5.- Conjecture or theorem ?
in [Math]
|
From: Ludovicus on 30 Jul 2010 13:38 On 28 jul, 16:32, Kermit Rose <ker...(a)POLARIS.NET> wrote: > "The set of least divisors of x^2 + x + A is the set of prime > numbers, (A = an odd integer.) > for each A testing x = 0,1,2,3...(Non negative integers)." > > The least divisor of a number > 1, is, by nature, > a prime number. > > x^2 + x + A > > For x = 0, (A + 0); x^2 + x + A = {1,3,5,7,....} > For x = 1, (A + 2): x^2 + x + A = {3,5,7,...} > For x = 2, (A + 3): x^2 + x + A = {4,6,8,10,...} > > For A ranging over the odd integers, > x^2 + x + A ranges over all integers. > > Your conjecture is trivially true. > > Kermit Rose Trivial because my wording was incorrect. This is the true wording: "Each odd prime is the smaller divisor of the polynom Y = X^2 + X + A for some A and for all X = 0, 1, 2, 3 ,... infinite." Given a prime p it is not easy to find the corresponding A that makes Y not divisible by primes < p. It is possible that an A be inexistent. For example: Which is the value of A that makes Y only divisible by primes => 331 ? Ludovicus
From: Gerry on 30 Jul 2010 18:57 On Jul 31, 3:38 am, Ludovicus <luir...(a)yahoo.com> wrote: > On 28 jul, 16:32, Kermit Rose <ker...(a)POLARIS.NET> wrote: > > > > > > > "The set of least divisors of x^2 + x + A is the set of prime > > numbers, (A = an odd integer.) > > for each A testing x = 0,1,2,3...(Non negative integers)." > > > The least divisor of a number > 1, is, by nature, > > a prime number. > > > x^2 + x + A > > > For x = 0, (A + 0); x^2 + x + A = {1,3,5,7,....} > > For x = 1, (A + 2): x^2 + x + A = {3,5,7,...} > > For x = 2, (A + 3): x^2 + x + A = {4,6,8,10,...} > > > For A ranging over the odd integers, > > x^2 + x + A ranges over all integers. > > > Your conjecture is trivially true. > > > Kermit Rose > > Trivial because my wording was incorrect. > This is the true wording: > > "Each odd prime is the smaller divisor of the polynom Y = X^2 + X + A > for some A and for all X = 0, 1, 2, 3 ,... infinite." > > Given a prime p it is not easy to find the corresponding A that > makes Y not divisible by primes < p. > It is possible that an A be inexistent. > For example: Which is the value of A that makes Y only > divisible by primes => 331 ? > Ludovicus There's no one value of A, there are infinitely many, and the methods given earlier in this thread enable you to find them (in principle). -- Gerry
From: Ludovicus on 31 Jul 2010 21:57 O > There's no one value of A, there are infinitely many, > and the methods given earlier in this thread enable you > to find them (in principle). > -- > Gerry In that case it is a theorem. Do you know the demonstration? Ludovicus
From: Gerry on 1 Aug 2010 02:22
On Aug 1, 11:57 am, Ludovicus <luir...(a)yahoo.com> wrote: > O > > > There's no one value of A, there are infinitely many, > > and the methods given earlier in this thread enable you > > to find them (in principle). > > -- > > Gerry > > In that case it is a theorem. Do you know the demonstration? > Ludovicus Have you tried putting together a proof from the discussions in this thread? -- GM |