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– Fermat –
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���� ��� �� – Fermat� ��� � ��� � – – p. 1
• Fermat n = 3Euler .
• Fermat
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, � �
.• Euler
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.
1 +1
4+
1
9+
1
16+ · · · =
π2
6
1 +1
16+
1
81+
1
256+ · · · =
π4
90
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���� ��� �� – Fermat� ��� � ��� � – – p. 2
Euler Kummer1630 �40 Fermat
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1753 Euler
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, n = 3, 4
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1816 Paris
� � , Fermat.
1823 Sophie Germain � .
1825 Legendre, Drichlet n = 5 .
1850 Paris
� � , Fermat3000 FF ( 1000 )
.
1857 � Kummer .
���� ��� �� – Fermat� ��� � ��� � – – p. 3
Sophie Germain (1/3)Sophie Germain �
�.1 p
�
.
xp + yp = zp
,xyz 6≡ 0 (mod p)
� � � �
, Fermat Case I
�
. ,
xyz ≡ 0 (mod p)
� �
Case II�
. ���� ��� �� – Fermat� ��� � ��� � – – p. 4
Sophie Germain (2/3)1 p, q
�
.• p mod q p
� �
.• x, y, z
�
,
xp + yp + zp ≡ 0 (mod q)
xyz ≡ 0 (mod q).
� �
p Fermat Case I.
���� ��� �� – Fermat� ��� � ��� � – – p. 5
Sophie Germain (3/3)
� �
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, Legendre ��
.2 p
2p + 1, 4p + 1, 8p + 1, 10p + 1, 14p + 1, 16p + 1
� � , Fermat Case Ip .
���� ��� �� – Fermat� ��� � ��� � – – p. 6
Fermat• Gauss
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���� ��� �� – Fermat� ��� � ��� � – – p. 7
(1/15)l
�
, ζ = exp(2pi/l)
�
.
Z[ζ] := {a0 + a1ζ + a2ζ2 + · · · + al−1ζ
l−1|ai ∈ Z}
( l )
�
.
�
, .
�
ζ
�
� , ζ l = 1.
���� ��� �� – Fermat� ��� � ��� � – – p. 8
(2/15)
�
l = 5
�
.
(ζ − 1)(ζ4 + ζ3 + ζ2 + ζ + 1) = ζ5 − 1 = 0
(1 + ζ + ζ3) + (2ζ + 4ζ3) = 1 + 3ζ + 5ζ3
f(ζ)�
f(x) ζ� � �
.
� �
.
2 f(ζ) , f(ζ i), i = 2, · · · l − 1f(ζ) (
�
� �)
� �
.
���� ��� �� – Fermat� ��� � ��� � – – p. 9
(3/15)1 l = 7, f(x) = x2 + 1
�
, f(ζ) = ζ2 + 1
ζ4 + 1, ζ6 + 1, ζ + 1, ζ3 + 1, ζ5 + 1
�
.3 f(ζ) ,
�
f(ζ)
� �
Nf(ζ).
� �� � �� �� – Fermat� �� � � �� � – – p. 10
(4/15)2 f(ζ)
�
Nf(ζ) = (ζ2+1)(ζ4+1)(ζ6+1)(ζ+1)(ζ3+1)(ζ5+1)
�
.
� �
.ζ i, i = 1, · · · 6 x6 + x5 + x4 + x3 + x2 + x + 1 = 0
,
�
.
x6 + x5 + x4 + x3 + x2 + x + 1
= (x − ζ)(x − ζ2)(x − ζ3)(x − ζ4)(x − ζ5)(x − ζ6)
�
−1
�
, .
Nf(ζ) = 1
� �� � �� �� – Fermat� �� � � �� � – – p. 11
(5/15)4 f(ζ) g(ζ)
f(ζ)g(ζ) = 1
� �
,
� �
. f(ζ)g(ζ)
�
,
f(ζ)h(ζ) = g(ζ)
h(ζ)
� �
.f(ζ)
� � �
.
f(ζ) g(ζ)h(ζ) ⇒
f(ζ) g(ζ) h(ζ)
� �� � �� �� – Fermat� �� � � �� � – – p. 12
(6/15)3 l
�
, ζ = exp(2πi/l)
� �.
