If there is a finite field $F=\{a_1, …, a_n\}$ with $|F|\geq 4$. The prove that
$$a_1^2 + a_2^2 + … + a_n^2 = 0$$
I have a proof but I am not sure of it.
We have that $F^* = F-\{0\}$ is a cyclic multiplicative group.
So there is a generator $a$ such that $F^*=\{a, a^2, …, a^{n-1}\}$
If $a=1$ then $F=\{0,1\}$ which contradicts $|F|>4$ and hence we can choose $a\in F^*, a\neq 1$.
Now assume WLOG that $a_n=0$ is the zero element in the field.
Then $\sum_{i=1}^n a_i ^2 = \sum_{i=1}^{n-1} a_i^2 = \sum_{i=1}^{n-1}a^{2i} = \frac{a^2(1-a^{2(n-1)})}{1-a}=\frac{a^2(1-1)}{1-a}=0$
However, I feel I have done something here that implicitly uses the fact that $F$ has more than $3$ elements, but I don't see where I used that.
Best Answer
If $\#F > 3$, then some $\alpha\in F^\times$ has $\alpha^2 \not= 1$. But the map $x \to \alpha x$ permutes $F$, so \begin{align*} \sum_{x\in F} x^2 = \sum_{x\in F} (\alpha x)^2 = \alpha^2 \sum_{x\in F} x^2. \end{align*} The given sum must therefore vanish.