Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
Drexel University or The Math Forum.



Re: Conjecture Prime
Posted:
Jun 26, 2009 2:01 PM


Musatov wrote: Musatov wrote: > Musatov wrote: > pubkeybreaker wrote: > > On Jun 25, 8:53 am, "Dik T. Winter" <Dik.Win...@cwi.nl> wrote: > > > In article <d0f06d1808444c109f84d20bfd704...@z20g2000prh.googlegroups.com> Musatov <marty.musa...@gmail.com> writes: > > > > On Jun 24, 10:39 pm, ab <hobk...@gmail.com> wrote: > > > ... > > > > > > > > The equation and opeation must be written as follows: > > > > > > > > (N*N)((N*N)+2)1 > > > ... > > > > > > 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 > > > > > > COMPOSITES. > > > > > > > > > > unique prime factors? or do you allow a prime repeated, for example > > > > > (N*N)((N*N)+2)  1 = p^2 > > > > > > > > It's not up to me, I am stating factual discovery. Or at least is is > > > > my intent. I cannot see how multiplying to even numbers with a > > > > difference of two will allow for a composite square with a identical > > > > prime factors. > > > > > > > > Do you? > > > > > > N = 46, (N*N)(N*N+2)1 = 4481687 = 7 * 7 * 91463 > > > > Indeed. Just apply Hensel's lemma to a solution mod 7. It will > > also have solutions mod 7^3, 7^4, ........ > > > > Note also that n^4 +2n^21 will be divisible by 7^2 for n = 46 + > > 49k > > for all integer k. Note the solution at n = 3. It also has > > solutions at > > 3 + 49k for all integer k...... (note that the function is even) > > The intent is to create a useful function, not win a Field's award. > Thanks for the information and feedback. > >  > Musatov
What I meant to say, is the same in the instance of two identical prime factors they will not be the only two prime factors in this case. This is to say the 'everything but the square of the prime function'.
 Musatov