�
(1) N(1 − ζ) = N(ζ − 1) = l.
(2) 1 − ζ .
(3) 1 − ζ l
� �
.
l = u(ζ)(1 − ζ)l−1
�
.
� �
u(ζ) .
� �� � �� �� – Fermat� �� � � �� � – – p. 13
(7/15)(1) . .
xl − 1 = (x − 1)(x − ζ) · · · (x − ζ l−1)
� �
xl − 1
x − 1= 1 + x + x2 + · · · + xl−1
�
1 + x + x2 + · · · + xl−1 = (x − ζ) · · · (x − ζ l−1)
, x = 1
� �
N(1 − ζ) = l.
� �� � �� �� – Fermat� �� � � �� � – – p. 14
(8/15)(2) . f(ζ) g(ζ)
f(ζ)g(ζ) ≡ 0 (mod 1 − ζ)
�
.
f(ζ) ≡ f(1) (mod 1 − ζ)
,
f(1)g(1) ≡ 0 (mod 1 − ζ).
f(1), g(1) , 1 − ζ
� � �
l� � � �
.
� �� � �� �� – Fermat� �� � � �� � – – p. 15
(9/15)
�
f(1)g(1) ≡ 0 (mod l)
l ,
f(1) ≡ 0 (mod l) g(1) ≡ 0 (mod l)
�
. �
f(1) ≡ 0 (mod 1−ζ) g(1) ≡ 0 (mod 1−ζ)
� �
,
f(ζ) ≡ 0 (mod 1−ζ) g(ζ) ≡ 0 (mod 1−ζ)
� �� � �� �� – Fermat� �� � � �� � – – p. 16
(10/15)(3) . j 6≡ 0 (mod l)
�
.
1 − ζj = (1 − ζ)(1 + ζ + · · · + ζj−1)
i ij ≡ 1 (mod l)
� �
.
�
�
,
1 − ζ = 1 − ζ ij = (1 − ζj)(1 + ζj + · · · + ζj(i−1))
1 − ζj ,1 − ζ
�
,
1 = (1 + ζ + · · · + ζj−1)(1 + ζj + · · · + ζj(i−1))
�
. � �� � �� �� – Fermat� �� � � �� � – – p. 17
(11/15)
� j 6≡ 0 (mod l) ,
1 + ζ + · · · + ζj−1
.
� �
N(1 − ζ)�
�
� � � ,
l = (1 − ζ)(1 − ζ2) · · · (1 − ζ l−1)
1 − ζj = (1 − ζ)(1 + ζ + · · · + ζj−1)
�
, ,
� �� � �� �� – Fermat� �� � � �� � – – p. 18
(12/15)
u(ζ) = (1 + ζ)(1 + ζ + ζ2) · · · (1 + ζ + · · · + ζ l−2)
=l−2∏
j=1
j∑
k=0
ζk
� � �
,
� �
,
l = u(ζ)(1 − ζ)l−1
. 2
� �� � �� �� – Fermat� �� � � �� � – – p. 19
(13/15)1 α(ζ)
� �
,
Nα(ζ) ≡ 0, 1 (mod l).
. ζj ≡ 1 (mod 1 − ζ)
�
,
α(ζk) ≡ α(1k) ≡ α(1) (mod 1 − ζ),
�
,
Nα(ζ) ≡ α(1)l−1 (mod 1 − ζ)
� �� � �� �� – Fermat� �� � � �� � – – p. 20
(14/15)Nα(ζ) α(1) ,
Nα(ζ) ≡ α(1)l−1 (mod l).
Fermat ,
α(1)l−1 ≡ 0, 1 (mod l)
�
, .
