(a+b)²+(b+c)²+(c+d)²=4(ab+bc+cd),
a²+2ab+b²+b²+2bc+c²+c²+2cd+d² = 4ab+4bc+4cd
Get 0 on the right by subtracting it from both sides:
a²-2ab+b²+b²-2bc+c²+c²-2cd+d² = 0
(a-b)²+(b-c)²+(c-d)² = 0
The three terms on the left are non-negative.
Thus they must all be 0.
Thus a-b=0, which means a=b.
b-c=0, which means b=c,
c-d=0, which means c=d
So a=b=c=d
Edwin