– fermat...– fermat – – p. 20 pell ii (20/30) i = 1 . [p n] = a a1 = n n2;b1 = 2n;c1 = 1. i...
TRANSCRIPT
�
– Fermat –
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)
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– Fermat
� ��� � ��� �
– – p. 1
Pell II (1/30)Pell .
• F0(X0, Y0) = X20 − NY 2
0
� �
.
•
�
F0(X0, Y0) Xi, Yi
�
Fi(Xi, Yi) (i = 1, 2, 3, . . .) .
• Fi(Xi, Yi) = (−1)i Xi = 1, Yi = 0
� �
,
�
Xi, Yi
.
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 2
Pell II (2/30)Fi(Xi, Yi) Fi+1(Xi+1, Yi+1)�
.• Fi(Xi, Yi) = AX2
i −BXiYi −CY 2i ,
At2 − Bt − C = 0 ,mi
�.
• Fi+1(Xi+1, Yi+1) = −Fi(miXi+1 + Yi+1, Xi+1)�
.
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 3
Pell II (3/30)F0(X0, Y0) = X2
0 − 23Y 20�
. F1(X1, Y1),
• t2 − 23 = 0 .• .
.
�m0 = 4
�
.
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 4
Pell II (4/30)F1(X1, Y1)
�
�
,
F1(X1, Y1) = −F0(4X1 + Y1, X1)
= −(4X1 + Y1)2+23X2
1
= 7X2
1 − 8X1Y1 − Y 2
1
� �
(X1, Y1) = (1, 0) F1(X1, Y1) = (−1)1
, �
� �
.
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 5
Pell II (5/30)F2(X2, Y2) ,7t2 − 8t − 1 = 0 ,
�
�
, m1 = 1
�
. �
F2(X2, Y2) = −F1(1 · X2 + Y2, X2)
= −7(X2 + Y2)2+8(X2 + Y2)X2+X2
2
= 2X22 − 6X2Y2 − 7Y 2
2
(X2, Y2) = (1, 0) F2(X2, Y2) = (−1)2
, �
� �
.
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 6
Pell II (6/30)F3(X3, Y3) ,2t2 − 6t − 7 = 0 ,
�
�
, m2 = 3
�
. �
F3(X3, Y3) = −F2(3 · X3 + Y3, X3)
= −2(3X3 + Y3)2+6(3X3 + Y3)X3+7X2
3
= 7X23 − 6X3Y3 − 2Y 2
3
(X3, Y3) = (1, 0) F3(X3, Y3) = (−1)3
, �
� �
.
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 7
Pell II (7/30)Fi(Xi, Yi) i = 4
� �
.
F0(X0, Y0) = X2
0 − 23Y 2
0 , m0 = 4
F1(X1, Y1) = 7X2
1 − 8X1Y1 − Y 2
1 , m1 = 1
F2(X2, Y2) = 2X22 − 6X2Y2 − 7Y 2
2 , m2 = 3
F3(X3, Y3) = 7X23 − 6X3Y3 − 2Y 2
3 , m3 = 1
F4(X4, Y4) = X2
4 − 8X4Y4 − 7Y 2
4 , m4 = 8
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 8
Pell II (8/30)Fi(Xi, Yi) i = 9
�
�
.
F4(X4, Y4) = X2
4 − 8X4Y4 − 7Y 2
4 , m4 = 8
F5(X5, Y5) = 7X2
5 − 8X5Y5 − Y 2
5 , m5 = 1
F6(X6, Y6) = 2X26 − 6X6Y6 − 7Y 2
6 , m6 = 3
F7(X7, Y7) = 7X27 − 6X7Y7 − 2Y 2
7 , m7 = 1
F8(X8, Y8) = X2
8 − 8X8Y8 − 7Y 2
8 , m8 = 8
F9(X9, Y9) = 7X2
9 − 8X9Y9 − Y 2
9 , m9 = 1
���� ��� ��
– Fermat
� ��� � ��� �
– – p. 9
Pell II (9/30)
• Fi(Xi, Yi) = (−1)i Xi = 1, Yi = 0
� �
,
�
Xi, Yi
.
, Fi(Xi, Yi) = (−1)i
�
, (Xi, Yi) = (1, 0)
�
i
i = 4, 8, 12, . . .
�
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 10
Pell II (10/30)
� (X4, Y4) = (1, 0)
�
,
Xi = miXi+1 + Yi+1, Yi = Xi+1
�
,
(X3, Y3) = (X4 + Y4, X4) = (1, 1)
(X2, Y2) = (3X3 + Y3, X3) = (4, 1)
(X1, Y1) = (X2 + Y2, X2) = (5, 4)
(X0, Y0) = (4X1 + Y1, X1) = (24, 5)
� �
, (24, 5)
�
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 11
Pell II (11/30)
�
,,
.1 x2 − 35y2 = 1
.. (x, y) = (6, 1).
