Let $\mathbb{F}_q$ be a finite field with $q$ element, and let $\mathbb{F}_{q^m}$ be an extension of $\mathbb{F}_q$. One of the classical results is, all the elements of $\mathbb{F}_q$ are squares in $\mathbb{F}_{q^m}$ if $m$ is even, and only the half of them are square if $m$ is odd. Let $\alpha \in \mathbb{F}_{q^m}$ be a generator of the multiplicative group $\mathbb{F}_{q^m}^*$, we define the following sets
$$C_i= \{\, 1+\alpha^ix \, : \,x \in \mathbb{F}_{q}^*\, \} \quad\quad i \in [1,\cdots,\frac{q^m-1}{q-1}-1],$$ and $C_0=\mathbb{F}_{q}^*$. We can see that the sets $C_i$ form a partition of the group multiplicative $\mathbb{F}_{q^m}^*$.
If $m$ is an odd number, then for $C_0$ half the elements are square and the other half are not square. My question is how many squares there are in $C_i$. More specifically, how many squares are in $C_2$.
Thank's in advance.