基礎數論
數
數論 基 數 數 ,
數 . 基 , 數論
. 數論 ( ) ,
Silverman A Friendly Introduction to Number Theory (Prentice Hall, Third Edition
2006). .
v
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
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) .
p 數 p̸= 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
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= c′1, . . . , xi= c′i, . . . , xn= c′n, i∈ {1,...,n} c′i< ci, x1= c′1, . . . , xi= c′i, . . . , xn= c′n 數 數 x1= c′′1, . . . , xi= c′′i, . . . , xn= c′′n
c′′i < c′i. 數數 ci> c′i> 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 = d1 數 x = c2, y = d2
數 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 .
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
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 x 數 y 數. Pythagorean triples
.
Definition 7.2.1. a, b, c∈ N a2+ b2= c2 gcd(a, b, c) = 1 a 數 b 數, a, b, c primitive Pythagorean triple.
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 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). y 數 x, z 數, y/2,(z + x)/2 (z− x)/2 數 (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
數 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′, n′ 數 m > 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, m′ 數 m = m′. n = n′.
.
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 數
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 數 .
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
數 c1/d2< c1.
x = a1, y = b1, z = c1 x4+ y4= z2 數 gcd(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 , a21 數 b21 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.
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= 2b′1 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= z2 數 c2< 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 數 . 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, z1 數 z1= 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) .
Lemma 7.3.1. m, n∈ N 數 , mn 數 .
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 數 , mn
數 ; m, n 數 , 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 數 .
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 數
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 數
(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 數
{−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,
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. , p 數 p≡ 1 (mod 4) , .
.
Proposition 7.3.4. p 數 p≡ 1 (mod 4), p 數 . 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 數 . p≡ 1 (mod 4) ,
Proposition 7.3.4 x2≡ −1 (mod p) p 數
. p≡ 3 (mod 4) , x2≡ −1 (mod p) p
數 .
Lemma 7.3.5. p 數 p≡ 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. p 數 p≡ 3 (mod 4), p 數 .
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
7.3. 99
數. pi n 數 ni pi 數. 2250 = 2× 32× 53,
2250 2 3 數 3 5 數. .
Theorem 7.3.7. n∈ N. n 數 n
4 3 數 數 數.
Proof. n 4 3 數 數 數. n≥ 2
. n 數
n = 2n0qn11···qnrr· p2m1 1··· p2ms s,
qi, pj 數 qi≡ 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) .
(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 數 , mn 數
.
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 數 .
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,
m 數 m 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 數
{−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.
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 )
=− (α
p )
,
(−α p
)
= 1
(α 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 , a 數 modulo p x2≡ a
(mod p) ; , a 數 modulo 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. p 數 p≡ 3 (mod 4), p 數 .
, 數 數 .
Proof. p 數 p≡ 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 S 數 x2≡α (mod p) ,
(α p
)
=−1.
(1 p
)
= 1, α > 1, α − 1 ∈ S, x2≡α − 1 (mod p) ( α S 數 x2≡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
數 n = pn11··· pnrr. pi= 2 pi≡ 1 (mod 4) pi 數 ,
數 . pi≡ 3 (mod 4), 前 pi 數 .
Lemma 7.3.8 n = pn11··· pnrr 數 .
基礎數論 基 , .