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基礎數論

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數論 基 數 數 ,

數 . 基 , 數論

. 數論 ( ) ,

Silverman A Friendly Introduction to Number Theory (Prentice Hall, Third Edition

2006). .

v

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Chapter 7

略談 Diophantine Equations

Diophantine equations . 言

f (x1, . . . , xn) 數 數 , f (x1, . . . , xn) = 0 數 Dio-

phantine equation . 數 , Diophantine equations

congruence equations . 前 論

論 Diophantine equations. 前

, 談論 Diophantine equations.

7.1. Diophantine Equations

Diophantine equations . Diophantine equations .

congruence . Diophantine equation

f (x1, . . . , xn) = 0 數 , m∈ N modulo m f (x1, . . . , xn)≡ 0 (mod m)

. m f (x1, . . . , xn)≡ 0 (mod m) , Diophantine

equation f (x1, . . . , xn) = 0 .

Proposition 7.1.1. f (x1, . . . , xn) n 數 數 . m∈ N f (x1, . . . , xn)≡ 0 (mod m) , f (x1, . . . , xn) = 0 數 .

Proof. x1 = c1, . . . , xn = cn f (x1, . . . , xn) = 0 數 . f (c1, . . . , cn) = 0, f (c1, . . . , cn)≡ 0 (mod m), x1 ≡ c1 (mod m), . . . , xn ≡ cn

(mod m) f (x1, . . . , xn)≡ 0 (mod m) . f (x1, . . . , xn)≡ 0 (mod m)

f (x1, . . . , xn) = 0 數 . 

89

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Proposition 7.1.1 m∈ N f (x1, . . . , xn)≡ 0 (mod m) , f (x1, . . . , xn) = 0 數 . m∈ N f (x1, . . . , xn)≡ 0 (mod m)

, f (x1, . . . , xn) = 0 數 . , .

Example 7.1.2. Diophantine equation 11x2− 7y2= 2. modulo 11 ,

−7y2≡ 2 (mod 11). −7 × 3 ≡ 1 (mod 11), −7y2≡ 2 (mod 11) 3 y2≡ 6 (mod 11). Legendre symbol

( 6 11

)

= ( 2

11 )( 3

11 )

. 11≡ 3 (mod 8) Theorem 5.4.3

( 2 11

)

=−1 Theorem 5.4.6

( 3 11

)

= (11

3 )

= (2

3 ) ( = 1.

6 11

)

=−1, y2≡ 6 (mod 11) . 言 , 11x2− 7y2− 2 ≡ 0 (mod 11) , Proposition 7.1.1 11x2− 7y2= 2 數 .

11x2− 7y2= 2 modulo 7 , 11x2≡ 2 (mod 7).

11× 2 ≡ 1 (mod 7), 11x2≡ 2 (mod 7) 2 x2≡ 4 (mod 7). x≡ 2

(mod 7) , 前 11x2− 7y2= 2 數 . ,

m∈ N f (x1, . . . , xn)≡ 0 (mod m) , 言 f (x1, . . . , xn) = 0 . , 數 m, f (x1, . . . , xn)≡ 0 (mod m) ,

f (x1, . . . , xn) = 0 數 ? , .

Example 7.1.3. f (x) = (x2−17)(x2−19)(x2−323) Diophantine equation f (x) = 0.

Diophantine equation 數 . , m∈ N, f (x) ≡ 0

(mod m) .

