$\mathbb F_9 = \mathbb F_3(i)$ Question about squares Is it true that $\forall a,b \in \mathbb F_9$: $a \cdot b$ is a square $\iff$ $a \cdot \overline b$ is a square ?
 A: Since your group of units is cyclic, you can argue very similarly to the case of $F_p$, and the "Legendre symbol" analogue here is still a homomorphism.
In particular, multiplying by $a$ doesn't matter. You can reduce to $$b \; \mathrm{square} \Leftrightarrow \overline{b} \; \mathrm{square}.$$ This is true, because for nonzero $b$ $$b \; \mathrm{square} \Leftrightarrow b^4 = 1 \Leftrightarrow (\overline{b})^4 = \overline{b^4} = 1 \Leftrightarrow \overline{b} \; \mathrm{square}.$$
A: Adding a different approach to the one in Cocopuffs +1 answer.
The lone conjugate of any element $b$ in $F=\mathbb{F}_{p^2}$ is the Frobenius conjugate $\overline{b}=b^p$ (Caveat: The overline may mean something other than changing the sign of the coefficient of $\sqrt{-1}$. If $p\equiv3\pmod4$ we get the quadratic extension by adjoining $i$, but if $p\equiv1\pmod4$ another square root needs to be used instead).
But then we see that
$$
\frac{\overline{b}}{b}=\frac{b^p}b=(b^{(p-1)/2})^2
$$
is always a square, and the claim follows trivially from this.
