If there are nonzero elements $a$ and $b$ in $A$ such that $(a+b)^2 = a^2 + b^2$, then $A$ has characteristic 2. Let $A$ be a finite integral domain. If there are nonzero elements $a$ and $b$ in $A$ such that $(a+b)^2 = a^2 + b^2$, then $A$ has characteristic 2.
I was thinking, if $(a+b)^2 = a^2 + 2ab + b^2 = a^2 + b^2$, wouldn't that mean that $2ab = 0$? But I thought that would be a contradiction since $A$ is a finite integral domain and has no zero divisors. I am not sure where I am confused. 
However, I do think I know how to prove the statement. We know that in any integral domain of characteristic $p$, then $(a+b)^p = a^p + b^p$. Thus 2 is either the characteristic or 2 divides the characteristic. Since the characteristic must be prime, then 2 is the characteristic for $A$. 

It is an if and only if, prove both ways.


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*If the characteristic is $2$, then $2 = 0$, so $(a + b)^2 = a^2 + 2 a b + b^2 = a^2 + b^2$

*If $a, b \ne 0$ are such that $(a + b)^2 = a^2 + b^2$, then $2 a b = (a + b)^2 - a^2 - b^2 = 0$. As $a b \ne 0$, and it is an integral domain, it has no zero divisors, so $2 = 0$.
