Fermat s Last Theorem Hajime Mashima November 19, 2018 Abstract About 380 years ago, Pierre de Fermat wrote the following idea to Diophantus s Arithme
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1 Fermat s Last Theorem Hajime Mashima November 19, 2018 Abstract About 380 years ago, Pierre de Fermat wrote the following idea to Diophantus s Arithmetica. Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et generaliter nullam in infinitum ultra quadratum potestatem in duos eiusdem nominis fas est dividere cuius rei demonstrationem mirabilem sane detexi. Hanc marginis exiguitas non caperet. Later, this proposition(fermat s Last Theorem) has continued to be a presence, such as the One Ring that appeared in J R R Tolkien s Lord of the Rings. Finally in 1994, it has been proven by Sir Andrew Wiles. However, interesting Fermat s proof is still unknown. Perhaps this is assumed to algebra category. Contents 1 introduction Fermat s Last Theorem Structure of the product Case 1 (p xyz) p = p = p Case 2 (p xyz) Comon(p 3) p = p introduction Fermat Fermat 1
2 1.1 Fermat s Last Theorem Theorem 1 (Fermat s Last Theorem) n, x, y, z x n + y n z n (0 < x < y < z, n 3) x p + y p z p (p 3 x, y, z ) 1.2 Structure of the product Theorem 2 (Fermat s little theorem) A p p A A p 1 1 (mod p) (1) x p + y p z p 0 mod p x p 1 x + y p 1 y z p 1 z 0 mod p (1) x + y z 0 mod p Definition 3 p xyz The Barlow-Abel Equations[1, p.45] x p + y p = (x + y) γ p z p y p = (z y) α p z p x p = (z x) β p L = {(x + y), (z y), (z x)}, R = {γ p, α p, β p } k Proposition 4 p xyz R 1 mod p Proof 5 p = 5 (y + (z y)) 5 = y 5 + 5y 4 (z y) + 10y 3 (z y) y 2 (z y) 3 + 5y(z y) 4 + (z y) 5 z 5 = y 5 + 5y 4 (z y) + 10y 3 (z y) y 2 (z y) 3 + 5y(z y) 4 + (z y) 5 z 5 y 5 = (z y)(5y y 3 (z y) + 10y 2 (z y) 2 + 5y(z y) 3 + (z y) 4 ) (2) ( y + (x + y)) 5 = y 5 + 5y 4 (x + y) 10y 3 (x + y) y 2 (x + y) 3 5y(x + y) 4 + (x + y) 5 x 5 = y 5 + 5y 4 (x + y) 10y 3 (x + y) y 2 (x + y) 3 5y(x + y) 4 + (x + y) 5 x 5 + y 5 = (x + y)(5y 4 10y 3 (x + y) + 10y 2 (x + y) 2 5y(x + y) 3 + (x + y) 4 ) 2
