## Re: [PrimeNumbers] On the representation of some even numbers as sums of two prime numbers

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• yes, because gaps between prime numbers grow arbitrarily large. Maximilian On Wednesday, February 2, 2011, Dimiter Skordev
Message 1 of 11 , Feb 2, 2011
yes, because gaps between prime numbers grow arbitrarily large.
Maximilian

On Wednesday, February 2, 2011, Dimiter Skordev
<skordev@...-sofia.bg> wrote:
> Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
>
>
• I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A
Message 2 of 11 , Feb 3, 2011
I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q > p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.

--- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@...> wrote:
>
> yes, because gaps between prime numbers grow arbitrarily large.
> Maximilian
>
> On Wednesday, February 2, 2011, Dimiter Skordev
> <skordev@...> wrote:
> > Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
> >
> >
>
• Where do you disagree with the following reasoning I mailed you privately: Let M = max F and P the set of prime numbers. The set P+F is the set of all numbers
Message 3 of 11 , Feb 3, 2011
Where do you disagree with the following reasoning I mailed you privately:

Let M = max F and P the set of prime numbers.
The set P+F is the set of all numbers that can be written as sum of a
number from F plus some arbitrary prime q from P.
It is contained in the set P + [0, M],
and this set has (infinitely many, arbitrarily large) holes, since in P there are infinitely many gaps larger than M.
So it is sufficient to take any number of the form 2p (with p in P)
which is inside such a hole.

I admit that I did not prove that in at least one of all these (infinitely many, arbitrarily large) holes there will be a number of the form 2p, but I think this should be not so difficult. If you think that this is a major problem, I'll give it a second thought.

Regards,
Maximilian

--- In primenumbers@yahoogroups.com, "Dimiter Skordev" <skordev@...> wrote:
>
> I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q > p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.
>
> --- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@> wrote:
> >
> > yes, because gaps between prime numbers grow arbitrarily large.
> > Maximilian
> >
> > On Wednesday, February 2, 2011, Dimiter Skordev
> > <skordev@> wrote:
> > > Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
> > >
• A concrete example, perhaps... Let the set F be {3,5}. All you need to do is find a prime number p such that 2p-3 and 2p-5 are not prime. Use the chinese
Message 4 of 11 , Feb 3, 2011
A concrete example, perhaps...

Let the set F be {3,5}. All you need to do is find a prime number p such that 2p-3 and 2p-5 are not prime.

Use the chinese remainder theorem to find N such that:

N == 2 mod 4.
N == 1 mod 3.
N == 1 mod 5.
N == 3 mod 7.
N == 5 mod 11.

In this case, N must be of the form 4620x + 346.

Let N be 2p, then p must be of the form 2310x + 173. According to Dirichlet, there are an infinite
number of primes of that form. For none of those primes is 2p ever the sum of 3 and a prime or the sum of
5 and a prime.

If the set F gets bigger, the numbers D and A in the arithmetic progression Dx + A get larger, but Dirichlet
always applies and there are an infinite number of primes of the right form.

On 2/3/2011 5:27 AM, maximilian_hasler wrote:
> Where do you disagree with the following reasoning I mailed you privately:
>
> Let M = max F and P the set of prime numbers.
> The set P+F is the set of all numbers that can be written as sum of a
> number from F plus some arbitrary prime q from P.
> It is contained in the set P + [0, M],
> and this set has (infinitely many, arbitrarily large) holes, since in P there are infinitely many gaps larger than M.
> So it is sufficient to take any number of the form 2p (with p in P)
> which is inside such a hole.
>
> I admit that I did not prove that in at least one of all these (infinitely many, arbitrarily large) holes there will be a number of the form 2p, but I think this should be not so difficult. If you think that this is a major problem, I'll give it a second thought.
>
> Regards,
> Maximilian
>
> --- In primenumbers@yahoogroups.com, "Dimiter Skordev"<skordev@...> wrote:
>>
>> I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q> p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.
>>
>> --- In primenumbers@yahoogroups.com, Maximilian Hasler<maximilian.hasler@> wrote:
>>>
>>> yes, because gaps between prime numbers grow arbitrarily large.
>>> Maximilian
>>>
>>> On Wednesday, February 2, 2011, Dimiter Skordev
>>> <skordev@> wrote:
>>>> Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
>>>>
>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
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>
>
• Thank you for your answer. It makes your statement rather plausible, but, anyway, the existence of a number of the form 2p with prime p in some of the holes of
Message 5 of 11 , Feb 3, 2011
Thank you for your answer. It makes your statement rather plausible, but, anyway, the existence of a number of the form 2p with prime p in some of the holes of the set P+[0,M} needs a proof.