Corollary 4.4.3 m∈ N, f (x) ≡ 0 (mod m) ,

數 p n∈ N, f (x) ≡ 0 (mod pn) .

p = 2, n = 1 , f (x)≡ (x2−1)3 (mod 2), f (x)≡ 0 (mod 2) . p = 2, n = 2 , f (x)≡ (x2− 1)(x2− 3)2 (mod 4), f (x)≡ 0 (mod 4) . p = 2, n≥ 3 , 17≡ 1 (mod 8), Proposition 5.2.1 x2≡ 17 (mod 2n) , f (x) = (x2− 17)(x2− 19)(x2− 323) ≡ 0 (mod 2n) .

p = 17 17≡ 1 (mod 8), Theorem 5.4.3 x2≡ 19 ≡ 2 (mod 17) . Proposition 5.2.4 n∈ N, x2 ≡ 19 (mod 17n) . f (x)≡ 0 (mod 17n) . p = 19 , 17≡ 1 (mod 8)

(17 19

)

= (19

17 )

= ( 2

17 )

= 1, x2≡ 17 (mod 19) . Proposition 5.2.4 n∈ N, x2≡ 17 (mod 19n) . f (x)≡ 0 (mod 19n) .

pp̸= 17,19 , x2≡ 17 (mod p) , Proposition 5.2.4 n∈ N, x2≡ 17 (mod pn) . f (x)≡ 0 (mod pn) .

x2≡ 19 (mod p) , n∈ N, f (x) ≡ 0 (mod pn) . x2≡ 17 (mod p) x2≡ 19 (mod p) ,

(17 p

)

= (19

p )

=−1,

(232 p

)

= (17

p )(19

p )

= 1

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7.1. Diophantine Equations 91

x2≡ 232 (mod p) , n∈ N, x2≡ 232 (mod pn) . f (x)≡ 0

(mod pn) .

, 數 p n∈ N, f (x) ≡ 0 (mod pn) .

m∈ N, f (x) ≡ 0 (mod m) . f (x) = 0 數 .

, congruence Diophantine equation .

Diophantine equation 數 , congruence

. m∈ N modulo m ,

Diophantine equation 數 . Diophantine equation ,

congruence , Diophantine equation

.

descent . Diophantine equation

數 . 數 well-ordering principle. :

Diophantine equation f (x1, . . . , xn) = 0 數 x1= c1, . . . , xi= ci, . . . , xn= cn . x1= c1, . . . , xi= ci, . . . , xn= cn 數 數 x1= c1, . . . , xi= ci, . . . , xn= cn, i∈ {1,...,n} ci< ci, x1= c1, . . . , xi= ci, . . . , xn= cn 數 數 x1= c′′1, . . . , xi= c′′i, . . . , xn= c′′n

c′′i < ci. 數數 ci> ci> c′′i >···

數 well-ordering principle , f (x1, . . . , xn) = 0 數 .

descent Diophantine equation 數 .

descent .

Example 7.1.4.

2 數, x2− 2y2= 0 diophantine

equation 數 . descent x2− 2y2= 0 數 .

x = c1, y = d1 x2− 2y2= 0 數 . c21= 2d12, c1

數, c2∈ N c1= 2c2. 4c22= 2d12, 2c22= d12. d1 數, d2∈ N d1= 2d2. 2c22= 4d22, c22= 2d22. x = c2, y = d2

x2− 2y2= 0 數 . x = c1, y = d1x = c2, y = d2

c1= 2c2> c2, descent x2− 2y2= 0 數 .

. descent Diophantine equation

數 “ ” 數 , Diophantine

equation 數 . 數 “ ” 數

論. x = 8, y = 6, z = 10 x2+ y2= z2 數 , x, y, z 2 x = 4, y = 3, z = 5 x2+ y2= z2 數 ,

, 論. x2+ y2= z2 數 ,

descent .

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7.2. Pythagorean Triple Fermat’s Last Theorem .

數 數 Pythagorean triple. 言 , Diophantine

equation x2+y2= z2 數 Pythagorean triple. Pythagorean

triples Fermat’s Last Theorem .

7.2.1. Pythagorean Triples. Pythagorean triples.

, Pythagorean triple

Pythagorean triple . Pythagorean triples

Pythagorean triples Pythagorean triples.

? x2+ y2= z2 數 . x = 0

y = 0 x2= z2 y2= z2 Diophantine equation ;

數 數 , 數 .

