> On Jun 25, 8:53 am, "Dik T. Winter" > <Dik.Win...@cwi.nl> wrote: > > In article > <d0f06d18-0844-4c10-9f84-d20bfd704...@z20g2000prh.goog > legroups.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^2-1 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) >
musatov thinks he is to cool for all of us, and does not need to explain himself clearly: we are the serfs that attend to the needs of the genius he believes himself to be. So he feels no need to clearly state a problem, or he is to ignorant to be able to really understand what he wants and what he's after, other than attention.
After a reply, musatov says, e.g:
Of course I mean by $# that you need to first multiply by 2.
Yet another buffoon, with his sorry P=NP, trashing this site.