3 (z y) p 1 (x + y) p 1 R 1 mod p Proposition 6 p xyz L R (3) Proof 7 x p + y p = L R (x + y) c L 0 mod c R py p 1 mod c c py L R py p 1 mod c z p x p z p y p Proposition 8 q R (q p ) q 1 mod p (q p) (4) Proof 9 q 1 mod p (q p) q x p q y, z g, h(< q) y g mod q z h mod q z y h g mod q (3) g h mod q (5) y p = (qn 1 + g) p z p = (qn 2 + h) p z p y p = x p q zy Fermat s little theorem (qn 1 + g) p (qn 2 + h) p mod q (6) (qn 1 + g) q 1 (qn 2 + h) q 1 mod q (7) 3
4 q 1 mod p q 1 = pn + k (0 < k < p) (q 1)k p 2 = pn k p 2 + k p 1 p k Fermat s little theorem (q 1)k p 2 1 mod p (7) (qn 1 + g) (q 1)kp 2 (qn 2 + h) (q 1)kp 2 mod q (q 1)k p 2 = pm + 1 (qn 1 + g) pm+1 (qn 2 + h) pm+1 mod q (8) (6) (8), (9) (qn 1 + g) pm (qn 2 + h) pm mod q (9) (qn 1 + g) (qn 2 + h) mod q g h mod q (5) 4
5 1.3 Case 1 (p xyz) Proposition 10 x p + y p = z p p 2 (x + y z) Proof 11 (4) R = q p 1 qp 2 qp 3 q p n = (pk + 1) p q p n = (pk) p + p 2 (...) + 1 q p n 1 mod p 2 R 1 mod p 2 (10) x p + y p z p 0 mod p 2 x p z p y p mod p 2 y p z p x p mod p 2 z p x p + y p mod p 2 (10) x p (z y) 1 mod p 2 y p (z x) 1 mod p 2 z p (x + y) 1 mod p 2 x p + y p z p (z y) + (z x) (x + y) mod p 2 0 2z (x + y) (x + y) mod p 2 0 2z 2(x + y) mod p 2 0 2(x + y z) mod p 2 0 (x + y z) mod p 2 5
6 1.3.1 p = 3 Proposition 12 x 3 + y 3 = z 3 3 xyz Proof 13 (x + (y z)) 3 = x 3 + 3x 2 (y z) + 3x(y z) 2 + (y z) 3 (x + y z) 3 = x 3 + 3x 2 y 3x 2 z + 3x(y 2 2yz + z 2 ) + y 3 3y 2 z + 3yz 2 z 3 x 3 + y 3 z 3 = 0 = x 3 + 3x 2 y 3x 2 z + 3xy 2 6xyz + 3xz 2 + y 3 3y 2 z + 3yz 2 z 3 = x 3 + 3x 2 y + 3xy 2 + 3xz 2 + y 3 + 3yz 2 3x 2 z 6xyz 3y 2 z z 3 = 3x 2 y + 3xy 2 + 3xz 2 + 3yz 2 3x 2 z 6xyz 3y 2 z = 3 ( x 2 y + xy 2 + xz 2 + yz 2 x 2 z 2xyz y 2 z ) = 3 ( xy (x + y) + z 2 (x + y) z ( x 2 + 2xy + y 2)) = 3 (xy (x + y) + z 2 (x + y) z (x + y) 2) = 3(x + y) ( xy + z 2 z (x + y) ) (x + y z) 3 = 3(x + y)(z x)(z y) 3 3 (x + y z) (x + y)(z x)(z y) x + y z 0 mod 3 x + y z mod 3 z x y mod 3 z y x mod 3 3 xyz 6
7 1.3.2 p = 5 Proposition 14 x 5 + y 5 z 5 Proof 15 (x + y z) 5 = 5x 4 y 5x 4 z + 10x 3 y 2 20x 3 yz + 10x 3 z x 2 y 3 30x 2 y 2 z + 30x 2 yz 2 10x 2 z 3 + 5xy 4 20xy 3 z + 30xy 2 z 2 20xyz 3 + 5xz 4 5y 4 z + 10y 3 z 2 10y 2 z 3 + 5yz 4 + x 5 + y 5 z 5 x 5 + y 5 z 5 = 0 (x + y z) 5 = 5x 4 y 5x 4 z + 10x 3 y 2 20x 3 yz + 10x 3 z x 2 y 3 30x 2 y 2 z + 30x 2 yz 2 10x 2 z 3 + 5xy 4 20xy 3 z + 30xy 2 z 2 20xyz 3 + 5xz 4 5y 4 z + 10y 3 z 2 10y 2 z 3 + 5yz 4 = 5 ( x 4 y x 4 z + 2x 3 y 2 4x 3 yz + 2x 3 z 2 + 2x 2 y 3 6x 2 