If $K$ finite field of order $p^8$ where $p\ne3$ then $\sum_{\alpha \in K}{\alpha^2} = 0$

Question

Let $K$ be finite field of order $p^8$ where $p\ne3$ is a prime.

Show that $\sum_{\alpha \in K}{\alpha^2} = 0$.

Answer 1

Hint for approach #1: Use the fact in a finite field the non-zero squares form a cyclic subgroup of the multiplicative group, so the sum (well, half of it) is a segment of a geometric series such that...

Hint for approach #2: Let $$ S=\sum_{\alpha\in K}\alpha^2=\sum_{\alpha\in K^*}\alpha^2. $$ Because $p^8>3$ (this is enough) there exists an element $\beta\in K^*$ such that $\beta^2\neq1$. Show that $$ \beta^2S=\sum_{\alpha\in K^*}(\beta\alpha)^2 $$ is actually equal to $S$, because it contains exactly the terms (but in a different order). Conclude that $S=0$.