Download 直角三角形與同餘數(Congurent Numbers)

直角三角形與同餘數
(Congruent Numbers)
台師大數學系
紀文鎮
2007.10.2.
三邊長都是有理數的直角三角形稱為「有理直角三
角形」(rational right triangle)
a，b，c是有理數
c
b
a2+b2=c2
a
有理直角三角形
Congruent numbers
area
6,5,…
5
3
4
41
6
3
2
20
3
Congruent Number Problem
Find all congruent numbers among squarefree positive integers.
在第十世紀，這個問題就備受數學家的重視。
為什麼稱之 congruent number 呢?
Fibonacci 1225
“Liber Quadratorum” (The Book of squares).
定義：An integer n is called a “congruum” if there is an integer x such that
x2±n are both squares.
i.e. x2-n , x2 , x2+n is a 3-term arithmetic progression of squares with
common difference n.
拉丁文 “Congruere”
Congruum
“to meet together”.
Congruent
定理： 設 n>0
1－1
{ right triangles with area n }
對應


0abc
(a, b, c) a 2  b 2  c 2 , 1 ab  n 


2
Pf:
3-term arithmetic progression of
squares with common difference n
1－1
對應

0  r  s  t


2
2
(r , s, t ) s  r  n 
2
2


t

s

n


ba c ba
(a, b, c)  (
, ,
)
2 2 2
(t  r , t  r ,2s)  (r , s, t )
根據上述定理，
n is a congruent number
存在一個有理平方 s2 使得 s2-n
和 s2+n 都是平方
尋找 Congruent numbers ：
Arab (10th Century) ： 5, 6
Fibonacci (13th Century)： 7
Is 1 a congruent number ?
Fibonacci said “no”
But the first acceptable proof due to Fermat.
定理(Fermat): 1 and 2 are not congruent numbers.
w2=u4±v4
Naïve algorithm:
(1)基礎數論： (尋找 integral right triangles)
Primitive Pythagorean triples:
(k 2  l 2 ,2kl, k 2  l 2 ), k  l  0, (k , l )  1, k  l (mod 2)
(2)Find an integral right triangle, then the square free part n
of its area is a congruent numbers.
背景定理：For n>0, there is a 1-1 correspondence
between the following two sets:

 1－1
0abc
(a, b, c) a 2  b 2  c 2 , 1 ab  n  對應


2
( x, y) y
2
2
2
nb 2n
(a, b, c)  (
,
)
ca ca
2
2
2
2
x  n 2nx x  n
(
,
,
)  ( x, y )
y
y
y
n is congruent
y 2  x3  n2 x

 x  n x, y  0
3
has a rational solution (x,y) with y≠0.
方程式
y 2  x3  n 2 x  x( x  n)( x  n), n  0
定義了一條橢圓曲線(elliptic curve)
En: y2=x(x+n)(x-n), n :squarefree positive integer.
定理： En(Q)tors = {(0,0), (n,0), (-n,0), ∞}
定理：n is congruent if and only if there is (x,y) in En(Q) with y≠0.
if and only if rank(En(Q)) ≧1.
In other words, En(Q) is infinite.
Corollary : If there is one rational right triangle with area n,
then there are infinitely many.
Corollary: If there is a right triangle with
rational sides and area n, then L(En, 1) = 0.
反之，若 B-SD conjecture 成立，則 L(En,1)=0 implies n is congruent.
定理(1983)
猜測：If n is positive, squarefree, and n≡ 5, 6, or 7 (mod 8), then
there is a rational right triangle with area n.
This has been verified for n <1,000,000
Serre’s Conjecture
Serre
Ribet
T-W conjecture
FLT
A. Wiles proved T-W conjecture, hence proved FLT.
Summer School on Serre's Modularity Conjecture
Luminy, July 9-20, 2007
今年7月在法國的學術會議證實：
印度人 Chandrashekhar khare,及法國人 Jean –Pierre Wintenberger
兩人已證明了 Serre’s conjecture.
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