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Topic: Conjecture Prime
Replies: 33   Last Post: Apr 26, 2013 7:21 AM

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Re: Conjecture Prime
Posted: Jun 25, 2009 1:34 AM
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On Jun 24, 10:17 pm, ab <> wrote:
> On Jun 25, 1:36 pm, "" <> wrote:

> > Even 'n' only.
> > The equation and opeation must be written as follows:
> > (N*N)((N*N)+2)-1
> >
> > "Better Google Search"

> > In instance of a composite number it will have at least two prime
> > factors.

> what does this last statement mean? are you asserting the equation is
> either prime or composite? well of course it is.
> or do you mean in the instance of a composite number it will have at
> least two *unique* primes factors? in other words the equation is
> anything except the square of a prime?

I mean it to say when "n" is even we plug it in the equation...

for example

4*6-1=23 Prime

Now when I say it produces prime number or composites, I am saying it
produces ALL primes, except when it does not
it will produce at least two prime factors from the exception

It is rational square building with primes.

Let me break it down:

1. I think this is the surest way to an understanding of prime #'s.
Basically thinking of primes as an "exception" to the composite of
all ...

2. The initial idea comes from the following observation: The number
that immediately precedes a perfect square is never prime with the
single exception of 4 ...

3. So there is a Pattern in the Primes Factors. I noticed that the
number of prime factors along each of the lines of the magic square
equaled 7 in every direction. The possible exception to the
exceptional set for the sum of a prime and a square.

Musatov's Theorem: Every large integer n is either a square or the sum
of a prime.


For all primitive characters z(mod q), q<: T, with the possible
exception of at most Prime numbers with the exception of 2, then, --
which is the only even prime -- a prime number is a kind ....when a
prime appears twice, that product is a square number. ...

Sum of a prime and a square is the possible exception of at most one
primitive character x mod r.

In this note, we shall remove the condition that 1+4m is square free.

All 14 Rabinowitsch polynomials fm(x)=x2+x-m with at most one possible

Because 1+4m=245=5·72 is not square free and 102+10-61=49=72 is not


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