Let $(R,+,*)$ be a ring so that $(x+y)^{2}=x^{2}+y^{2}$ $\forall\ x, y \in R$. Prove that

A)$xy=-yx$ $\forall\ x, y \in R$

B)$x^{2}+x^{2}=0$ $\forall\ x, y \in R$ and $x+x=0\ \forall\ x, y \in R$

I have already proved A) and $x^{2}+x^{2}=0$ $\forall\ x, y \in R$ and for the last part of B I had that:

$x^{2}+x^{2}=0 \Rightarrow\ (x+x)^{2}=0$ but I don´t know if this implies that $x+x=0$

So I would really appreciate your help for the last part of B

• If your ring has a unit $1$, then you can use part (A), setting $y=1$. – ajd Mar 4 '14 at 5:16
• If it doesn't have a unit, the result is false (try a zero ring on three elements). – Jack Schmidt Mar 4 '14 at 5:18
• But how can I know that the ring has unit 1? Do I need to prove it? – user128422 Mar 4 '14 at 5:39
• They bring up the question of $1$, because there are two different conventions about what the word "ring" means. Check your definition. – user14972 Mar 4 '14 at 6:02

Since $(x+y)^2=x^2+y^2+xy+yx$ , then by our assumption $x^2+y^2=(x+y)^2\Rightarrow xy+yx=0$ That is exactly what you wanted to show.note that taking $x=y$ immediately implies second part.it seems to me that ring $R$ ought to have multiplicative identity $1_R$.assuming that let $y=1_R$ then by first part $1x=-x1$ thus $x+x=0$