y 2 z +6x 2 yz 2 2x 2 z 3 + xy 4 4xy 3 z + 6xy 2 z 2 4xyz 3 + xz 4 y 4 z + 2y 3 z 2 2y 2 z 3 + yz 4) = 5 ( x 4 y x 4 z + 2x 3 z 2 + 2x 2 y 3 4xy 3 z + 6xy 2 z 2 4xyz 3 + xz 4 +2x 3 y 2 4x 3 yz + 6x 2 yz 2 + xy 4 y 4 z + 2y 3 z 2 2y 2 z 3 + yz 4 2x 2 z 3 6x 2 y 2 z ) = 5 ( x 4 y x 4 z + 2x 3 z 2 + x 2 y 3 xy 3 z + 2xy 2 z 2 2xyz 3 + xz 4 +x 3 y 2 x 3 yz + 2x 2 yz 2 + xy 4 y 4 z + 2y 3 z 2 2y 2 z 3 + yz 4 +x 2 y 3 3xy 3 z + 4xy 2 z 2 2xyz 3 + x 3 y 2 3x 3 yz + 4x 2 yz 2 2x 2 z 3 6x 2 y 2 z ) = 5 ( x ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 ) +y ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 ) +x 2 y 3 + x 3 y 2 3x 3 yz + 4xy 2 z 2 3xy 3 z + 4x 2 yz 2 2xyz 3 2x 2 z 3 6x 2 y 2 z ) = 5 ( (x + y) ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 ) +x 2 y 2 (x + y) 3x 3 yz 6x 2 y 2 z + 4x 2 yz 2 3xy 3 z + 4xy 2 z 2 2xyz 3 2x 2 z 3) 7
8 +xy 3 + xyz 2 xz 3 xz 3) 8 = 5 ( (x + y) ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 ) +x 2 y 2 (x + y) 3x 2 yz(x + y) 3x 2 y 2 z + 4x 2 yz 2 3xy 3 z + 4xy 2 z 2 2xz 3 (x + y) ) = 5 ( (x + y) ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 ) +x 2 y 2 (x + y) 3x 2 yz(x + y) 3xy 2 z(x + y) + 4x 2 yz 2 + 4xy 2 z 2 2xz 3 (x + y) ) = 5 ( (x + y) ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 ) +x 2 y 2 (x + y) 3x 2 yz(x + y) 3xy 2 z(x + y) + 4xyz 2 (x + y) 2xz 3 (x + y) ) = 5 (x + y) ( x 3 y x 3 z + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 2yz 3 + z 4 +x 2 y 2 3x 2 yz 3xy 2 z + 4xyz 2 2xz 3) = 5 (x + y) ( x 3 (z y) + 2x 2 z 2 + xy 3 y 3 z + 2y 2 z 2 yz 3 + z 3 (z y) +x 2 y 2 3x 2 yz 3xy 2 z + 4xyz 2 2xz 3) = 5 (x + y) (( z 3 x 3) (z y) + 2x 2 z 2 + xy 3 y 3 z + y 2 z 2 + y 2 z 2 yz 3 +x 2 y 2 x 2 yz 2x 2 yz 3xy 2 z + 4xyz 2 2xz 3) = 5 (x + y) (( z 3 x 3) (z y) + xy 3 y 3 z + y 2 z 2 + y 2 z 2 yz 3 +x 2 y 2 x 2 yz + 2x 2 z(z y) 3xyz(y z) + xyz 2 2xz 3) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz ) (z y) + xy 3 y 2 z(y z) +y 2 z 2 yz 3 + x 2 y 2 x 2 yz + xyz 2 2xz 3) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z ) (z y) + xy 3 +yz 2 (y z) + x 2 y(y z) + xyz 2 2xz 3) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z yz 2 x 2 y ) (z y) +xy 3 + xyz 2 2xz 3) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z yz 2 x 2 y ) (z y)
