Fermat – · Euler Kummer 1630 40 Fermat... 1753 Euler , n = 3;4 1816 Paris , Fermat. 1823 Sophie...

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  • – Fermat –

    �� � � � �� � �� � �(

    )

    ���� ��� �� – Fermat� ��� � ��� � – – p. 1

  • • Fermat n = 3Euler .

    • Fermat

    , � �

    .• Euler

    .

    1 +1

    4+

    1

    9+

    1

    16+ · · · =

    π2

    6

    1 +1

    16+

    1

    81+

    1

    256+ · · · =

    π4

    90

    � � �

    ( ).

    ���� ��� �� – Fermat� ��� � ��� � – – p. 2

  • Euler Kummer1630 �40 Fermat

    � � � �

    ...

    1753 Euler

    , n = 3, 4

    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)

    � �

    � �

    , Legendre ��

    .2 p

    2p + 1, 4p + 1, 8p + 1, 10p + 1, 14p + 1, 16p + 1

    � � , Fermat Case Ip .

    ���� ��� �� – Fermat� ��� � ��� � – – p. 6

  • Fermat• Gauss

    ... �� �

    Fermat

    �.

    , ,

    � � � �,

    ,

    , �

    ...

    • Kummer

    ...Fermat

    � � �

    ,� �

    ...

    ......

    ���� ��� �� – 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)