數學問題 : 四元方程

By (1) , a = 2 - bcd , substitute into (2) , (3) , (4) : b + (2 - bcd)cd = 2 ...... (2)

c + (2 - bcd)bd = 2 ...... (3)

d + (2 - bcd)bc = 2 ...... (4)

⇒

b(1 - c²d²) = 2(1 - cd) ...... (2)

c(1 - b²d²) = 2(1 - bd) ...... (3)

d(1 - b²c²) = 2(1 - bc) ...... (4)

⇒

b(1 + cd) = 2 ...... (2)

c(1 + bd) = 2 ...... (3)

d(1 + bc) = 2 ...... (4)

⇒

b = 2 - bcd ...... (2)

c = 2 - bcd ...... (3)

d = 2 - bcd ...... (4)

⇒

b = c = dSubstitute into (1) , (2) :a + b³ = 2 ...... (1)

b + ab² = 2 .....(2)

By (1) , a = 2 - b³ , substitute into (2) :b + (2 - b³)b² = 2

b⁵ - 2b² - b + 2 = 0

(b - 1)² (b + 1) (b² + b + 1) = 0

b = ± 1

∴

a = b = c = d = 1

or

a = 3 , b = c = d = - 1 Similarly ,

b = 3 , a = c = d = - 1

or

c = 3 , a = b = d = - 1

or

d = 3 , a = b = c = - 1

2012-12-03 18:23:29 補充：

The above result is Case 1 : None of (1 - cd , 1 - bd , 1 - bc) = 0

Sorry for missing Case 2 : If any one of (1 - cd , 1 - bd , 1 - bc) = 0

b(1 - c²d²) = 2(1 - cd) .. (2)

c(1 - b²d²) = 2(1 - bd) .. (3)

d(1 - b²c²) = 2(1 - bc) .. (4)

Say (1 - cd) = 0 , then cd = 1 , gives

2012-12-03 18:24:08 補充：

b + a = 2 ..... (2)

c + abd = 2...(3)

d + abc = 2 ..(4)

(3) - (4) :

(c - d) - ab(c - d) = 0

(c - d)(1 - ab) = 0

When c = d , then c = d = ± 1 since cd = 1

When ab = 1 , by (3) , c + d = 2 , then c = d = 1 since cd = 1

All in all , c = d = ± 1 , so we have

b + a = 2 ..... (2)

1 + ab = ± 2 .(3)

2012-12-03 18:24:53 補充：

Substitute (2) into (3) :

1 + a(2 - a) = ± 2

a² - 2a + 1 = 0 for c = d = 1 or a² - 2a - 3 = 0 for c = d = - 1

a = 1 for c = d = 1 or a = - 1 or a = 3 for c = d = - 1

⇒

a = b = c = d = 1 or b = 3 , a = c = d = - 1 or a = 3 , b = c = d = - 1

2012-12-03 18:25:22 補充：

Similarly ,

1 - bd = 0 gives

a = b = c = d = 1 or a = 3 , b = c = d = - 1 or c = 3 , a = b = d = - 1

1 - bc = 0 gives

a = b = c = d = 1 or a = 3 , b = c = d = - 1 or d = 3 , a = b = c = - 1

2012-12-03 18:26:05 補充：

∴

a = b = c = d = 1

a = 3 , b = c = d = - 1

b = 3 , a = c = d = - 1

c = 3 , a = b = d = - 1

d = 3 , a = b = c = - 1

2012-12-03 18:28:08 補充：

b = 3 , a = c = d = - 1

or

c = 3 , a = b = d = - 1

or

d = 3 , a = b = c = - 1

是據 a , b , c , d 的對等地位得出。

2012-12-05 18:53:48 補充：

但得出這些Case1的解後要全部捨棄(雖然它們真是解),

因為它們都使(1 - cd , 1 - bd , 1 - bc) = (0 , 0 , 0)。

2012-12-05 18:56:38 補充：

解法二 :

明顯 a , b , c , d ≠ 0

a² + abcd = 2a … (1)*a

b² + abcd = 2b … (2)*b

c² + abcd = 2c … (3)*c

d² + abcd = 2d … (4)*d

⇒

a² - 2a = b² - 2b = c² - 2c = d² - 2d = - abcd

令 a² - 2a = b² - 2b = c² - 2c = d² - 2d = k(常數)

則 x² - 2x = k 的 x 有 1 或 2 個實根值。

2012-12-05 18:59:01 補充：

故 a = b = c = d 或 不妨設 (a = b) ≠ (c = d) 或 a = b = c ≠ d。

當 a = b = c = d ,

得 a + a³ = 2 ... (1)

(a - 1)(a² + a + 2) = 0

a = 1

∴ (a , b , c , d) = (1 , 1 , 1 , 1)

當 (a = b) ≠ (c = d) , 得

a + ac² = 2 ... (1)

c + a²c = 2 ... (3)

2012-12-05 18:59:47 補充：

(1) - (3) :

(a - c) - ac(a - c) = 0

(a - c)(1 - ac) = 0

ac = 1 因 a ≠ c , 得

a + c = 2 ... (1) , 代入(3) :

(2 - a) + a²(2 - a) = 2

a³ - 2a² + a = 0

a(a² - 2a + 1) = 0

a = 1 因 a ≠ 0

故 c = 1

則 a = c 茅盾!

當 a = b = c ≠ d , 得

a + a²d = 2 ... (1)

d + a³ = 2 ... (4)

2012-12-05 19:00:23 補充：

代(4) 入(1) :

a + a²(2 - a³) = 2

a⁵ - 2a² - a + 2 = 0

(a - 1)² (a + 1) (a² + a + 2) = 0

a = 1 , d = 1 , a = d 不合

或

a = - 1 , d = 3

2012-12-05 19:00:46 補充：

∴ (a , b , c , d) = (- 1 , - 1 , - 1 , 3)

同理可得

(a , b , c , d) = (- 1 , - 1 , 3 , - 1)

(a , b , c , d) = (- 1 , 3 , - 1 , - 1)

(a , b , c , d) = (3 , - 1 , - 1 , - 1)

綜上 , (a , b , c , d) 共 5 組解。

b(1 - c²d²) = 2(1 - cd) ⇒ b(1 + cd) = 2 ???

also b = c = d cannot lead to

b = 3 , a = c = d = - 1

or

c = 3 , a = b = d = - 1

or

d = 3 , a = b = c = - 1

略欠工整???