9 = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z yz 2 x 2 y ) (z y) +xy 3 + xz 2 (y z) xz 3) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z yz 2 x 2 y ) (z y) +xy 3 xz 3 + xz 2 (y z) ) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z yz 2 x 2 y ) (z y) +x(y 3 z 3 ) + xz 2 (y z) ) = 5 (x + y) (( z 3 x 3 + 2x 2 z + 3xyz + y 2 z yz 2 x 2 y xz 2) (z y) x(z y)(y 2 + yz + z 2 ) ) = 5 (x + y) (z y) ( (z x)(x 2 + xz + z 2 ) +2x 2 z + 3xyz + y 2 z yz 2 x 2 y xz 2 x(y 2 + yz + z 2 ) ) = 5 (x + y) (z y) ( (z x)(x 2 + xz + z 2 ) +2x 2 z + 3xyz + y 2 z yz 2 x 2 y xz 2 xy 2 xyz xz 2) = 5 (x + y) (z y) ( (z x)(x 2 + xz + z 2 ) +2x 2 z 2xz 2 + 2xyz + y 2 z yz 2 x 2 y xy 2) = 5 (x + y) (z y) ( (z x)(x 2 + xz + z 2 ) +2xz(x z) + 2xyz + y 2 (z x) yz 2 x 2 y ) = 5 (x + y) (z y) ( (z x)(x 2 xz + z 2 + y 2 ) +xyz + xyz yz 2 x 2 y ) = 5 (x + y) (z y) ( (z x)(x 2 xz + z 2 + y 2 ) +xyz + yz(x z) x 2 y ) = 5 (x + y) (z y) ( (z x)(x 2 xz + z 2 + y 2 yz) +xyz x 2 y ) = 5 (x + y) (z y) ( (z x)(x 2 + y 2 + z 2 xz yz) + xy(z x) ) 9
10 (x + y z) 5 = 5(x + y)(z y)(z x)(x 2 + y 2 + z 2 + xy xz yz) (x + y z) 2 = x 2 + y 2 + z 2 + 2xy 2xz 2yz (x + y z) 2 (xy xz yz) = x 2 + y 2 + z 2 + (xy xz yz) (11) (x + y z) 5 = 5(x + y)(z y)(z x) ( (x + y z) 2 (xy xz yz) ) x + y z 0 mod 5 x + y z mod 5 z x y mod 5 z y x mod (x + y z) 5, 5 (x + y)(z y)(z x) 5 4 ( (x + y z) 2 (xy xz yz) ) 5 2 (xy xz yz) (11) x 2 + y 2 + z 2 0 mod 5 (12) 5 xyz k = {x, y, z} k 4 1 mod 5 k 2 ±1 mod 5 (12) x 2 + y 2 + z 2 ±1 mod 5 x 2 + y 2 + z 2 ±3 mod 5 10
11 1.3.3 p 7 Proposition 16 x p + y p z p Definition 17 θ x + y z θ xyz Proof 18 x p + y p z p = 0 x p + y p z p x p + yy p 1 zz p 1 (13) x + y z x p + yx p 1 zx p 1 y p 1 z p 1 θ yz θ y p 1, z p 1 k k = k x p 1 ky p 1 x p 1 kz p 1 x p 1 (13) kx p + yky p 1 zkz p 1 kx p + yx p 1 zx p 1 k 1 y p 1 x p 1 z p 1 x p 1 11
12 y p 1 z p 1, y p 1 z p 1 θ yz θ y p 1 z p 1 k k = k x p 1 ky p 1 z p 1 x p 1 (14) x p + yx p 1 zx p 1 x p + yky p 1 z p 1 zky p 1 z p 1 x p + y p kz p 1 z p ky p 1 (15) z p y p + y p kz p 1 z p ky p 1 y p kz p 1 y p z p ky p 1 z p y p (kz p 1 1) z p (ky p 1 1) (16) yy p 1 (kz p 1 1) zz p 1 (ky p 1 1) yky p 1 (kz p 1 1) zkz p 1 (ky p 1 1) 0 < (ky p 1 ) < θ 0 < (kz p 1 ) < θ ky p 1 1, kz p 1 1 (ky p 1 ) (ky p 1 ) 1 (kz p 1 ) (kz p 1 ) 1 ky p skz p 1 (17) kz p tky p 1 (18) y p sz p 1 (19) z p ty p 1 (20) x p + y p z p x p + sz p 1 ty p 1 (19) (20) yz st (21) (15) y p k z p k st k 2 y p z p yz k 2 y p 1 z p 1 1 (14) kx p 1 1 (22) 12
