pq - (p+q) is prime
Thudan...@googlemail.com
+q) is also prime.
quasi
wrote:
>Prove that for every prime p there exists a prime q such that pq - (p
>+q) is also prime.
Well, p is the larger prime of a pair of twin primes, use q = 2.
Of course, that hardly makes any progress.
The claim is almost certainly true, but I don't think you will be able
to prove it.
Where did you get the problem from?
quasi
Thudan...@googlemail.com
> Where did you get the problem from?
Another forum.
quasi
wrote:
Forget it -- the problem is out of range.
quasi
galathaea
> On Thu, 6 Dec 2007 22:41:13 -0800 (PST), ThudanBlun...@googlemail.com
> wrote:
>
> >On Dec 7, 6:37 am, quasi <qu...@null.set> wrote:
> >> On Thu, 6 Dec 2007 22:15:51 -0800 (PST), ThudanBlun...@googlemail.com
> >> wrote:
>
> >> Where did you get the problem from?
>
> >Another forum.
>
> Forget it -- the problem is out of range.
it seems to scream out for dirichlet's theorem
but i can't get it work
-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-=-
galathaea: prankster, fablist, magician, liar
quasi
<gala...@gmail.com> wrote:
>On Dec 6, 10:46 pm, quasi <qu...@null.set> wrote:
>> On Thu, 6 Dec 2007 22:41:13 -0800 (PST), ThudanBlun...@googlemail.com
>> wrote:
>>
>> >On Dec 7, 6:37 am, quasi <qu...@null.set> wrote:
>> >> On Thu, 6 Dec 2007 22:15:51 -0800 (PST), ThudanBlun...@googlemail.com
>> >> wrote:
>>
>> >> Where did you get the problem from?
>>
>> >Another forum.
>>
>> Forget it -- the problem is out of range.
>
>it seems to scream out for dirichlet's theorem
> but i can't get it work
If q wasn't required to be prime, Dirichlet's theorem would clinch it
instantly.
It needs a kind of "double-Dirichlet" theorem.
The truth of the following conjecture would suffice:
If a,b are coprime positive integers, not both odd, there are
infinitely many primes q such that a*q + b is also prime.
quasi
quasi
Note -- using a = 1, b =2, the above conjecture reduces to the Twin
Prime Conjecture.
quasi
philippe
"quasi" <qu...@null.set> a écrit dans le message de
news:c9qhl3pd6hc9fff39...@4ax.com...
It's equivalent to say that {phi(p,q)+1} is prime but ....
philippe
"quasi" <qu...@null.set> a écrit dans le message de
news:c9qhl3pd6hc9fff39...@4ax.com...
Without making mistake it's equivalent to say that
phi(p*q)-1 is prime!
But, ...
quasi
>On Fri, 07 Dec 2007 02:15:18 -0500, quasi <qu...@null.set> wrote:
>
>>On Thu, 6 Dec 2007 22:52:36 -0800 (PST), galathaea
>><gala...@gmail.com> wrote:
>>
>>>On Dec 6, 10:46 pm, quasi <qu...@null.set> wrote:
>>>> On Thu, 6 Dec 2007 22:41:13 -0800 (PST), ThudanBlun...@googlemail.com
>>>> wrote:
>>>>
>>>> >On Dec 7, 6:37 am, quasi <qu...@null.set> wrote:
>>>> >> On Thu, 6 Dec 2007 22:15:51 -0800 (PST), ThudanBlun...@googlemail.com
>>>> >> wrote:
>>>>
>>>> >> Where did you get the problem from?
>>>>
>>>> >Another forum.
>>>>
>>>> Forget it -- the problem is out of range.
>>>
>>>it seems to scream out for dirichlet's theorem
>>> but i can't get it work
>>
>>If q wasn't required to be prime, Dirichlet's theorem would clinch it
>>instantly.
>>
>>It needs a kind of "double-Dirichlet" theorem.
For contrast, let's make 2 versions. The first one is weaker, but if
it's true, it's still strong enough to instantly prove the OP's claim.
Conjecture (1):
If a,b are coprime positive integers, not both odd, there is
at least one prime q such that a*q + b is also prime.
Conjecture (2):
If a,b are coprime positive integers, not both odd, there are
infinitely many primes q such that a*q + b is also prime.
Remarks:
As noted in my prior reply, the truth of Conjecture (2) would already
imply the Twin Prime Conjecture. Just use a = 1, b = 2.
quasi
philippe
<Thudan...@googlemail.com> a écrit dans le message de
news:302bf2bc-6d7c-49a9...@w34g2000hsg.googlegroups.com...
Can you post some link related to the original forum?
BR
fjm...@gmail.com
> Prove that for every prime p there exists a prime q such that pq - (p
> +q) is also prime.
p=2, q=3 your formula gives 1, which is not usually treated as a
prime. p=2, q=2 gives 0 (worse surely)
p=2, q=11 appears to give 9 (unless my sums are wrong) which is even
less of a prime than 1 or 0.
Francis
Denis Feldmann
I think the OP knew all this. You, on the other hand, seem not very
familiar with the meaning of "for every x, there exists y such..."
>
> Francis
Robert Israel
Did you miss the "there exists"?
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
ThudanBlunder
>
> BR
Here you are:
http://tinyurl.com/24q8xr