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From: david on 31 Jul 2010 21:38
Consider the following equation for the given condition.

2^[(k-1)(k-2)] = 1 mod k^2 (1)

Condition: prime k > 3.

Question: For what value of k (1) is valid?

A helpful reply will be appreciated.

David
From: Rob Johnson on 1 Aug 2010 00:48
In article <92e181c5-a6b2-4280-8892-90d3ef0f931c(a)q22g2000yqm.googlegroups.com>,
david <dlee753(a)yahoo.com> wrote:
>Consider the following equation for the given condition.
>
> 2^[(k-1)(k-2)] = 1 mod k^2 (1)
>
>Condition: prime k > 3.
>
>Question: For what value of k (1) is valid?
>
>A helpful reply will be appreciated.

If k is prime, then phi(k^2) = k(k-1). If k is odd, gcd(2,k) = 1.
These two conditions imply

k(k-1) 2
2 = 1 mod k [1]

Using this, we get that

(k-1)(k-2) -2(k-1) 2
2 = 2 mod k [2]

So, your question reduces to

k-1 2
4 = 1 mod k [3]

Since 2^{k-1} = 1 mod k^2 for Wieferich primes, [3] must be true
also, but there could be other primes that satisfy [3]. I have run
a check on the first 5,000,000 primes, and the only k less than or
equal to 86,028,121 which satisfy [3] are the Wieferich primes:
1093 and 3511.

Rob Johnson <rob(a)trash.whim.org>
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