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 12
Pell II (12/30)Fi(Xi, Yi)
� � �.
F0(X0, Y0) = X2
0 − 23Y 2
0 , m0 = 4
F1(X1, Y1) = 7X2
1 − 8X1Y1 − Y 2
1 , m1 = 1
F2(X2, Y2) = 2X22 − 6X2Y2 − 7Y 2
2 , m2 = 3
F3(X3, Y3) = 7X23 − 6X3Y3 − 2Y 2
3 , m3 = 1
F4(X4, Y4) = X2
4 − 8X4Y4 − 7Y 2
4 , m4 = 8
F5(X5, Y5) = 7X2
1 − 8X1Y1 − Y 2
1 , m5 = 1
F6(X6, Y6) = 2X2
6 − 6X6Y6 − 7Y 2
6 , m6 = 3
F7(X7, Y7) = 7X27 − 6X7Y7 − 2Y 2
7 , m7 = 1
F8(X8, Y8) = X28 − 8X8Y8 − 7Y 2
8 , m8 = 8
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 13
Pell II (13/30)√
23
�
,�
. (
�
,
�.)
√23 = 4 +
1
1+
1
3+
1
1+
1
8+
1
1+· · ·
�
,
� � �
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 14
Pell II (14/30)Fi(Xi, Yi)
�
mi,� � �
.• Fi(Xi, Yi) 4 .
• mi
�
.
� �
Fi(Xi, Yi),mi� �
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 15
Pell II (15/30)
� �
.
Fi(Xi, Yi) = AX2
i − BXiYi − CY 2
i
,
Gi(Xi, Yi) = CX2i − BXiYi − AX2
i
�
. ,
F1(X1, Y1) = 7X21 − 8X1Y1 − 1Y 2
1
G1(X1, Y1) = 1X21 − 8X1Y1 − 7Y 2
1
F2(X2, Y2) = 2X2
2 − 6X2Y2 − 7Y 2
2
G2(X2, Y2) = 7X2
2 − 6X2Y2 − 2Y 2
2� �� � �� ��
– Fermat
� �� � � �� �
– – p. 16
Pell II (16/30)
� �
Gi(Xi, Yi) = CX2i − BXiYi − AY 2
i ,, Ct2 − Bt − A = 0 ,
ni�
.
�
�
, .
Gi−1(Xi−1, Yi−1) = −Gi(niXi−1 + Yi−1, Xi−1)
�
�
. ,
G1(X1, Y1) = X21 − 8X1Y1 − 7Y 2
1 , n1 = 8
G2(X2, Y2) = 7X2
2 − 6X2Y2 − 2Y 2
2 , n2 = 1
G3(X3, Y3) = 2X2
3 − 6X3Y3 − 7Y 2
3 , n3 = 3
G4(X4, Y4) = 7X2
4 − 8X4Y4 − Y 2
4 , n4 = 1
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 17
Pell II (17/30)
−G4(n4X3 + Y3, X3) = −7(X3 + Y3)2
+8(X3 + Y3)X3 + X2
3
= 2X2
3 − 6X3Y3 − 7Y 2
3
= G3(X3, Y3)
−G3(n3X2 + Y2, X2) = −2(3X2 + Y2)2
+6(3X2 + Y2)Y2 + 7X2
2
= 7X2
2 − 6X2Y2 − 2Y 2
2
= G2(X2, Y2)
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 18
Pell II (18/30)Fi(Xi, Yi)
�
Gi(Xi, Yi)
� �
�
.
Fi = AX2
i − BXiYi − CY 2
i
mi = At2 − Bt − C = 0
Gi = CX2
i − BXiYi − AY 2
i
ni = Ct2 − Bt − A = 0
Fi+1 = Fi(miXi+1 + Yi+1, Xi+1)
Gi−1 = Gi(niXi−1 + Yi−1, Xi−1)
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 19
Pell II (19/30)1 N
�
. Pellx2 −Ny2 = 1 , Fi(Xi, Yi)�
, Fi(Xi, Yi) p
. ,
F1 = Fp+1, F2 = F2+p . . .
.. .
Fi(Xi, Yi) = AiX2i − BiXiYi − CY 2
i
�
�
. i ≥ 1 Ai, Bi, Ci�
, |Ai − Ci| < Bi.B2
i + 4AiCi = 4N .
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 20
Pell II (20/30)i = 1 . [
√N ] = a
� �
A1 = N − n2, B1 = 2n,C1 = 1
�
.
i = k
�
.� �
, Fi+1
,
Ai+1 = −(Aim2i − Bmi − Ci), Bi+1 = 2Aimi − Bi
Ci+1 = Ai
�
.
�
B2i+1 + 4Ai+1Ci+1 = 4N
. mi Ait2 − Bit − C = 0
�
, Ai+1 > 0.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 21
Pell II (21/30)
Bi+1 − (Ai+1 − Ci+1) = Ai(mi + 1)2
−Bi(mi + 1) − Ci
Bi+1 + (Ai+1 − Ci+1) = −Ai(mi − 1)2
+Bi(mi − 1) + Ci
�
. mi Ait2 − Bit − C = 0� � � � �
�
> 0�
, |Ai+1 − Ci+1| < Bi+1
Bi+1 > 0 ..
