Number of non-zero solutions of an equation in $F \times F$ where $F$ is a field Let $F$ be a field of order $32$.
Then I need to find the number of non-zero solutions $(a,b)$ $\in$ $F\times F$ of the equation $x^2+xy+y^2 = 0$.
I know that characteristic of a ring with unity $1$ is the order of $1$ in the group $(R,+).
Since $F$ has unity being a field, the order of unity i.e. $1$ divides $2^5$.
Since characteristic of a field is either $0$ or prime.
Since we have a finite field, hence characteristic can't be $0$ and hence it is $2$.
So the characteristic of $F \times F$ is $lcm(2,2)$ which is $2$.
Since order of $F \times F$ is $2^{32}$. So from here, can I say that $F \times F$ is a field?
If I can, then if $x = 0$ then $y$ will be $0$ and also the other way round.
So to get the required solutions, $x$ and $y$ both should be non-zero.
From here, how could I proceed?
 A: You are correct that the characteristic of $F$ is $2$.
I think you misunderstood the role of $F \times F$ here, though. $(a,b) \in F \times F$ is just saying that $(a,b)$ is a pair of elements, each in $F$. That is, $a \in F$ and $b \in F$, just that we want to count the number of such pairs, where order matters. $F \times F$ is being used here simply as a set of pairs, not really as a field or even as a ring.
We're looking for solutions of $x^2 + xy + y^2 = 0$ where $x$ and $y$ are elements of $F$.
The answer is, there are no solutions (other than $(0,0)$).
As you noticed, the field properties give that if $x=0$ then $y=0$, and vice versa. So suppose $x$ and $y$ are elements in $F$ with $x \neq 0$, $y \neq 0$, and $x^2+xy+y^2=0$.
In any field, polynomial multiplication gives
$$ (x-y)(x^2 + xy + y^2) = x^3 - y^3 $$
So we must also have
$$x^3 - y^3 = (x-y) \cdot 0 $$
$$x^3 = y^3$$
$$(xy^{-1})^3 = 1$$
The multiplicative group of $F$ has $31$ elements ($0$ is not a member). Since $31$ is prime, that group is cyclic, so the multiplicative order of every element other than identity $1$ is $31$. So the only way the cube of $xy^{-1}$ can be $1$ is if $xy^{-1} = 1$. Then we must have $x=y$, so
$$0 = x^2+xy+y^2 = 3x^2 = x^2$$
But this would require $x=0$, contradicting the assumption. There cannot be any nonzero solutions.
