一个自然数a若恰好等于另一个自然数b的平方,则称自然数a为完全平方数.已知a=2002²+2002²

a=2002^2+2002^2×2003^2＋2003^2
=2002^2+2002^2×(2002+1)^2＋(2002+1)^2
=2002^2+2002^2×(2002^2+2×2002+1)+2002^2+2×2002+1
=2002^2+2002^4+2×2002^3+2002^2+2002^2+2×2002+1
=2002^4+2×2002^3+3×2002^2+2×2002+1
=2002^4+2×2002^3+2002^2+2×2002^2+2×2002+1
=(2002^2+2002)^2+2×(2002^2+2002)+1
=(2002^2+2002+1)^2
=【2002（2002+1）+1】^2
=(2002x2003+1)²
得证
如果本题有什么不明白可以追问,

设A＝2002
a=A^2+A^2*(A+1)^2+(A+1)^2
=A^2*(A+1)^2+A^2+A^2+2A+1
=[A(A+1)]+2A(A+1)+1
=[A(A+1)+1]^2
=(2002*2003+1)^2
a的平方根=+-(2002*2003+1)