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 22
Pell II (22/30)Fi = AiX
2i −BiXiYi −CiY
2i
Ai, Bi, Ci .• Ai, Bi, Ci .
• B2i + 4AiCi = 4N
�
(Ai, Bi, Ci)
� � �
. �
i, p ,
Fi = Fi+p
�
.� �
Gi = Gi+p
. � �� � �� ��
– Fermat
� �� � � �� �
– – p. 23
Pell II (23/30)Gi Gi−1
� �
Gi+p Gi+p−1� � � �
,
Gi = Gi+p ⇒ Gi−1 = Gi+p−1 ⇒ Gi−2 = Gi+p−2
� � � �
,
G1 = Gp+1
.
�
F1 = Fp+1
�
,
�
� . 2
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 24
Pell II (24/30)1 Pell
Fi(Xi, Yi)
�
, i
Fi = (−1)i (1, 0)
�
. , Pell
�
.. p ,
F1 = Fp+1
�
.
� �
Fp�
Fp+1�
,
Ap = Cp+1
.
� �
,
Fp+1 = F1 = (N − m20)X
2 − 2m0XY − Y 2
�
, Ap = Cp+1 = 1. � �� � �� ��
– Fermat
� �� � � �� �
– – p. 25
Pell II (25/30)
�
Fp = X2 − BpXY − CpY2
�
, p Fp = (−1)p (1, 0)
�
. p , F2p
�
� �
F2p (1, 0)
�
. 2
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 26
Pell II (26/30)2 N
�
.
�
� √N .
. x2 − Ny2 ,
�Fi, mi .
�
Fi
�
ξi =Bi + 2
√N
2Ai
� �
.� �
.
√N = m0 +
1
ξ1
, ξi = mi +1
ξi+1
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 27
Pell II (27/30)A1 = N − m2
0, B1 = 2m0 ,
1
ξ1
=N − m2
0√N + m0
=√
N − m0
. Bi+1 = 2Aimi − Bi�
1
ξi − mi
=2Ai
Bi − 2Aimi + 2√
N
= − 2Ai
Bi+1 − 2√
N
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 28
Pell II (28/30)
� �
4N 2 = B2i + 4AiCi Ai+1 = Ci
�
1
ξi − mi
=2Ai(Bi+1 + 2
√N)
B2i+1
− 4N
=Bi+1 + 2
√N
2Ai+1
= ξi+1
�
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 29
Pell II (29/30)√
N = m0 +1
ξ1
, ξi = mi +1
ξi+1
√N
√N = m0 +
1
ξ1
= m0 +1
m1+
1
ξ2
= m0 +1
m1+
1
m2+
1
ξ3
= m0 +1
m1+
1
m2+
1
m3+
1
ξ4� �� � �� ��
– Fermat
� �� � � �� �
– – p. 30
Pell II (30/30)
� �
,
√N = m0 +
1
m1+
1
m2+
1
m3+
1
m4+· · ·
�
. pm1 = mp+1 ,
� � �
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 31
Fermat (1/2)
�
Fermat
� � �
� �
,• Fermat .• p = n2 + m2
�
.•
�
Pythagoras .• Pell .
�
,
� � �
,� � � �
.
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 32
Fermat (2/2)
� � � �
� �
,
� � �
� .Fermat
, Fermat .� � �
.• Fermat
� �
� �
� � � �
.•
� �
.• , � .• ,
�
� �
� � � �
� �
� .
� �� � �� ��
– Fermat
� �� � � �� �
– – p. 33
足立恒雄 (1994,2006), フェルマーの大定理―整数論の源流,日本評論社/ちくま学芸文庫,より引用
とこ
ヨーロッパ
足立恒雄、三宅 克哉 (1987), 数論―歴史からのアプローチ,日本評論社/ちくま学芸文庫にて復刊予定,より引用
Fermat (1/2)Fermat , Fermat
Huygens .
, ,,
Fermat. .
– Fermat – – p. 34
りとう
と とこっうう とこ
っェ ユ っ
Fermat (2/2),. ,
,
., ,
...( )...,
,,
,.
. ,. – Fermat – – p. 35
足立恒雄、三宅 克哉 (1987), 数論―歴史からのアプローチ,日本評論社/ちくま学芸文庫にて復刊予定,より引用
こう と つここ
こ とこ とくう
と り とく こ く
くつ と く
っ こと と とこ
ょう とことっ
う とこ ょうう
こ
とっ ょう く り
足立恒雄 (1994,2006), フェルマーの大定理―整数論の源流,日本評論社/ちくま学芸文庫,より引用
とこ
ヨーロッパ
こ
こ
く