Đáp án:
Ta thấy : a+b=c+d => (a+b)2=(c+d)2(a+b)2=(c+d)2
<=> a2+2ab+b2=c2+2cd+d2a2+2ab+b2=c2+2cd+d2(1)
Mà a2+b2=c2+d2a2+b2=c2+d2(2)
Từ (1)(2) => 2ab=2cd => ab=cd => ad=cb=kad=cb=k
=> a=dk; c=bk
Ta xét : a2+b2=c2+d2a2+b2=c2+d2
<=> (dk)2+b2=(bk)2+d2(dk)2+b2=(bk)2+d2
<=> d2(k2−1)=b2(k2−1)d2(k2−1)=b2(k2−1)
<=> (d2−b2)(k2−1)=0(d2−b2)(k2−1)=0
=>[d2−b2=0k2−1=0[d2−b2=0k2−1=0<=> [d=±bk=±1[d=±bk=±1
Th1 :d=±b±b mà ad=cbad=cb=> a=±c±c
=> d2002=b2002;a2002=c2002d2002=b2002;a2002=c2002
=> a2002+b2002=c2002+d2002a2002+b2002=c2002+d2002(3)
Th2: k=±1±1 => a=±d;c=±b=±d;c=±b
=> a2002=d2002;c2002=b2002a2002=d2002;c2002=b2002
=> a2002+b2002=c2002+d2002a2002+b2002=c2002+d2002(4)
Từ (3)(4)=> đpcm