� �� � �� �� – Fermat� �� � � �� � – – p. 21
(15/15)
� �
.1 �(ζ)
�
, η := �/�̄
� �
. η 1 l.
. |η| = 1
�
, η �(ζ i)/�(ζ−i)
�
, 1 .
�
η1 l
� � �
. 2� �� � �� �� – Fermat� �� � � �� � – – p. 22
(1/14)
� �
Z[ζ]
� �
�
, Fermat Case I
� � �
.
.
� �
, l
�
,
xl + yl = zl
xyz 6≡ 0
�
�
. , x, y
�
� �
. .
xl + yl = (x + y)(x + ζy) · · · (x + ζ l−1y) = zl
� �
.
� �� � �� �� – Fermat� �� � � �� � – – p. 23
(2/14)2 x + ζ iy, x + ζ i+k ..
�
.
x + ζ i+ky − (x + ζ iy) = −ζ i(1 − ζk)y
x + ζ i+ky − ζk(x + ζ i)y = (1 − ζk)x
, ζ
1 − ζk = (1 + ζ + · · · + ζk−1)(1 − ζ)
�
1 + · · · + ζk , x, y
� � � �
,
� �
, (1 − ζ) .
� �� � �� �� – Fermat� �� � � �� � – – p. 24
(3/14)
�
1 − ζ x + ζ iy x + ζ i+ky
�
.
� �
,
x + ζ i+2ky − (x + ζ i+ky) = ζ i+k(1 − ζk)y
�
, x + ζ i+2ky (1 − ζ)
�
.
�
� � �
,
x + ζ i+mky,m = 0, 1, · · ·
1 − ζ
� � �
. ζζ l = 1 , ζ i+mk
ζj, j = 0, 1, · · · l − 1
�
.
� �� � �� �� – Fermat� �� � � �� � – – p. 25
(4/14)
�
�
,
xl + yl = (x + y)(x + ζy) · · · (x + ζ l−1y) = zl
(1 − ζ)l
�
. 1 − ζ, z 1 − ζ
� � �
�
z = (1 − ζ)α(ζ)
�
.
� � � �
z,
Nz = z = N(1 − ζ)Nα(ζ)
�
. N(1− ζ) = l , z l
� �
�
.�
xyz 6≡ 0 (mod l) . 2
� �� � �� �� – Fermat� �� � � �� � – – p. 26
(5/14). , Z[ζ]
� � ,
xl + yl = (x + y)(x + ζy) · · · (x + ζ l−1y) = zl
f(ζ)
�
,
x + ζy = �(ζ)f(ζ)l
�
.
� �
�(ζ) .
�
, .
x + ζ−1y = �(ζ−1)f(ζ−1)l
� �� � �� �� – Fermat� �� � � �� � – – p. 27
(6/14)
�
,
�(ζ−1) = ζr�(ζ)
�
,
f(ζ)l ≡ f(ζ−1)l (mod l)
. �
x + ζ−1y = �(ζ−1)f(ζ−1)l
= ζ−r�(ζ)f(ζ−1)l
≡ ζ−r�(ζ)f(ζ)l (mod l)
≡ ζ−r(x + ζy)l (mod l)� �� � �� �� – Fermat� �� � � �� � – – p. 28
(7/14),
ζrx + ζr−1y ≡ x + ζy (mod l)
.
� �
λ = 1 − ζ
�
,
�
,
xλr + (rx + y)λr−1 + · · · ≡ 0 (mod l)
�
.
� �
r �
�
.
� �� � �� �� – Fermat� �� � � �� � – – p. 29
(8/14)• r = 0
• 1 < r < l − 1
• r = l − 1
r = 0 .
� �
,
x + ζ−1y ≡ x + ζy (mod l)
�
,
ζ(1 − ζ)(1 + ζ)y ≡ 0 (mod l)
ζ, 1 + ζ ,
� �� � �� �� – Fermat� �� � � �� � – – p. 30
(9/14)
(1 − ζ)y ≡ 0 (mod l)
� � �
(1 − ζ)y = α(ζ)l
(α ). 1− ζ l, y 1 − ζ
�
.
y , z l
� �
�
�
, y l
�
.