, x2+ y2= z2

數 . x = 3, y = 4, z = 5 數 , λ ∈ N, x = 3λ,y = 4λ,z =

5λ 數 . Pythagorean triples.

Pythagorean triple . Pythagorean

triple “ ” Pythagorean triple. 數 1

Pythagorean triple. Pythagorean triple 數

1 Pythagorean triple. x = a, y = b, z = c Pythagorean triple gcd(a, b, c) = d, a, b, c∈ N a = da, b = db, c = dc gcd(a, b, c) = 1.

a2+ b2= c2 a′2+ b′2= c′2, x = a, y = b, z = c 數 1

Pythagorean triple. 數 1 Pythagorean triple .

x = a, y = b, z = c 數 1 Pythagorean triple,

x = b, y = a, z = c 數 1 Pythagorean triple, x, y

. x, y .

x > y . .

. 數 Pythagorean triple x, y

數, z2= x2+ y2 z 數, x, y, z 數 2.

x, y 數. x, y 數, x2≡ y2≡ 1 (mod 4).

x2+ y2≡ 2 (mod 4), x2+ y2 數, z2= x2+ y2 z 數. z2≡ 0 (mod 4). 0≡ z2≡ x2+ y2≡ 2 (mod 4) . Pythagorean

triple 數 1, x y . Pythagorean

triple xy 數. Pythagorean triples

.

Definition 7.2.1. a, b, c∈ N a2+ b2= c2 gcd(a, b, c) = 1 ab 數, a, b, c primitive Pythagorean triple.

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7.2. Pythagorean Triple Fermat’s Last Theorem 93

primitive Pythagorean triples. primitive Pythagorean

triples , primitive Pythagorean triples

, primitive Pythagorean triples .

Theorem 7.2.2. primitive Pythagorean triple x, y, z m, n∈ N

m > n, gcd(m, n) = 1 m, n 數 數

x = m2− n2, y = 2mn, z = m2+ n2.

m, n∈ N m > n, gcd(m, n) = 1 m, n數,

x = m2− n2, y = 2mn z = m2+ n2, x, y, z primitive Pythagorean triple.

Proof. x, y, z primitive Pythagorean triple. x2+ y2= z2, y2= (z + x)(z− x). yx, z 數, y/2,(z + x)/2 (z− x)/2(y/2)2= ((z + x)/2)((z−x)/2). (z + x)/2 (z−x)/2 . 數 p (z + x)/2 (z−x)/2 數, p (z + x)/2 + (z−x)/2 = z (z + x)/2−(z−x)/2 = x

數. p|y2= z2− x2, p|y, gcd(x, y, z) = 1 .

(z + x)/2 (z− x)/2 , (y/2)2= ((z + x)/2)((z− x)/2) (z + x)/2

(z− x)/2 數 , m, n∈ N m2= (z + x)/2 n2= (z− x)/2.

x = m2− n2, y = 2mn, z = m2+ n2.

m n, x > 0, m2− n2> 0, m > n. (z + x)/2 (z− x)/2 , gcd(m2, n2) = 1, gcd(m, n) = 1. x = m2− n2 數, m n

數 數.

m, n∈ N m > n, gcd(m, n) = 1 m, n 數 數,

x = m2− n2, y = 2mn z = m2+ n2, x, y, z∈ N x2+ y2= z2, x, y, z Pythagorean triple. primitive, gcd(x, y, z) = 1, x

y 數. y = 2mn y 數, m, n 數 數

x = m2− n2 數. gcd(x, y, z) = 1 gcd(x, y, z) > 1, gcd(x, y, z)

數 ( x 數) 數 p x, y, z 數. p z + x = 2m2

p z− x = 2n2 p 數, p|m p|n. m, n ,

gcd(x, y, z) = 1. 

Theorem 7.2.2 primitive Pythagorean triple

數 m, n , 數 primitive Pythagorean triple.

m, n primitive

Pythagorean triples, m, n Pythagorean triple.

m, n m, nm > n m> n, m, n m, n

primitive Pythagorean triple. m2−n2= m′2−n′2 m2+ n2= m′2+ n′2. 2m2= 2m′2, m, mm = m. n = n.

