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Aug 12, 2016 at 20:48 comment added user44191 $k \equiv 1 (mod n)$ didn't affect the proof; however, it would be necessary for any such partition, as the $s$-cosets must all have size $n$. Therefore, stating it is unnecessary. The only assumptions I really used about $p, k, n$ are that $1 < n, 1 < k < p, p$ prime.
Aug 12, 2016 at 19:05 comment added Sávio I mean, $2 \le k \le p-1$
Aug 12, 2016 at 4:14 comment added Sávio Indeed, the hypothesis $k \equiv 1 \pmod n$ is useless. You just used $1 \le k \le q-1$, right? Thanks!
Aug 12, 2016 at 2:06 vote accept Sávio
Aug 12, 2016 at 1:58 history answered user44191 CC BY-SA 3.0