Suppose that $F$ is a field with $27$ elements. Show that for every element $a \in F$, $5a = −a$ 
Suppose that $F$ is a field with $27$ elements. Show that for every
  element $a \in F$, $5a = −a$.

I am not able to understand how to approach this problem.
 A: Recall that a field $F$ has a characteristic, which is defined as follows:
Define the homomorphism of rings
\begin{align}
\psi:\mathbb{Z}&\to F
\end{align}
That sends $1$ to $1_F$. Its kernel is principal, say, generated (in particular) by a prime number $p$ or $0$ (if it is injective). That number is called the characteristic of the field $F$.
Now, you have to know that every finite field, has positive characteristic $p$ (i.e. the characteristic is not zero) and have $p^n$ elements. This is because as it is finite, via $\psi$ we can not have infinitely many sums of $1_F$ that are different pairwise, and if $\mathbb{F}_p$ is the field with $p$ elements, $F$ can be seen as a non-trivial $\mathbb{F}_p$ vector space, so is isomorphic to $\mathbb{F}^n$, for some $n>1$.
Because of the previous results, your field $F$ with $3^3$ elements, has characteristic $3$, so that $3a=0$ for all $a\in F$, in particular $6a=0$, so that $5a=-a$.
A: An alternative answer is the following.
Since (F,+) is a finite Abelian group with 27 elements it is isomorphic to either $\mathbb{Z}_{27},\mathbb{Z}_3 \times \mathbb{Z}_9$ or $\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$. Any Integral domain, (I'm defining this here to be a commutative unital ring that has no 0 divisors) has either prime or 0 characteristic. The proof for this can be found in Gallian's contemporary abstract Algebra in chapter 13. Since F is finite, the characteristic is prime. Our characteristic can't be larger than 27, for if it were $\left\{ n1 \right\}_{n \in \mathbb{N}}\cup\{ 0 \}$ would form a subset with larger cardinality than F, a contradiction.  Thus, our characteristic must be one of the following naturals,
\begin{equation}
S=\left\{ 2,3,5,7,11,13,17,19,23 \right\}
\end{equation}
. Let $p \in S \setminus \{3 \}$. Convince yourself $\boldsymbol{1} \in \mathbb{Z}_{27},\mathbb{Z}_3 \times \mathbb{Z}_9,\mathbb{Z}_3 \times \mathbb{Z}_3 \times \mathbb{Z}_3$ has the property that $p \boldsymbol{1} \neq 0$ in any group. It follows for our group isomorphism $\varphi$ that $\varphi(p \boldsymbol{1})=p\varphi(\boldsymbol{1}) \neq 0$. Since $\varphi(\boldsymbol{1})\in F$, $p$ can't be the characteristic for $F$.   Thus our field must have characteristic $3$. Our desired result follows.