.

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Corollary 7.2.3. primitive Pythagorean triple.

primitive Pythagorean triple m, n∈ N m > n, gcd(m, n) = 1 m, n 數 數 x = m2− n2, y = 2mn, z = m2+ n2.

7.2.2. Fermat’s Last Theorem. x2+ y2= z2 數 ,

x3+ y3= z3 數 , 3 數 n, xn+ yn= zn 數 .

Fermat n≥ 3 xn+ yn= zn 數 .

, Fermat’s Last Theorem.

Fermat’s Last Theorem conjecture ( )

. 數 , 1995

. 數 論, Fermat

. Diophantine equation 論 數 ,

Diophantine equation 數 .

Fermat’s Last Theorem 3 數. n

p, n = pm, x = a, y = b, z = c xn+ yn= zn 數 , apm+ bpm= cpm x = am, y = bm, z = cm xp+ yp= zp 數 . 言 xp+ yp= zp

數 , n = pm, xn+ yn= zn 數 . n 數, n = 2r,

r≥ 2 4|n, x4+ y4= z4 數 , n = 2r> 2, xn+ yn= zn

數 . Fermat’s Last Theorem, 數 p, xp+ yp= zp

數 , x4+ y4= z4 數 . 前 數 ,

descent x4+ y4= z4 數 .

x4+ y4= z4 Diophantine equation.

Proposition 7.2.4. x4+ y4= z2.

Proof. descent x4+ y4= z2 數 . x = a1, y = b1, z = c1

x4+ y4= z2 數 , 數 x = a2, y = b2, z = c2

c1> c2. 數 well-ordering principle , 數 .

x = a1, y = b1, z = c1 x4+ y4= z2 數 . gcd(a1, b1) = d > 1, d|a1 d|b1 d4|a41+ b41= c21, d2|c1. x = a1/d, y = b1/d, z = c1/d2 x4+ y4= z2

c1/d2< c1.

x = a1, y = b1, z = c1 x4+ y4= z2gcd(a1, b1) = 1.

gcd(a21, b21, c1) = 1 ( gcd(a21, b21) = 1) x = a21, y = b21, z = c1 x2+ y2= z2. 前

論 primitive Pythagorean triple , a21b21 數,

x = a21, y = b21, z = c1 primitive Pythagorean triple. Theorem 7.2.2 m, n∈ N m > n gcd(m, n) = 1

a21= m2− n2, b21= 2mn, c1= m2+ n2.

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7.3. 95

gcd(a1, m, n) = 1 ( gcd(m, n) = 1), x = a1, y = n, z = m x2+ y2= z2, a1

數 前 論 primitive Pythagorean triple n 數 ( m

數), x = a1, y = n, z = m primitive Pythagorean triple.

Theorem 7.2.2 u, v∈ N u > v gcd(u, v) = 1 a1= u2− v2, n = 2uv, m = u2+ v2.

, b1 n 數, b1= 2b1 n = 2n. b21= 2mn

b′21 = mn. gcd(m, n) = 1 m n 數 , c2, e∈ N m = c22 n= e2. 2e′2= 2n= n = 2uv gcd(u, v) = 1, u v

數 , a2, b2∈ N u = a22 v = b22. m = u2+ v2 c22= (a22)2+ (b22)2 x = a2, y = b2, z = c2 x4+ y4= z2 數 . c1= m2+ n2> m2= c42, x = a2, y = b2, z = c2 x4+ y4= z2c2< c1. descent

. 

Proposition 7.2.4 x4+ y4= z2 數 ,

x4+ y4= z4 數 . x = a, y = b, z = c x4+ y4= z4 數 , x = a, y = b, z = c2 x4+ y4= z2 數 . Proposition 7.2.4 ,

論.

