$p$ is an odd prime. $S_p =\{ {p \choose k} /p \pmod p : 0$S_p$ is the set of integers in the $p$-th row of pascals triangle, excluding the $1$’s on either end, each divided by $p$, and then modulo by $p$, where $p$ is an odd prime.
My question is as follows:
Prove that $\forall$ odd prime $p$ $\forall p>x>0$, we have $$x \in S_p \iff p-x \not\in S_p.$$
I have verified the statement holds by hand for $p$ up to $19$, but I don’t have any intuition for why it should be true. It may very well be false for some large enough prime, but I don’t believe it to be the case.
I do know that $S_p$ can be generated by $(-1)^{k-1}/k \in \mathbb{F}_p$, but I don’t know how to turn this into a proof.
 A: You have bumped onto the key clue, $(-1)^{k-1}/k$. Here is a solution that uses counting argument (and, of course, modular arithemetic).

Fix an arbitrary odd prime $p$. We will be working in either $\Bbb Z$ or $\Bbb F_p$, depending on the context. In particular, $S_p$ can be viewed as a subset of $\{1,2,\cdots,p-1\}\subset\Bbb Z$ or a subset of $\Bbb F_p$. For $0<k<p$,
$${p\choose k}/p=_{\Bbb Z}\frac{(p-1)(p-2)\cdots(p-k+1)}{1\cdot2\cdots (k-1)\cdot\ k}=_{\Bbb F_p}(-1)^{k-1}\frac1k.$$
In particular, we have verified that $0\not\in S_p$.
Claim. The sum of any two elements in $S_p$ is not $0\in{\Bbb F_p}$.
Proof.
Consider two arbitrary elements in $S_p\subset\Bbb F_p$, $(-1)^{k-1}\frac1k$ and $(-1)^{j-1}\frac1j$. WLOG suppose $0<k<j<p$. Their sum is
$$\frac{(-1)^{k-1}}{kj}(j+(-1)^{j-k}k).\tag{*}\label{*}$$
There are two cases.

*

*$j$ and $k$ have the same parity,  $(-1)^{j-k}=1$.
Since $0<j+k<2p$ and as an even number, $j+k\not=p$, we have $p\not\mid(j+k)$.

*$j$ and $k$ have different parity, $(-1)^{j-k}=-1$.
Since $0<j-k<p$, $p\not\mid(j-k)$.

In all cases, $p\not\mid j+(-1)^{j-k}k$. Hence the sum $\eqref{*}$ is not $0\in\Bbb F_p$.  $\quad\checkmark$
The claim means, when we view $S_p$ as a subset of $\{1,2,\cdots, p-1\}$, it contains at most one element among each of the following $\frac{p-1}2$ pairs, $\{1,p-1\}, \{2,p-2\}, \{3,p-3\},\cdots$. To show the wanted conclusion, it is enough to show that $S_p$ contains exactly $\frac{p-1}2$ element.
For $0<k\not=j<p$, we have $(-1)^{k-1}\frac1k=_{\Bbb F_p}(-1)^{j-1}\frac1j$ $\iff$ $k=_{\Bbb F_p}-j$ $\iff$ $k+j=_{\Bbb Z}p$. So, each element of the multi-set $[(-1)^{k-1}\frac1k\in\Bbb F_p \mid 0<k<p]$ is the same as exactly another element. That means the set $S_p$ contains exactly $\frac{p-1}2$ element.
