Problem

Source: Poland - Second Round 2020 P1

Tags: algebra, Poland



Assume that for pairwise distinct real numbers $a,b,c,d$ holds: $$ (a^2+b^2-1)(a+b)=(b^2+c^2-1)(b+c)=(c^2+d^2-1)(c+d).$$Prove that $ a+b+c+d=0.$