Corollary 7.2.5. x4+ y4= z4. 7.3.

數論 , 數 數

. 數 數 ,

數 數 .

7.3.1. Sum of Two Squares. 數 數 ,

數 . n 數 , m∈ N n = m2,

n n = m2+ 02. 數 數

數 .

:

(a2+ b2)(c2+ d2) = (ac + bd)2+ (ad− bc)2. (7.1)

, 數 .

z1= a + bi, z2= d + ci∈ C ( C 數 , i∈ C i2=−1). z1, z1z1= a− bi,z2= d−ci |z1|2= z1z1= a2+ b2 |z2|2= z2z2= c2+ d2.

(a2+ b2)(c2+ d2) = z1z1z2z2= z1z2z1z2=|(ad − bc) + (ac + bd)i|2= (ac + bd)2+ (ad− bc)2.

(7.1) .

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Lemma 7.3.1. m, n∈ N, mn.

Proof. m = a2+ b2 n = c2+ d2, a, b, c, d∈ Z, (7.1) mn = (ac +

bd)2+ (ad− bc)2. ac + bd, ad− bc ∈ Z mn 數 . 

Lemma 7.3.1 m, n, mn

數 ; m, n, mn

數 .

1 數 數 , Lemma 7.3.1

數 數 數 . 2 = 12+ 12, 2

數 , 數 . Lemma 7.3.1

數 數 .

Lemma 7.3.2. p 數. a, b∈ Z a2+ b2=λ p, λ ∈ N

λ < p, p.

Proof. S ={s ∈ N | u, v∈ Z u2+ v2= sp}. S , p

數 1∈ S. 1∈ S ? S

( λ ∈ S) S 數, 1∈ S S 1 (

數 well-ordering principle, S S ). m∈ S

S , m = 1.

, m̸= 1. λ ∈ S λ < p 1 < m < p. S m

數 . m∈ S, u, v∈ Z u2+ v2= mp, m

m 數 論.

(I) m 數: u2+ v2= mp 數, u, v ( u2+ v2

數), u + v u− v 數. (u + v)/2 (u− v)/2

(u + v

2 )2+ (u− v

2 )2= u2 2 +v2

2 =m 2p.

m/2∈ S m/2 < m, m S .

(II) m 數: m

{−m + 1

2 ,−m + 1

2 + 1, . . . , 0, 1, . . . ,m− 1

2 − 1,m− 1 2 }

complete residue system modulo m. c, d∈ Z c≡ u (mod m)

d≡ v (mod m), −(m − 1)/2 ≤ c,d ≤ (m − 1)/2. c d 0,

c = d = 0 u≡ v ≡ 0 (mod m), m|u m|v. m2|u2+ v2= mp, m|p. 1 < m < p , c d 0. c2+ d2≡ u2+ v2 (mod m)

u2+ v2 = mp, c2+ d2≡ 0 (mod m). k∈ Z c2+ d2= km.

c d 0, k̸= 0. −(m − 1)/2 ≤ c,d ≤ (m − 1)/2,

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7.3. 97

c2+ d2≤ (m − 1)2/4 + (m− 1)2/4 = (m− 1)2/2 < m2, 0 < k < m. k∈ N

k < m. : u2+ v2= mp c2+ d2= km. (7.1)

(uc + vd)2+ (ud− vc)2= m2k p.

u≡ c (mod m) v≡ d (mod m),

uc + vd≡ u2+ v2≡ 0 (mod m) and ud − vc ≡ uv − uv ≡ 0 (mod m).

(uc + vd)/m∈ Z (ud− vc)/m ∈ Z.

(uc + vd

m )2+ (ud− vc

m )2= k p,

k p 數 , k∈ N k∈ S. k < m,

m S .

m̸= 1 m 數 數 .

m̸= 1 , m = 1. p 數 . 

Lemma 7.3.2 descent

. 論 , ; .

descent 論 , 數

數 , 論.

m 數 , m > 1 . ,

m = 1, . , .

