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From: recoder on 29 Jun 2010 09:32
what is 2^(p-2) mod (p^2)
and what is the name of the topic in Number theory which handles such
congruences?
Thanx
From: Pubkeybreaker on 29 Jun 2010 13:15
On Jun 29, 9:32 am, recoder <kurtulmeh...(a)gmail.com> wrote:
> what is 2^(p-2) mod (p^2)
> and what is the name of the topic in Number theory which handles such
> congruences?
> Thanx

2^(p-2) mod p is just 2^-1 mod p == (p+1)/2
Now apply Hensel's Lemma to lift to p^2.

I don't know of any particular name.
From: spudnik on 29 Jun 2010 19:31
Hensel's lemma -- yeah, team!

anyway, what is it called, that one can use "mod p"
on either side of the equation or inequality?

> 2^(p-2) mod p   is just  2^-1 mod p == (p+1)/2

--BP's cap&trade plus free beer points on the CO2 creds!
http://wlym.com
From: Gerry Myerson on 29 Jun 2010 19:52
In article
<f9b42ac4-e976-4409-ace9-5b51bc37f246(a)q12g2000yqj.googlegroups.com>,
Pubkeybreaker <pubkeybreaker(a)aol.com> wrote:

> On Jun 29, 9:32�am, recoder <kurtulmeh...(a)gmail.com> wrote:
> > what is 2^(p-2) mod (p^2)
> > and what is the name of the topic in Number theory which handles such
> > congruences?
> > Thanx
>
> 2^(p-2) mod p is just 2^-1 mod p == (p+1)/2
> Now apply Hensel's Lemma to lift to p^2.

Hensel's Lemma?

2^(p-1) = 1 (mod p) (provided that p is prime)
but 2^(p-1) (mod p^2) is another kettle of fish.
Very little is known about it.
Only 2 primes are known for which 2^(p-1) = 1 (mod p^2).
The same would hold, mutatis mutandis, for 2^(p-2).

And if p is not prime (and OP may well have meant it to be,
but didn't say so), then all bets are off.

--
Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Bill Dubuque on 29 Jun 2010 20:57
Gerry Myerson <gerry(a)maths.mq.edi.ai.i2u4email> writes:
> Pubkeybreaker <pubkeybreaker(a)aol.com> wrote:
>>
>> On Jun 29, 9:32 am, recoder <kurtulmeh...(a)gmail.com> wrote:
>>> what is 2^(p-2) mod (p^2)
>>> and what is the name of the topic in Number theory which handles such
>>> congruences?
>>
>> 2^(p-2) mod p is just 2^-1 mod p == (p+1)/2
>> Now apply Hensel's Lemma to lift to p^2.
>
> Hensel's Lemma?
>
> 2^(p-1) = 1 (mod p) (provided that p is prime)
> but 2^(p-1) (mod p^2) is another kettle of fish.
> Very little is known about it.
> Only 2 primes are known for which 2^(p-1) = 1 (mod p^2).
> The same would hold, mutatis mutandis, for 2^(p-2).

Google "Wieferich Prime"
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