13 yky p 1 (kz p 1 1) zkz p 1 (ky p 1 1) ky p (kz p 1 1) z(k 2 y p 1 z p 1 kz p 1 ) ky p (kz p 1 1) z(kx p 1 kz p 1 ) (22) ky p (kz p 1 1) z(kz p 1 1) y(ky p 1 1) kz p (ky p 1 1) ky p 1 1, kz p 1 1 ky p z kz p y (15) s z t y x p + yx p 1 zx p 1 x p + yy p 1 zz p 1 yy p 1 yx p 1 zz p 1 zx p 1 y(y p 1 x p 1 ) z(z p 1 x p 1 ) (22) y(ky p 1 1) z(kz p 1 1) (23) (16) z p (ky p 1 1) y p (kz p 1 1) z p (y(ky p 1 1)) y p+1 (kz p 1 1) (23) z p (z(kz p 1 1))) y p+1 (kz p 1 1) z p+1 (kz p 1 1) y p+1 (kz p 1 1) (z p+1 y p+1 )(kz p 1 1) 0 (z p+1 y p+1 )(ky p 1 1) 0 13
14 ky p 1 1, kz p 1 1 z p+1 y p+1 (24) s z, t y (19)(20) y p z p z p y p z p+1 zy p (24) y p+1 zy p y z y z s y, t z (17) ky p 1 kz p 1 x p + y p z p x p ky p 1 + y p ky p 1 z p ky p 1 x p ky p 1 + y p kz p 1 z p ky p 1 (15) ky p 1 1 kz p 1 1 z + y 0 z + x 0 (25) 14
15 x p 1 y p 1, x p 1 y p 1 θ xy θ x p 1 y p 1 k k = k z p 1 kx p 1 y p 1 z p 1 (26) xz p 1 + yz p 1 z p xkx p 1 y p 1 + ykx p 1 y p 1 z p x p ky p 1 + y p kx p 1 z p (27) x p ky p 1 + y p kx p 1 x p + y p y p kx p 1 y p x p x p ky p 1 y p (kx p 1 1) x p (ky p 1 1) (28) yy p 1 (kx p 1 1) xx p 1 (ky p 1 1) yky p 1 (kx p 1 1) xkx p 1 (ky p 1 1) 0 < (ky p 1 ) < θ 0 < (kx p 1 ) < θ kx p 1 1, ky p 1 1 (ky p 1 ) (ky p 1 ) 1 (kx p 1 ) (kx p 1 ) 1 ky p skx p 1 (29) kx p tky p 1 (30) y p sx p 1 (31) x p ty p 1 (32) x p + y p z p ty p 1 + sx p 1 z p (31) (32) xy st (33) (27) x p k y p k st k 2 x p y p xy k 2 x p 1 y p 1 1 (26) kz p 1 1 (34) 15
16 yky p 1 (kx p 1 1) xkx p 1 (ky p 1 1) ky p (kx p 1 1) x(k 2 x p 1 y p 1 kx p 1 ) ky p (kx p 1 1) x(kz p 1 kx p 1 ) (34) ky p (kx p 1 1) x(kx p 1 1) y(ky p 1 1) kx p (ky p 1 1) kx p 1 1, ky p 1 1 ky p x kx p y (27) s x t y xz p 1 + yz p 1 z p xx p 1 + yy p 1 z p xz p 1 xx p 1 yz p 1 + yy p 1 x(x p 1 z p 1 ) y(y p 1 z p 1 ) (34) x(kx p 1 1) y(ky p 1 1) (35) (28) y p (kx p 1 1) x p (ky p 1 1) y p ( x(kx p 1 1)) x p+1 (ky p 1 1) (35) y p (y(ky p 1 1)) x p+1 (ky p 1 1) y p+1 (ky p 1 1) x p+1 (ky p 1 1) (y p+1 x p+1 )(ky p 1 1) 0 (x p+1 y p+1 )(kx p 1 1) 0 16