Lemma 7.3.2 , .

. .

Example 7.3.3. p = 89. 89≡ 1 (mod 4), a∈ Z a2≡ −1 (mod 89). a = 34 , a2= 1156≡ −1 (mod 89), (34)2+ 1 = 13× 89.

13 < 89 Lemma 7.3.2 89 數 . Lemma 7.3.2

89 數 .

Lemma 7.3.2 , 13∈ S. 13̸= 1, 13, S

. c, d 34≡ c (mod 13), 1 ≡ d (mod 13) −6 ≤ c,d ≤ 6.

c =−5 d = 1. c2+ d2= 25 + 1 = 26 = 2× 13. (7.1) (34× (−5) + 1)2+ (34− (−5))2= 1692+ 392= 2× 132× 89.

169 = 13× 13 39 = 3× 13, 132+ 32= 2× 89, 2∈ S.

13∈ S 2∈ S. 數 , 2 .

(13 + 3

2 )2+ (13− 3

2 )2= 82+ 52= 89.

Example 7.3.3 89 ≡ 1 (mod 4) a∈ Z a2 ≡ −1

(mod 89). a ( a ) a1+ 1 =λ p, 0 <λ < p,

Lemma 7.3.2. , pp≡ 1 (mod 4) , .

.

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Proposition 7.3.4. pp≡ 1 (mod 4), p. Proof. p≡ 1 (mod 4), Theorem 5.4.1 x2≡ −1 (mod p) .

a∈ N a2≡ −1 (mod p). {1,2,..., p − 1} reduced residue system modulo

p, 1≤ a ≤ p − 1 a2≡ −1 (mod p) ( 1 < a < p/2).

λ ∈ N a2+ 1 =λ p. a≤ p − 1, λ p = a2+ 1≤ (p − 1)2+ 1 =

p2− 2(p − 1) < p2. λ < p, Lemma 7.3.2 p

. 

數 數 . p≡ 1 (mod 4) ,

Proposition 7.3.4 x2≡ −1 (mod p) p

. p≡ 3 (mod 4) , x2≡ −1 (mod p) p

數 .

Lemma 7.3.5. pp≡ 3 (mod 4) n∈ N p|n. a, b∈ Z a2+ b2= n, p|a p|b.

Proof. . , p- a. a2+ b2= n p|n

p- b, a2= n− b2 p|a2 p- a . a2+ b2= n p|n,

a2≡ −b2 (mod p). a, b p , Legendre symbol

. Legendre symbol (Lemma 5.3.2) 1 =

(a2 p

)

= (−b2

p )

= (−1

p )(b2

p )

= (−1

p )

. p≡ 3 (mod 4) Theorem 5.4.1

(−1 p

)

=−1. p|a,

b2= n− a2 p|b. 

Lemma 7.3.5 Proposition 7.1.1 , x2+ y2= n

Diophantine equation modulo p Diophantine equation .

Lemma 7.3.5 .

Proposition 7.3.6. pp≡ 3 (mod 4), p.

Proof. . a, b∈ Z a2+ b2= p. p 數, a, b

0. a, b∈ N 1≤ a ≤ p − 1 1≤ b ≤ p − 1. a, b p

Lemma 7.3.5 . p 數 . 

數 數 , 數 數 .

數 數 . 數 n. n = 1

數 . n≥ 2, n 數 . 2

4 1 數 略, 數 . n 4 3

數, 略, 數 .

數 數 , : n = pn11··· pnrr, pi

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7.3. 99

數. pi nni pi 數. 2250 = 2× 32× 53,

2250 2 3 數 3 5 數. .

Theorem 7.3.7. n∈ N. nn

4 3 數 數 數.

Proof. n 4 3 數 數 數. n≥ 2

. n

n = 2n0qn11···qnrr· p2m1 1··· p2ms s,

qi, pjqi≡ 1 (mod 4) pj≡ 3 (mod 4). n

n = 2n0qn11···qnrr· (pm11··· pmss)2,

2 數 , q1, . . . , qr 數 (Proposition 7.3.4)

(pm11··· pmss)2 數 ( 數), Lemma 7.3.1

n 數 .

