finite field multiplicative group squares I'm working through a proof on t-designs and I'm stuck at this part. Let $\mathbb{F}_{p^k}$ be a field of order $p^k$ with $p^k$ = $4n + 3$. Let $\alpha$ be the cyclic generator. Then the set of all the distinct even powers of alpha are $$S = \left \{ 1, \alpha^2, \alpha^4, \alpha^6, ....., \alpha^{p^{k} - 3} \right \}$$
If we have $S_m = \left \{S + m\right \}$ where $m$ is some element of ${F}_{p^k}$, how do we show that there are $p^k$ of these sets (that are distinct)?
In other words for $m,n \in {F}_{p^k}$, with $m \neq n$, how do we show $S_m \neq S_n$?
My thoughts on it this far are that if $S_m = S_n$, then for each $\alpha^{2i} + m \in S_m$ there is an $\alpha^{2j} + n \in S_n$ equal to it.
So   $\alpha^{2i} + m = \alpha^{2j} + n$ or  
$\alpha^{2i} + (m - n) = \alpha^{2j}$ or  
$\alpha^{2i} + d = \alpha^{2j}$ with $m - n = d$. 
So for every element $\alpha^{2i} \in  S$ then ${\alpha^{2i}} + d \in  S$ as well.
So I wanted to show that $d=0$ is the only solution, but I'm not getting anywhere with that. If anyone could show why $d=0$ or another way of proving why $S_m \neq S_n$, it would be much appreciated.
 A: You made a good start. The first way of showing that we must have $d=0$ that occured to me is the following.
Lemma. Assume $d\neq0$. The number $N(d)$ of solutions $(x,y)\in\Bbb{F}_{p^k}^2$ of the equation
$$
x^2=y^2+d\qquad(*)
$$
is $N(d)=p^k-1$.
Proof. The equation $(*)$ is equivalent to
$$
d=x^2-y^2=(x+y)(x-y).
$$
Write $u=x+y, v=x-y$, so we want $d=uv$. We see that any non-zero $u$ determines $v$ uniquely, so there are $p^k-1$ pairs $(u,v)$ such that $uv=d$. Because $(u,v)$ determine $(x,y)$ uniquely as
$x=(u+v)/2$, $y=(u-v)/2$ the claim follows.
The main claim follows from the Lemma as follows. Assume that there exists $d\neq0$ such that $S=S+d$. Because $S$ is the set of squares, it follows that $y^2+d\in S$ for any $y\in\Bbb{F}_{p^k}$. This means that to each $y$ there exists 
$x=x(y,d)\in\Bbb{F}_{p^k}$ such that $x^2=y^2+d$. The solution $x$ can be zero for only two choices of $y$ (namely $y=\pm\sqrt{d}$), but for $p^k-2$
choices of $y$ we get two distinct solutions: $x=\pm x(y,d)$. 
Therefore $S=S+d$ implies that the
equation $(*)$ has $2p^k-2$ solutions $(x,y)$ altogether. This contradicts the Lemma.
A: Consider the set of elements 
$$G=\{d\in\Bbb{F}_{p^k}\mid S+d=S\}.$$
Clearly:


*

*$0\in G$,

*If $d_1\in G$ and $d_2\in G$ then $d_1+d_2\in G$. In other words $G$ is a subgroup of the additive group of $\Bbb{F}_{p^k}$.

*If $d\in G$, and $z\in\Bbb{F}_{p^k}^*$, then $dz^2\in G$. This is because if $d+x^2$ is a square for all $x\in \Bbb{F}_{p^k}$, then $$dz^2+x^2=z^2\left(d+(\frac xz)^2\right)$$ is a square for all x as a product of two squares.

*By item 3, if $G$ contains a non-zero element, it contains at least $|S|=(p^k-1)/2$ non-zero elements.

*So if $G$ contains a non-zero element, then $|G|>p^{k-1}$. By Lagrange's theorem $|G|\mid p^k$, so we must have $G=\Bbb{F}_{p^k}$.

*But this is absurd because then $S$ would be all of $\Bbb{F}_{p^k}$. Therefore $$G=\{0\}.$$