17 kx p 1 1, ky p 1 1 x p+1 y p+1 (36) s x, t y (31)(32) y p x p x p y p (36) x p+1 xy p y p+1 xy p y x (29) (27) s y, t x ky p 1 kx p 1 kx p 1 1 ky p 1 1 (25)(37) x y 0 (37) (x + y z) + (x y) + (z + x) 0 (x + y z) + (2x y + z) 0 3x 0 (x + y z) + (y x) + (z + y) 0 (x + y z) + ( x + 2y + z) 0 3y 0 (x + y z) + (z + x) + (z + y) 0 (x + y z) + (x + y + 2z) 0 3z 0 θ = p ( 5) p xyz p xyz 17
18 1.4 Case 2 (p xyz) Proposition 19 Proof 20 p x, p yz p n x (n 2), p pn 1 L p x, p (z y) (2) ( x p = (z y) py p 1 + R = py p 1 + (p 2)!2! yp 2 (z y) + + ) 1!(p 1)! y(z y)p 2 + (z y) p 1 (p 2)!2! yp 2 (z y) + + 1!(p 1)! y(z y)p 2 + (z y) p 1 p 2 R p y p 1 p 1 R (38) p x p L R p abc Definition 21 z y = a p p p 1 z x = b p x + y = c p p (x + y z) (z x) (x + y) = b p c p (x + y z) x = b p c p 0 mod p (z y) 2x = b p c p 0 mod p p L p R b p c p x p 2 a p p p 1 2x = b p c p 0 mod p 2 (x (z y)) p = x p (p 1)!1! xp 1 (z y) + + 1!(p 1)! x(z y)p 1 (z y) p x p = (z y)pα p ( (x + y z) p = (z y) pα p p 2 x (39) (p 1)!1! xp (p 2)!2! xp 2 (z y) 2 (p 3)!3! xp 3 (z y) 3 + ) 1!(p 1)! x(z y)p 2 (z y) p 1 18
19 K = pα p (p 1)!1! xp !(p 1)! x(z y)p 2 (z y) p 1 (40) (39) x = ap 2 α (x (z y)) p = (z y) K (ap 2 α a p p p 1 ) p = a p p p 1 K a p p 2p (α a p 1 p p 3 ) p = a p p p 1 K p p+1 (α a p 1 p p 3 ) p = K p p+1 K (40), p α p p 1 K p 2 x p 2p 1 (z y) p n x (n 2) p pn x p p pn 1 L x + y z = x (z y) x + y z = p n aα p pn 1 a p x + y z = p n (aα p n(p 1) 1 a p ) p n x + y z (41) 19
20 1.4.1 Comon(p 3) Proposition 22 3 x + y z Proof 23 3 x, 3 yz x p + y p z p 0 mod 3 Fermat s little theorem k 2n 1 mod 3 (k 3) (42) y p z p 0 mod 3 yy p 1 zz p 1 0 mod 3 yy 2n zz 2n 0 mod 3 y z 0 mod 3 x + y z 0 mod 3 3 z, 3 xy 3 xyz x p + y p z p 0 mod 3 (42) x p + y p z p 0 mod 3 xx p 1 + yy p 1 zz p 1 0 mod 3 x + y z 0 mod 3 20
21 Proposition 24 Proof 25 x p + y p = z p (p 3) x + y z = 3 m p n abc x + y z = p n abct θ (41) p T θ T θ x + y z θ = p θ 3 θ 3 Case 1 θ xyz θ z, θ x + y θ xyz ((x + y) z) p = z p + (p 1)!1! zp 1 (x + y) ( x p + y p = (x + y) py p 1 1!(p 1)! z(x + y)p 1 + (x + y) p (p 2)!2! yp 2 (x + y) + (p 2)!2! zp 2 (x + y) 2 + (p 3)!3! zp 3 (x + y) 3 ) 1!(p 1)! y(x + y)p 2 + (x + y) p 1 z p = x p + y p (44) (43) ( (x + y z) p = (x + y) py p 1 (p 2)!2! yp 2 (x + y) + 1!(p 1)! y(x + y)p 2 + (x + y) p 1 (p n abt ) p = (p 1)!1! zp 1 + (p 2)!2! zp 2 (x + y) ) + 1!(p 1)! z(x + y)p 2 (x + y) p 1 ( py p 1 (p 1)!1! zp z y p = 3 (p 2)!2! yp 2 (x + y) + (p 2)!2! zp 2 (x + y) ) 1!(p 1)! z(x + y)p 2 (x + y) p 1 θ py p 1 θ y Leonhard Euler (p 3)!3! zp 3 (x + y) 2 1!(p 1)! y(x + y)p 2 + (x + y) p 1 (p 3)!3! zp 3 (x + y) 2 (43) (44) 21
