令x=2003
则a=x²+x²(x+1)²+(x+1)²
=(x+1)²-2x(x+1)+x²+2x(x+1)+x²(x+1)²
=[(x+1)-x]²+2x(x+1)+x²(x+1)²
=1²+2x(x+1)+x²(x+1)²
=[1+x(x+1)]²
=(1+2003*2004)²
令x=2003
则a=x²+x²(x+1)²+(x+1)²
=(x+1)²-2x(x+1)+x²+2x(x+1)+x²(x+1)²
=[(x+1)-x]²+2x(x+1)+x²(x+1)²
=1²+2x(x+1)+x²(x+1)²
=[1+x(x+1)]²
=(1+2003*2004)²