Best regards,
Dimiter

--- In primenumbers@yahoogroups.com, "maximilian_hasler" <maximilian.hasler@...> wrote:
>
> Where do you disagree with the following reasoning I mailed you privately:
>
> Let M = max F and P the set of prime numbers.
> The set P+F is the set of all numbers that can be written as sum of a
> number from F plus some arbitrary prime q from P.
> It is contained in the set P + [0, M],
> and this set has (infinitely many, arbitrarily large) holes, since in P there are infinitely many gaps larger than M.
> So it is sufficient to take any number of the form 2p (with p in P)
> which is inside such a hole.
>
> I admit that I did not prove that in at least one of all these (infinitely many, arbitrarily large) holes there will be a number of the form 2p, but I think this should be not so difficult. If you think that this is a major problem, I'll give it a second thought.
>
> Regards,
> Maximilian
>
> --- In primenumbers@yahoogroups.com, "Dimiter Skordev" <skordev@> wrote:
> >
> > I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q > p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.
> >
> > --- In primenumbers@yahoogroups.com, Maximilian Hasler <maximilian.hasler@> wrote:
> > >
> > > yes, because gaps between prime numbers grow arbitrarily large.
> > > Maximilian
> > >
> > > On Wednesday, February 2, 2011, Dimiter Skordev
> > > <skordev@> wrote:
> > > > Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
> > > >
>
• My previous message gave the outline of a proof of your original statement which doesn t explicitly use the concept of prime gaps, but rather Dirichlet s
Message 6 of 11 , Feb 3, 2011
My previous message gave the outline of a proof of your original statement which doesn't explicitly use the concept of prime gaps, but rather Dirichlet's existence of primes in an arithmetic
progression.

Of course, they are somewhat related, as Dirichlet can be used to show that there are arbitrarily long prime gaps leading up to numbers of the form 2p.

On 2/3/2011 7:50 AM, Dimiter Skordev wrote:
> Thank you for your answer. It makes your statement rather plausible, but, anyway, the existence of a number of the form 2p with prime p in some of the holes of the set P+[0,M} needs a proof.
>
> Best regards,
> Dimiter
>
> --- In primenumbers@yahoogroups.com, "maximilian_hasler"<maximilian.hasler@...> wrote:
>>
>> Where do you disagree with the following reasoning I mailed you privately:
>>
>> Let M = max F and P the set of prime numbers.
>> The set P+F is the set of all numbers that can be written as sum of a
>> number from F plus some arbitrary prime q from P.
>> It is contained in the set P + [0, M],
>> and this set has (infinitely many, arbitrarily large) holes, since in P there are infinitely many gaps larger than M.
>> So it is sufficient to take any number of the form 2p (with p in P)
>> which is inside such a hole.
>>
>> I admit that I did not prove that in at least one of all these (infinitely many, arbitrarily large) holes there will be a number of the form 2p, but I think this should be not so difficult. If you think that this is a major problem, I'll give it a second thought.
>>
>> Regards,
>> Maximilian
>>
>> --- In primenumbers@yahoogroups.com, "Dimiter Skordev"<skordev@> wrote:
>>>
>>> I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q> p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.
>>>
>>> --- In primenumbers@yahoogroups.com, Maximilian Hasler<maximilian.hasler@> wrote:
>>>>
>>>> yes, because gaps between prime numbers grow arbitrarily large.
>>>> Maximilian
>>>>
>>>> On Wednesday, February 2, 2011, Dimiter Skordev
>>>> <skordev@> wrote:
>>>>> Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
>>>>>
>>
>
>
>
>
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> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
>
>
• Dear Jack, Your example is nice. But why you do not look for a prime number p of the simpler form 77X+19 ? Best regards, Dimiter
Message 7 of 11 , Feb 3, 2011
Dear Jack,

