Re: 2(p^2)+1 = 0 mod 3 for any prime p != 3

Tim Dierks (timd@consensus.com)
Wed, 25 Mar 1998 12:41:35 -0800

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At 7:35 PM +0100 3/25/98, Hamdi Tounsi wrote:
>Hi,
>can someone give me a proof for that ?
>

This is true not only for all primes != 3, but for all numbers which are
not multiples of 3.

It's a trivial result if you do all the computations modulo 3. In this field,

        2(p^2)+1 = 0
        2(p^2) = 2
        p^2 = 1

Since 1^2 = 1 mod 3 and 2^2 = 1 mod 3, any number which isn't an even
multiple of 3 meets the last condition. Thus, all numbers which are not
even multiples of three pass the test. Since all primes other than three
are not even multiples of three, the statement is true.

 - Tim

Tim Dierks - timd@consensus.com - www.consensus.com
  Director of Engineering - Consensus Development
  Developer of SSL Plus: SSL 3.0 Integration Suite

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