= Solution
\b[No such ring exists.] Characteristic two gives $r+r=0$ for every $r$. Choose a nonzero element $a$, which exists in a five-element set. Translation $r\mapsto r+a$ is an <involution> without fixed points: applying it twice gives $r$, while $r+a=r$ would force $a=0$ by additive cancellation. Its disjoint pairs would partition all five elements, which is impossible because five is <odd>.
This argument applies even to a <nonunital ring>. With the unital convention, it is also the case $n=2$ of <finite ring characteristic divides cardinality>.
Back to article page
