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• ## Important prime number relationship

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• Looking at prime number you can see this relationship: Conjeture: Any prime number can be written as the product of two primes plus or minus the difference
Message 1 of 4 , Apr 9, 2006
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Looking at prime number you can see this relationship:

Conjeture:

"Any prime number can be written as the product of two primes plus or
minus the difference between them."

This means that at one(at least one) or more of these four
relationships must be hold for every prime number:

(Pb*Pa)+(Pb+Pa)
(Pb*Pa)+(Pb-Pa)
(Pb*Pa)-(Pb+Pa)
(Pb*Pa)-(Pb-Pa)

Pb>Pa And both primes,(To Be consecutive primes is not a requierement)

This relationship explain some patterns about prime numbers. I think
yhat it is important. Tell me what do you think about it. Thank you

Jose.C Martinez
• ... Pb+Pa is not the difference between Pa and Pb. ... Ditto. ... Care to disambiguate what you really mean before we invest effort in the wrong one. Did you
Message 1 of 4 , Apr 9, 2006
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--- jcmtnez90 <jcmtnez90@...> wrote:
> Looking at prime number you can see this relationship:
>
> Conjeture:
>
> "Any prime number can be written as the product of two primes plus or
> minus the difference between them."
>
> This means that at one(at least one) or more of these four
> relationships must be hold for every prime number:
>
> (Pb*Pa)+(Pb+Pa)

Pb+Pa is not the difference between Pa and Pb.

> (Pb*Pa)+(Pb-Pa)
> (Pb*Pa)-(Pb+Pa)

Ditto.

> (Pb*Pa)-(Pb-Pa)
>
> Pb>Pa And both primes,(To Be consecutive primes is not a requierement)

Care to disambiguate what you really mean before we invest effort in the wrong
one. Did you actually mean "plus or minus the sum of, or difference between,
them"

I don't know if you've noticed that
Pa*Pb+Pa+Pb = (Pa+1)*(Pb+1)-1
and similar expressions for +-, -+, --.
So you can check your conjecture by looking at product of numbers which are
primes+/-1.

Phil
P

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• ... Putting aside that there is no solution for the prime 2, the first few cases where it does not hold are p = 101, 173 and 367. (Nice try though!) Mark
Message 1 of 4 , Apr 9, 2006
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--- In primenumbers@yahoogroups.com, "jcmtnez90" <jcmtnez90@...> wrote:
>
>
> This means that at one(at least one) or more of these four
> relationships must be hold for every prime number:
>
> (Pb*Pa)+(Pb+Pa)
> (Pb*Pa)+(Pb-Pa)
> (Pb*Pa)-(Pb+Pa)
> (Pb*Pa)-(Pb-Pa)
>

Putting aside that there is no solution for the prime 2, the first few
cases where it does not hold are p = 101, 173 and 367. (Nice try
though!)

Mark
• You are right, therefore I will change the conjecture to every prime greater than 2.
Message 1 of 4 , Apr 9, 2006
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You are right, therefore I will change the conjecture to every prime
greater than 2.
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