數學知識交流---求證

abc + a + b + c = 1 .....(1)
ab + ac + bc = -1 ........(2)(1) + (2) :abc + a + b + c + ab + ac + bc = 1 - 1(abc + ab + ac + a) + bc + b + c + 1 = 1a(bc + b + c + 1) + bc + b + c + 1 = 1(a + 1) (bc + b + c + 1) = 1(a + 1) (b(c + 1) + c + 1) = 1(a + 1) (b + 1) (c + 1) = 1不妨 a ≥ b ≥ c ,則 a + 1 = b + 1 = c + 1
==>
a = b = c = 0 (不合 (1) , (2) 式 , 捨去)或 a + 1 = 1 , b + 1 = c + 1 = - 1
==>
a = 0 , b = c = - 2 (不合 (1) , (2) 式 , 捨去)綜上方程組沒有整數解。

2011-06-29 23:46:36 補充：
不妨 a ≥ b ≥ c ,

則 a + 1 = b + 1 = c + 1 (= 1)

Suppose there exist an integral solution (a,b,c)
Let
S1 = a+b+c
S2 =ab + ac + bc
S3 = abcThen from the given conditions, S2 = -1, S3 = 1-S1The equation:
x^3-S1x^2+S2x-S3=0
x^3-S1x^2-x-1+S1=0
x^2(x-S1)-(x-S1)=1
(x-S1)(x^2-1)=1
has three roots a,b,cAs a is a root,
(a-S1)(a^2-1)=1
(-b-c)(a^2-1)=1
(b+c)(a^2-1)=-1Since b+c and a^2-1 are integers,
b+c = -1 and a^2-1=1
a^2 = 2 and b+c =-1
a= ±√2 (reject) and b+c=-1
OR
b+c = 1 and a^2-1=-1
a=0 and b+c=1By similar reasons,
b=0 and a+c=1
c=0 and a+b=1

As
a=0 and b+c=1
b=0 and a+c=1
c=0 and a+b=1
obvously cannot occur at the same time,
contradiction occurs.
So, there does not exist integral solution (a,b,c) satifying the given conditions.