Finite fields of even cardinality have characteristic $2$ Computer science studies-
as for now we have dealt with the set-theory and residue class rings with modular arithmetic, although we did not really define the connection between the two so rigorously
So I have to prove the following property for every $x \in \mathbb{F}$, whereas $\mathbb{F}$ is a finite field with even cardinality:
$$ x+x = 0$$
As far as I considered it could be satisfied only with modular arithmetic, but apart of that I do not have a concrete approach..
Thank you for your help
 A: First, let's consider $F$ as an abelian group under addition.  Since $\vert F \vert$ is even, let $P$ be a $2$-Sylow subgroup of $F$.  Then $P$ is non-trivial and any non-zero element $z \in P$ has order $2^k$ for some $k \in \Bbb N$.  If $k \gt 1,$ let $y = 2^{k-1}z$.  If $k=1$, let $y = z$.  In either case, $F$ must have some element $y$ with order $2$; in other words, for that specific $y, y \neq 0 \text{ and } y+y=0$.
There's another way to see this.  Assume that $F$ has no element of additive order $2$.  Then you'd be able to partition $F$ by pairing up its non-zero elements,$\{x_n, -x_n\}$.  But $F$ has an odd number of non-zero elements, so that's not possible.  Thus, $F$ must have an element of order $2$.
Now choose an arbitrary $x \in F$ with $x \neq 0$.  Then $\frac xy$ exists and is nonzero and $x+x=y \frac xy+y \frac xy=(y+y) \frac xy=0$.
Alternatively, $0=y+y=1y+1y=(1+1)y \Rightarrow 1+1=0.$  Thus, for any $x \in F$ we have $x+x=1x+1x=(1+1)x=0x=0$.
Incidentally, this also proves that $\vert F \vert$ must be a power of $2$ because if an odd prime divided $\vert F \vert$, then $F$ would have to have an element of odd (additive) order, which we just proved can't happen.
