# How Many Solutions for $x+x=1$ in a Ring?

## Problem 204

Is there a (not necessarily commutative) ring $R$ with $1$ such that the equation
$x+x=1$ has more than one solutions $x\in R$?

## Solution.

We claim that there is at most one solution $x$ in the ring $R$.
Suppose that we have two solutions $r, s \in R$. That is, we have
$r+r=1 \text{ and } s+s=1.$

Then we have $r+r=s+s$. Putting $a=r-s\in R$, we have
$a+a=0.$ Now we compute
\begin{align*}
0&=1\cdot 0 =(r+r)(a+a)\\
&=ra+ra+ra+ra\\
&=(r+r)a+r(a+a)\\
&=1\cdot a+r\cdot 0\\
&=a.
\end{align*}

Therefore we obtain $a=0$ and thus $r=s$.
It follows that the equation $x+x=1$ has only one solution (at most).

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