Question 42
Let $A$ be a ring. Prove that the following statements are equivalent:
a) For every elements $a$ and $b$ from $A$ there exists two odd integers $p$ and $q$ such that $a^2 + b^2 = (pa + qb)^2$;
b) The ring $A$ is commutative and $1 + 1 = 0$.
a) For every elements $a$ and $b$ from $A$ there exists two odd integers $p$ and $q$ such that $a^2 + b^2 = (pa + qb)^2$;
b) The ring $A$ is commutative and $1 + 1 = 0$.