## Integers then Equals

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• 1. Integers then Equals Posted by: Sebastian Martin Ruiz s_m_ruiz@yahoo.es s_m_ruiz Date: Sat Mar 14, 2009 12:11 pm ((PDT)) Hello all: Hello Sebastián Let
Message 1 of 7 , Mar 15, 2009
1. Integers then Equals
Posted by: "Sebastian Martin Ruiz" s_m_ruiz@... s_m_ruiz
Date: Sat Mar 14, 2009 12:11 pm ((PDT))

Hello all:

Hello Sebastián

Let P(n)
the n-th prime number. Let Sqrt(P)=p^(1/2)

Kermit:
****
p1 = 2
p2 = 3
p3 = 5
2 + 7 = 9
5 - 1 = 4
sqrt(p1+7) = 3
sqrt(p3-1) = 2
****

If
Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:

Kermit:
*****
If sqrt(p(n-1) + 7) is an integer, then p(n-1) = 0 or 2 mod 3.
If sqrt(p(n+1) -1) is an integer, then p(n+1) = 1 or 2 mod 3.
p(n-1) = 0 mod 3 implies p = 3.

sqrt(3 + 7) is not an integer,

Thus

If sqrt(p(n-1) + 7) is an integer, then p(n-1) = 2 mod 3
if sqrt(p(n+1) -1) is an integer, then p(n+1) = 1 or 2 mod 3.

If sqrt(p(n-1) + 7) is an integer > 3,
and
sqrt(p(n+1) - 1) is an integer, then

p(n+1) = p(n-1) + 8

and p(n) might be p(n-1) + 2 or p(n-1) + 6.

sqrt(p1+7) = 3
sqrt(p3-1) = 2
*****

1)
Sqrt(P(n-1)+7)=Sqrt(P(n+1)-1)

Kermit:
****
Provided n is sufficiently large.
3**2 - 7 = 2
4**2 - 7 = 9
5**2 - 7 = 19
2,3,5,7,11,13,17,19
p8 = 19
p9 = 23
p10 = 29
sqrt(28) > sqrt(25)

******

2)
P(n)=P(n-1)+2 (Twin Primes)

Kermit:
***
I expect there to be exceptions to this also.

Because

If p(n+1) = p(n-1) + 8,

p(n) might be p(n-1) + 2,
or
p(n) might be p(n-1) + 6

******

3) P(n+1)=P(n-1)+8
Sincerely
Sebastián Martín Ruiz
• ... no: p(n+1)=m²+1 = p(n-1)+6 = m²-1 = (m+1)(m-1) can t be prime. M.
Message 2 of 7 , Mar 15, 2009
--- In primenumbers@yahoogroups.com, Kermit Rose <kermit@...> wrote:
> If
> Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:
> 2)
> P(n)=P(n-1)+2 (Twin Primes)
> I expect there to be exceptions to this also.
> Because
> If p(n+1) = p(n-1) + 8,
> p(n) might be p(n-1) + 2,
> or
> p(n) might be p(n-1) + 6

no:
p(n+1)=m²+1 => p(n-1)+6 = m²-1 = (m+1)(m-1) can't be prime.

M.
• In words: If N is a square (N=m²) and N+1 and N-7 are primes, then N-5 is a prime, too. 3) is superfluous, because it follows from 1) by simple
Message 3 of 7 , Mar 18, 2009
In words: If N is a square (N=m²) and N+1 and N-7 are primes, then N-5 is a prime, too.

3) is superfluous, because it follows from 1) by simple transformation.

Proof: Let N be a square and p1=N-7 and p3=N+1 primes. Let p2 be the single prime between p1 and p3: p1 < p2 < p3. Then the gap p3-p1=8. Then there are 3 possibilities for p2:
1.) p2=p1+2 (gaps 2,6)
2.) p2=p1+4 (gaps 4,4)
3.) p2=p1+6 (gaps 6,2)

2.) cannot occur, for there are no equal neighbouring gaps besides 6n,6n.
3.) From p1=N-7 and p2=p1+6 follows p2=N-1=m²-1=(m+1)*(m-1), not prime.

Thus 1.) is the only possibility.

Btw: The final digits of p1,p2,p3 are 9,1,7. Who proves?

Werner D. Sand
• ... Stated like this, it is indeed more or less trivial. As I see it, the nontrivial part of the assertion is: There is NO m such that * m²-7 is prime *
Message 4 of 7 , Mar 18, 2009
--- In primenumbers@yahoogroups.com, "Werner D. Sand" wrote:
>
> In words: If N is a square (N=m²) and N+1 and N-7 are primes,
> then N-5 is a prime, too.

Stated like this, it is indeed more or less trivial.

As I see it, the nontrivial part of the assertion is:

There is NO m such that
* m²-7 is prime
* (m+k)²+1 is prime for some k>0 (i.e. k>=2)
* there is only one prime between m²-7 and (m+k)²+1

I think that one would need to prove something like Cramer's conjecture to disprove this statement.

> 3) is superfluous, because it follows
> from 1) by simple transformation.

Of course.

> 3.) From p1=N-7 and p2=p1+6 follows p2=N-1=m²-1=(m+1)*(m-1),
> not prime.

As I wrote some days ago, cf. http://tech.groups.yahoo.com/group/primenumbers/message/19901/.

> Thus 1.) is the only possibility.
>
> Btw: The final digits of p1,p2,p3 are 9,1,7. Who proves?

m²-7 can't be 1 mod 10.

Maximilian
• ... er... I read but you didn t write: ...and there is a prime between N-7 and N+1, ... else we have counter-examples for m =
Message 5 of 7 , Mar 18, 2009
> --- In primenumbers@yahoogroups.com, "Werner D. Sand" wrote:
> >
> > In words: If N is a square (N=m²) and N+1 and N-7 are primes,
> > then N-5 is a prime, too.
>
> Stated like this, it is indeed more or less trivial.

er... I read but you didn't write:
"...and there is a prime between N-7 and N+1, ..."

else we have counter-examples for m =
54,66,90,156,240,270,306,474,570,576,636,750,780,1080,1320,1350,2034,
2154,2406,2700,2760,3204,3240,3306,3480,3516,3756,3774,3984,4056,4086,
4140,4146,4176,4716,4734,4794,5154,5370,5424,5550,5664,5700,5850,5856,
5970,6030,6060,6120,6366,6576,6714,6786,7050,7164,8196,8424,8454,8940,
9180,9246,9486,9696,9804...
• ... O.k., put it in.
Message 6 of 7 , Mar 20, 2009
--- In primenumbers@yahoogroups.com, "Maximilian Hasler" <maximilian.hasler@...> wrote:
>
> > --- In primenumbers@yahoogroups.com, "Werner D. Sand" wrote:
> > >
> > > In words: If N is a square (N=m²) and N+1 and N-7 are primes,
> > > then N-5 is a prime, too.
> >
> > Stated like this, it is indeed more or less trivial.
>
> er... I read but you didn't write:
> "...and there is a prime between N-7 and N+1, ..."
>
> else we have counter-examples for m =
> 54,66,90,156,240,270,306,474,570,576,636,750,780,1080,1320,1350,2034,
> 2154,2406,2700,2760,3204,3240,3306,3480,3516,3756,3774,3984,4056,4086,
> 4140,4146,4176,4716,4734,4794,5154,5370,5424,5550,5664,5700,5850,5856,
> 5970,6030,6060,6120,6366,6576,6714,6786,7050,7164,8196,8424,8454,8940,
> 9180,9246,9486,9696,9804...
>

O.k., put it in.
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