, p≡ 3 (mod 4) n 數 數 2k + 1, n = p2k+1n, p- n.

n 數 . a, b∈ Z a2+ b2= n.

Lemma 7.3.5 p|a p|b. a = prc b = psd, c, d p r, s∈ N.

2r≤ 2s, 2k + 1 > 2r. 2k + 1 < 2r, p2rc2+ p2sd2= p2k+1n n= p2r−2k−1c2+ p2s−2k−1d2. 2s− 2k − 1 ≥ 2r − 2k − 1 > 0 p|n. p- n , 2k + 1 > 2r. p2rc2+ p2sd2= p2k+1n

c2+ p2s−2rd2= c2+ (ps−rd)2= p2k+1−2rn,

p2k+1−2rn c ps−rd . p≡ 3 (mod 4), p|p2k+1−2rn p- c,

Lemma 7.3.5 , n 數 . 

2250 = 2× 32× 53 4 3 數 3, 3 數 2 數,

2250 數 . 2250 = 452+ 152. 6174 = 2×32×73

4 3 數 3 7, 7 數 3 數, 6174 數

.

7.3.2. Sum of Four Squares. 數 數

, 數 數 . ,

7 數 . 數 數

數 4m(8n + 3) . 數

Lemma 7.3.1 , 數 .

, 談 數 ,

談論 數 .

數 數 .

(7.1) .

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(a2+ b2+ c2+ d2)(e2+ f2+ g2+ h2) = (ae + b f + cg + dh)2+ (a f− be + ch − dg)2 +(ag− bh − ce + d f )2+ (ah + bg− c f − de)2.

(7.2)

. ,

數 quaternion algebra . quaternion

algebra , .

(7.2) .

Lemma 7.3.8. m, n∈ N, mn

.

1 數 數 , Lemma 7.3.8

數 數 . 2 4 1 數 數

, 數 ( 0), 論

4 3 數. Lemma 7.3.2 數

數 .

Lemma 7.3.9. p 數. a, b, c, d∈ Z a2+ b2+ c2+ d2=λ p,

λ ∈ N λ < p, p.

Proof. S ={s ∈ N | t, u, v, w∈ Z t2+ u2+ v2+ w2= sp}. S ,

p 數 1∈ S. 1∈ S ?

S ( λ ∈ S) S 數, 1∈ S S

1. m∈ S S , m = 1.

, m̸= 1. λ ∈ S λ < p 1 < m < p. S m

數 . m∈ S, t, u, v, w∈ Z t2+ u2+ v2+ w2= mp,

mm 數 論.

(I) m 數: t2+ u2+ v2+ w2= mp 數, t, u, v, w 數;

數 ; 數 數. t, u, v, w

. , t, u v, w , t + u, t− u, v + w

v− w 數. (t + u)/2, (t− u)/2, (v + w)/2 (v− w)/2 數 (t + u

2 )2+ (t− u

2 )2+ (v + w

2 )2+ (v− w 2 )2=m

2p.

m/2∈ S m/2 < m, m S .

(II) m 數: m

{−m + 1

2 ,−m + 1

2 + 1, . . . , 0, 1, . . . ,m− 1

2 − 1,m− 1 2 }

complete residue system modulo m. e, f , g, h∈ Z e≡ t (mod m), f ≡ u (mod m), g≡ v (mod m) h≡ w (mod m), −(m−1)/2 ≤ e, f ,g,h ≤ (m−1)/2.

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7.3. 101

e, f , g h 0, m|p 1 < m < p . e2+ f2+ g2+ h2 t2+ u2+ v2+ w2 (mod m) t2+ u2+ v2+ w2= mp, e2+ f2+ g2+ h2≡ 0 (mod m).

k∈ Z e2+ f2+ g2+ h2= km. e, f , g h 0, k̸= 0.

