Question 742010
{{{(a+b)^2 + (b+c)^2 + (c+d)^2 = 4(ab+bc+cd)}}}


{{{a^2+2ab+b^2+b^2+2bc+c^2+c^2+2cd+d^2=4ab+4bc+4cd}}}


{{{a^2+2ab+b^2+b^2+2bc+c^2+c^2+2cd+d^2-4ab-4bc-4cd=0}}}


{{{a^2-2ab+b^2+b^2-2bc+c^2+c^2-2cd+d^2=0}}}


{{{(a-b)^2+(b-c)^2+(c-d)^2=0}}}

so (a-b)^2=0
a-b=0
a=b

similarly b=c

c=d

a=b=c=d