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- From: goddard@NeXTwork.Rose-Hulman.Edu (Bart Goddard)
- Subject: Re: Can you prove it?
- Message-ID: <1992Aug20.144618.15637@cs.rose-hulman.edu>
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- Organization: Rose-Hulman Institute of Technology
- References: <1992Aug20.130921.3437@linus.mitre.org>
- Date: Thu, 20 Aug 1992 14:46:18 GMT
- Lines: 16
-
-
- In article <1992Aug20.130921.3437@linus.mitre.org> bs@gauss.mitre.org
- (Robert D. Silverman) writes:
- > :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.
-
- Well, 2 has order d which divides p-1. Case 1 is OK with p-1
- replaced by d. Case 2 shows d=q-1, so by definition of d
- we have 2^(q-1) == 1 (mod q^2)
-
- >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.
- This is only true if d=p.
-