Prime numbers and primality testing Yahoo Group Integers then Equals =============================================== Sebastian Martin Ruiz Message 1 of 7 Mar 14, 2009 ----------------------------------------------- Hello all:   Let P(n) the n-th prime number. Let Sqrt(P)=p^(1/2)  If Sqrt(P(n-1)+7) and Sqrt(P(n+1)-1) are both integers then:  1) Sqrt(P(n-1)+7)=Sqrt(P(n+1)-1)  2) P(n)=P(n-1)+2  (Twin Primes)  3) P(n+1)=P(n-1)+8 Sincerely Sebastián Martín Ruiz  [Non-text portions of this message have been removed] =============================================== Kermit Rose Message 2 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 =============================================== Maximilian Hasler Message 3 of 7 Mar 15, 2009 ----------------------------------------------- --- In primenumbers@yahoogroups.com, Kermit Rose 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. =============================================== Werner D. Sand Message 4 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 =============================================== Maximilian Hasler 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. 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 =============================================== Maximilian Hasler Message 6 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... =============================================== Werner D. Sand Message 7 of 7 Mar 20, 2009 ----------------------------------------------- --- In primenumbers@yahoogroups.com, "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. =============================================== Cached by Georg Fischer at Nov 14 2019 12:47 with clean_yahoo.pl V1.4