Your example is nice. But why you do not look for a prime number p of the simpler form 77X+19 ?

Best regards,
Dimiter

--- In primenumbers@yahoogroups.com, Jack Brennen <jfb@...> wrote:
>
> My previous message gave the outline of a proof of your original statement which doesn't explicitly use the concept of prime gaps, but rather Dirichlet's existence of primes in an arithmetic
> progression.
>
> Of course, they are somewhat related, as Dirichlet can be used to show that there are arbitrarily long prime gaps leading up to numbers of the form 2p.
>
>
>
> On 2/3/2011 7:50 AM, Dimiter Skordev wrote:
> > Thank you for your answer. It makes your statement rather plausible, but, anyway, the existence of a number of the form 2p with prime p in some of the holes of the set P+[0,M} needs a proof.
> >
> > Best regards,
> > Dimiter
> >
> > --- In primenumbers@yahoogroups.com, "maximilian_hasler"<maximilian.hasler@> wrote:
> >>
> >> Where do you disagree with the following reasoning I mailed you privately:
> >>
> >> Let M = max F and P the set of prime numbers.
> >> The set P+F is the set of all numbers that can be written as sum of a
> >> number from F plus some arbitrary prime q from P.
> >> It is contained in the set P + [0, M],
> >> and this set has (infinitely many, arbitrarily large) holes, since in P there are infinitely many gaps larger than M.
> >> So it is sufficient to take any number of the form 2p (with p in P)
> >> which is inside such a hole.
> >>
> >> I admit that I did not prove that in at least one of all these (infinitely many, arbitrarily large) holes there will be a number of the form 2p, but I think this should be not so difficult. If you think that this is a major problem, I'll give it a second thought.
> >>
> >> Regards,
> >> Maximilian
> >>
> >> --- In primenumbers@yahoogroups.com, "Dimiter Skordev"<skordev@> wrote:
> >>>
> >>> I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q> p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.
> >>>
> >>> --- In primenumbers@yahoogroups.com, Maximilian Hasler<maximilian.hasler@> wrote:
> >>>>
> >>>> yes, because gaps between prime numbers grow arbitrarily large.
> >>>> Maximilian
> >>>>
> >>>> On Wednesday, February 2, 2011, Dimiter Skordev
> >>>> <skordev@> wrote:
> >>>>> Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
> >>>>>
> >>
> >
> >
> >
> >
> > ------------------------------------
> >
> > Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> > The Prime Pages : http://www.primepages.org/
> >
> >
> >
> >
> >
> >
>
• Okay, yeah, that would work. :) If p is of the form 77x+19, then 2p is of the form 154x+38, and 2p-3 is always divisible by 7 and 2p-5 is always divisible by
Message 8 of 11 , Feb 3, 2011
Okay, yeah, that would work. :)

If p is of the form 77x+19, then 2p is of the form 154x+38, and
2p-3 is always divisible by 7 and 2p-5 is always divisible by 11.
So if 2p is a sum of two primes, neither of them is equal to 3 or 5.

I guess I was over-specifying the arithmetic progression to make
sure that none of its terms are divisible by 2, 3, or 5. None of
which affects the proof of its existence, of course.

