dangerousliri wrote:
Find all real numbers $a,b$ and $c$ such that $a+bc=b+ca=c+ab$.
Subtracting the first from the two others, system is equivalent to :
$(a-b)(c-1)=0$ and $(a-c)(b-1)=0$
Hence solutions $\boxed{(a,b,c)\in\{(u,u,u),(1,1,u),(1,u,1),(u,1,1)\}}$, whatever is $u\in\mathbb R$