pq. Finally using Lemma 2.4, we have (1 ? p)2 (1 ? q)2 (1 + pq) > 1. (1 ? pq)3 Theorem 1.1 is now completely proved.
13

References
[1] Harold M.Edwards, Fermat’s Last Theorem,Springer-Verlag,1977. [2] Kenneth Ireland and Michael Rosen,A Classical Introduction to Modern to Modern Number Theory, Springer-Verlag,New York, Heidenberg, Berlin,1972. [3] Kenneth H. Rosen, Elementary Number Theory and Its Applications, AddisonWesley Publishing Company,1984. [4] M.Loeve,Probability I,4th edition,Van Nostrant,1963.

14


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