## Re: [PrimeNumbers] Important prime number relationship

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