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- From: bs@gauss.mitre.org (Robert D. Silverman)
- Newsgroups: sci.math
- Subject: Re: Can you prove it?
- Keywords: Number theory
- Message-ID: <1992Aug20.130921.3437@linus.mitre.org>
- Date: 20 Aug 92 13:09:21 GMT
- References: <1992Aug20.055633.19046@nuscc.nus.sg>
- Sender: news@linus.mitre.org (News Service)
- Organization: Research Computer Facility, MITRE Corporation, Bedford, MA
- Lines: 37
- Nntp-Posting-Host: gauss.mitre.org
-
- In article <1992Aug20.055633.19046@nuscc.nus.sg> bhonsle@bhonsle.iss.nus.sg (Shailendra K Bhonsle) writes:
- :Hi
- :I have a simple problem, can you prove or disprove it? I will appreciate all replies:
- :
- :Problem:
- :
- :Prove that for primes p & q, q<p
- :
- :if 2^(p-1) == 1 (mod q^2)
- :
- :then either 2^(q-1) == 1 (mod q^2) or q | p-1 { q divides (p-1}
-
- We have that 2 has order p-1 in the group of units of Z/q^2Z.
- Hence, p-1 | phi(q^2) by Lagrange's theorem.
-
- Consequently, p-1 | q(q-1), or q(q-1) = k(p-1) for some integer k.
- Case 1.
- k != q
-
- We must have k | q-1, hence p-1 | q(q-1)/k or p-1 | q k' for some
- integer k' and hence q | p-1 as required.
-
- Case 2.
- k = q.
-
- Therefore p-q = q-1 and we have p=q. The first known case where 2^p-1 = 1 mod p^2 is p=1093.
-
- Since p =q in this case and we were given 2^(p-1) = 1 mod q^2, then we
- also have 2^(q-1) = 1 mod q^2.
-
- QED
-
- --
- Bob Silverman
- These are my opinions and not MITRE's.
- Mitre Corporation, Bedford, MA 01730
- "You can lead a horse's ass to knowledge, but you can't make him think"
-