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.