Re: Simple answer, surrogate factoring

From: David C. Ullrich (ullrich_at_math.okstate.edu)
Date: 03/05/05 

Date: Sat, 05 Mar 2005 06:29:49 -0600

On 4 Mar 2005 14:49:14 -0800, jstevh@msn.com wrote:

>[...]
>
>Well, the proof is actually not hard, and it is perfect, so that means
>in this case I can say definitely that you made some mistake.

Well, as long as the proof is finally perfect.

How many times has it happened that you explained that someone
must be wrong because your proof of something was correct?
Then how many times did it turn out that your proof was wrong.

Not that those numbers have any relevance _here_, since the
current proof is perfect.

>First off, you *will* get a factor of M, which has to be divided off,
>and maybe you can get more than one factor of M, but in any event, you
>have to divide off all factors of M, first.
>
>What's left over will have a single prime factor of M for at least one
>case, as the mathematical proof is in this time.
>
>If you do it right, it will work.
>
>
>James Harris

************************

David C. Ullrich

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