a^2b^2+a^2+b^2=2004
= & gt(a^2+ 1)(b^2+ 1)=2005=5*40 1
=>A 2+ 1 = 5, B 2+ 1 = 40 1 or A+ 1 = 40 1, B+ 1 = 5.
= & gt(a, b) combination can be any of the following:
(2,20),(-2,20),(2,-20),(-2,-20)
(20,2),(-20,2),(-20,-2),(20,-2)
Question 2:
a^2- 16b^2-c^2+6ab+ 10bc=0,
a^2+6ab+9b^2-25b^2+ 10bc-c^2=0
(a+3b)^2-(5b-c)^2=0
(a+3b)^2=(5b-c)^2
a+3b=5b-c
a+c=2b
Or a+3b=c-5b.
a+8b=c
Because the sum of two sides of a triangle is greater than the third side, that is, A+B >; c
So a+8b cannot be equal to c, so it is discarded.
So a+c=2b.
Question 3: Let 2001= x.
So a = x2+x2 (x+1) 2+(x+1) 2 = x4+2x3+3x2+2x+1= (x2+x+65438
So a is a complete square number.