−(m − 1)/2 ≤ e, f ,g,h ≤ (m − 1)/2, e2+ f2+ g2+ h2≤ (m − 1)2= (m− 1)2< m2, 0 < k < m. k∈ N k < m. : t2+ u2+ v2+ w2= mp e2+ f2+ g2+ h2= km. (7.2)

(te + u f + vg + wh)2+ (t f−ue+vh−wg)2+ (tg−uh−ve+w f )2+ (th + ug−v f −we)2= m2k p.

e≡ t (mod m), f ≡ u (mod m), g ≡ v (mod m) h≡ w (mod m),

te + u f + vg + wh≡ t f − ue + vh − wg ≡ tg − uh − ve + w f ≡ th + ug − v f − we ≡ 0 (mod m).

T =te + u f + vg + wh

m , U =t f− ue + vh − wg

m ,

V =tg− uh − ve + w f

m and W =th + ug− v f − we

m ,

T,U,V,W∈ Z

T2+U2+V2+W2= k p.

k p 數 , k∈ N k∈ S. k < m,

m S .

m̸= 1 m 數 數 . m̸= 1

, m = 1. p 數 . 

Lemma 7.3.9 數 數 .

4 3 數 數 . x2≡ −1

(mod p) , α ∈ N x2≡ −α (mod p) .

(−α p

)

= (−1

p )(α

p )

=

p )

,

(−α p

)

= 1

p

)

=−1.

α ∈ N x2≡α (mod p) . , S ={1,2,..., p − 1} modulo p reduced residue system, p- a, x2≡ a (mod p) S modulo

p . x2≡ a (mod p) c∈ S c2≡ a (mod p).

S , amodulo p x2≡ a

(mod p) ; , amodulo p x2≡ a (mod p)

. c∈ S p− c ∈ S (p− c)2≡ (−c)2≡ c (mod p), p 數,

c̸≡ p − c (mod p). S modulo p (p− 1)/2

. S (p− 1)/2 a x2≡ a (mod p) , (p− 1)/2

a x2≡ a (mod p) .

Theorem 7.3.10. pp≡ 3 (mod 4), p.

, 數 數 .

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Proof. pp≡ 3 (mod 4). a, b, c, d∈ Z a2+ b2+ c2+ d2=

λ p, λ ∈ N λ < p, Lemma 7.3.9 p 數 .

S ={1,2,..., p − 1} modulo p reduced residue system. α ∈ S Sx2≡α (mod p) ,

p

)

=−1.

(1 p

)

= 1, α > 1, α − 1 ∈ S, x2≡α − 1 (mod p) ( α Sx2≡α (mod p)

). p≡ 3 (mod 4),

(−1 p

)

=−1,

(−α p

)

= (−1

p )(α

p )

= 1,

x2≡ −α (mod p) . a∈ S x2≡α −1 , a 1≤ a ≤ (p−1)/2.

(p + 1)/2≤ a ≤ p − 1, p− a, (p− a)2≡ (−a)2≡α − 1 (mod p) x2≡α − 1 (mod p) 1≤ p − a < (p − 1)/2. b∈ S x2≡ −α (mod p) 1≤ b ≤ (p − 1)/2.

a2+ b2+ 1≡α − 1 + (−α) + 1 ≡ 0 (mod p), λ ∈ N a2+ b2+ 1 =λ p.

λ p = a2+ b2+ 1≤ (p− 1

2 )2+ (p− 1

2 )2+ 1 < p2

2 + 1 < p2,

λ < p. Lemma 7.3.9 p 數 .

n∈ N. n = 1, n 數 . n > 1, n

n = pn11··· pnrr. pi= 2 pi≡ 1 (mod 4) pi 數 ,

數 . pi≡ 3 (mod 4),pi 數 .

Lemma 7.3.8 n = pn11··· pnrr 數 . 

基礎數論 基 , .

參考文獻

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