22 1.4.3 p 5 Proposition 26 3 xyz, x + y z = 3 m p n abc x p + y p z p m > 1 3 z, 3 xy Proof 27 ( (x + y z) p = (x + y) py p 1 (p 1)!1! zp (3 m p n abc) p = c p ( py p 1... ) (3 m p n ab) p = ( py p 1... ) 3 pm (p n ab) p = ( py p 1... ) p 5 3 y 3 z 3 p (x + y) (p 2)!2! yp 2 (x + y) + (p 2)!2! zp 2 (x + y) ) 1!(p 1)! z(x + y)p 2 (x + y) p 1 3 py p 1 3 y p 1 3 y 1!(p 1)! y(x + y)p 2 + (x + y) p 1 (p 3)!3! zp 3 (x + y) 2 m = 0 x + y z = p n abc (45) Proposition x + y z Case.1 3 xyz x + y z = 3p n abc 3x 0 3x 0 mod x 3 xyz 3 xyz x + y z = 3p n abc (46) 22
23 z (x + y) = k > 0 z = x + y + k z p = x p + y p + k p + p(...) z p = x p + y p 0 = k p + p(...) p(...) > 0 0 k p + p(...) Kp n abc = x + y z > 0 (47) (x + y z) 3 = 3(x + y)(z x)(z y) + x 3 + y 3 z 3 (Kp n abc) 3 = 3(x + y)(z x)(z y) + x 3 + y 3 z 3 K 3 p 3n (abc) 3 = 3p pn 1 (abc) p + x 3 + y 3 z 3 K 3 p 3n (abc) 3 3p pn 1 (abc) p = x 3 + y 3 z 3 p 3n (abc) 3 (K 3 3p n(p 3) 1 (abc) p 3 ) = x 3 + y 3 z 3 (47) abc > 0 p 5, (45) K 3 = 1 p 3n (abc) 3 (1 3p n(p 3) 1 (abc) p 3 ) < 0 (48) p 5, (46) K 3 = 3 3 3p 3n (abc) 3 (3 2 p n(p 3) 1 (abc) p 3 ) < 0 (49) X < Y < Z, X n + Y n = Z n X n Y m + Y n+m < Z n+m X n+m + Y n+m < X n Y m + Y n+m < Z n+m X n+m + Y n+m < Z n+m x p + y p = z p (p 5) x 3 + y 3 z 3 > 0 (50) (48)(49) (50) References [1] Laubenbacher R, Pengelley D (2007). Voici ce que j ai trouvé: Sophie Germain s grand plan to prove Fermat s Last Theorem 23
L211 数学と論理学 第１回
L211 数学と論理学第 1 回 プライニングノルベルト preining@jaist.ac.jp http://www.preining.info/jaist/l211/2015j/ 数学と論理学 目的数学的な考え方は現代の科学技術の奥深くまで浸透している そして高度な抽象性を持つ数学の概念が思いもかけない形で応用されることもある 数学と論理学の発展を振り返りこれらのことを明らかにするとともに 現代の数学の流れについてもふれてみたい
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学術俯瞰講義 ~ 数学を創る ~ 第 2 回 Mathematics On Campus ことばを創り 世界を創る 2009.10.15 : このマークが付してある著作物は 第三者が有する著作物ですので 同著作物の再使用 同著作物の二次的著作物の創作等については 著作権者より直接使用許諾を得る必要があります 東京大学教養学部英語部会 ( 編集 ) 東京大学教養学部英語部会 ( 編集 ) On Campus
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