On 2/3/2011 9:10 AM, Dimiter Skordev wrote:
> Dear Jack,
>
> Your example is nice. But why you do not look for a prime number p of the simpler form 77X+19 ?
>
> Best regards,
> Dimiter
>
>
> --- In primenumbers@yahoogroups.com, Jack Brennen<jfb@...> wrote:
>>
>> My previous message gave the outline of a proof of your original statement which doesn't explicitly use the concept of prime gaps, but rather Dirichlet's existence of primes in an arithmetic
>> progression.
>>
>> Of course, they are somewhat related, as Dirichlet can be used to show that there are arbitrarily long prime gaps leading up to numbers of the form 2p.
>>
>>
>>
>> On 2/3/2011 7:50 AM, Dimiter Skordev wrote:
>>> Thank you for your answer. It makes your statement rather plausible, but, anyway, the existence of a number of the form 2p with prime p in some of the holes of the set P+[0,M} needs a proof.
>>>
>>> Best regards,
>>> Dimiter
>>>
>>> --- In primenumbers@yahoogroups.com, "maximilian_hasler"<maximilian.hasler@> wrote:
>>>>
>>>> Where do you disagree with the following reasoning I mailed you privately:
>>>>
>>>> Let M = max F and P the set of prime numbers.
>>>> The set P+F is the set of all numbers that can be written as sum of a
>>>> number from F plus some arbitrary prime q from P.
>>>> It is contained in the set P + [0, M],
>>>> and this set has (infinitely many, arbitrarily large) holes, since in P there are infinitely many gaps larger than M.
>>>> So it is sufficient to take any number of the form 2p (with p in P)
>>>> which is inside such a hole.
>>>>
>>>> I admit that I did not prove that in at least one of all these (infinitely many, arbitrarily large) holes there will be a number of the form 2p, but I think this should be not so difficult. If you think that this is a major problem, I'll give it a second thought.
>>>>
>>>> Regards,
>>>> Maximilian
>>>>
>>>> --- In primenumbers@yahoogroups.com, "Dimiter Skordev"<skordev@> wrote:
>>>>>
>>>>> I do not see how the existence of arbitrarily long gaps between prime numbers would imply the existence of a prime number with the indicated property. A straight-forward reasoning could start with considering a number m greater than all numbers in F, and the observation that the equality 2*p=q+r, where q belongs to F, implies r-p = p-q> p-m. But it is not clear to me how to go further, since p-m can be arbitrarily large.
>>>>>
>>>>> --- In primenumbers@yahoogroups.com, Maximilian Hasler<maximilian.hasler@> wrote:
>>>>>>
>>>>>> yes, because gaps between prime numbers grow arbitrarily large.
>>>>>> Maximilian
>>>>>>
>>>>>> On Wednesday, February 2, 2011, Dimiter Skordev
>>>>>> <skordev@> wrote:
>>>>>>> Let F be a finite set of prime numbers. Is it sure that a prime number p exists such that, whenever 2p is the sum of two prime numbers, none of them belongs to F?
>>>>>>>
>>>>
>>>
>>>
>>>
>>>
>>> ------------------------------------
>>>
>>> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
>>> The Prime Pages : http://www.primepages.org/
>>>
>>>
>>>
>>>
>>>
>>>
>>
>
>
>
>
> ------------------------------------
>
> Unsubscribe by an email to: primenumbers-unsubscribe@yahoogroups.com
> The Prime Pages : http://www.primepages.org/
>
>
>
>
>
>
• ... Why bother using 7 and 11 in the first place? We can use one prime in F = {3,5} to wipe out the other: p = 15*x + 19. Then 3 divides 38-5 and 5 divides
Message 9 of 11 , Feb 3, 2011
Jack Brennen <jfb@...> wrote:

> If p is of the form 77x+19

Why bother using 7 and 11 in the first place?
We can use one prime in F = {3,5} to wipe out the other:
p = 15*x + 19. Then 3 divides 38-5 and 5 divides 38-3:
mutual annihilation :-)

David
• ... More generally, if F = {q,r} contains two distinct odd primes, then any one of the infinity of primes of the form p = q*r*x + (q+r)/2 solves the problem,
Message 10 of 11 , Feb 3, 2011