Lemma used to prove Von Staudt's Theorem

  • #1
274
1

Main Question or Discussion Point

I am reading Hardy's Intro to the Theory of Numbers and I am currently trying to work through the proof of Von Staudt's Theorem. Hardy first proves the following lemma.
$$
\sum\limits_{1}^{p-1}m^{k}\equiv -\epsilon_{k}(p) (\mod p).
$$
Proof: If ##(p-1)|k## then ##m^{k}\equiv 1## by Fermat's Theorem and
$$
\sum m^{k}\equiv p-1\equiv -1\equiv -\epsilon_{k}(p) (\mod p).
$$
If ##(p-1)\not|k## and ##g## is a primitive root of ##p## then ##g^{k}\not\equiv 1 (\mod p)##.

Everything makes sense so far, but then Hardy goes on to say:

The sets ##g,2g,...,(p-1)g## and ##1,2,...,p-1## are equivalent (mod p).

I don't see how these two sets are equivalent. I believe these two sets are residues but I am very confused about what residues are in general and so I cannot understand what Hardy means when he says that the two sets are equivalent. Any help to understand the last statement that Hardy makes or just any conversation about residues in general to help me understand them better would be greatly appreciated. Thanks!
 

Answers and Replies

  • #2
274
1
In case anyone else ever has this problem, I was able to figure out why the sets are equivalent. By a previous theorem about 100 pages or so before this lemma, Hardy proved that if (g,p)=1 and 1,2,...,p-1 are a set of incongruent residues mod p then g,2g,...,g(p-1) is also such a set.

I would still love to talk about anything that has to do with number theory and residues if anybody is interested.
 

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