� � �
y 6≡ 0 (mod l).
� �� � �� �� – Fermat� �� � � �� � – – p. 31
(10/14)0 < r < l − 1 .
xλr + (rx + y)λr−1 + · · · ≡ 0 (mod l)
. λ = 1− ζ
� �
.
�
,
xλr + (rx + y)λr−1 + · · · =r∑
k=0
arλk = α(ζ)l
�
. l λ� � �
,a0 λ
� �
. a0,
�
l
� � �
.
a0 ≡ 0 (mod l).
� �� � �� �� – Fermat� �� � � �� � – – p. 32
(11/14)a1
�
arλr + · · · a2λ
2 + a1λ + a0 = α(ζ)l = α(ζ)u(ζ)λl−1
, a0 λl−1
�
.λ2
�
, a1λ
�
,�
a1 ≡ 0 (mod l)
.
� � �
,
ar = x ≡ 0 (mod l)
,�.
� �� � �� �� – Fermat� �� � � �� � – – p. 33
(12/14)r = 1, l − 1 .
ζrx + ζr−1y ≡ x + ζy (mod l)
r = 1 ,�
(1 − ζ)(x − y) ≡ 0 (mod l)
�
. � x − y 1 − ζ
�
, x − y
� �
,
x − y ≡ 0 (mod l)
. r = l− 1 x− y ≡ 0 .
� �� � �� �� – Fermat� �� � � �� � – – p. 34
(13/14),
� �
,
xl + yl = zl
xyz 6≡ 0 (mod l) x, y, z,
x ≡ y (mod l)
� � �
� .� �
xl + yl = zl
.
�
xl + (−z)l = (−y)l
�
,
x ≡ −z (mod l)
.
� �� � �� �� – Fermat� �� � � �� � – – p. 35
(14/14)
�
xl ≡ x (mod l) ,
xl + yl = zl
x + x ≡ −x (mod l)
3x ≡ 0 (mod l)
x 6≡ 0 (mod l)
�
,
3 ≡ 0 (mod l)
� �
, � l = 3�
.
x3 + y3 = z3
� �
� ,
�
. 2
� �� � �� �� – Fermat� �� � � �� � – – p. 36
(1/2)•
�
, �,
� �
Fermat Case I.
• , Z[ζ] � �
!• .
� �� � �� �� – Fermat� �� � � �� � – – p. 37
(2/2)• Kummer Liouville
....
�
�
,
�
,
� �
� �, a0 + a1ζ + · · · an−1ζn−1
�
�
, �
�
� � � � � �
. ,
�
�
, �
�
,
� � �
�
...
� �� � �� �� – Fermat� �� � � �� � – – p. 38
先週の復習EulerからKummerまでSophie Germain の結果(1/3)Sophie Germain の結果(2/3)Sophie Germain の結果(3/3)Fermat の大定理に関する評価円分整数(1/15)円分整数(2/15)円分整数(3/15)円分整数(4/15)円分整数(5/15)円分整数(6/15)円分整数(7/15)円分整数(8/15)円分整数(9/15)円分整数(10/15)円分整数(11/15)円分整数(12/15)円分整数(13/15)円分整数(14/15)円分整数(15/15)ニセの大定理の証明(1/14)ニセの大定理の証明(2/14)ニセの大定理の証明(3/14)ニセの大定理の証明(4/14)ニセの大定理の証明(5/14)ニセの大定理の証明(6/14)ニセの大定理の証明(7/14)ニセの大定理の証明(8/14)ニセの大定理の証明(9/14)ニセの大定理の証明(10/14)ニセの大定理の証明(11/14)ニセの大定理の証明(12/14)ニセの大定理の証明(13/14)ニセの大定理の証明(14/14)まとめ(1/2)まとめ(2/2)