- ...and I meant:

"

Proposition: If p,q,r,s are consecutive primes,

then valuation( ps+qr, P )=1 for some prime P.

Corollary: If p,q,r,s are consecutive primes,

then ps+qr cannot be a square.

"

(sorry)

Maximilian

> --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@> wrote:

> >

> >

> > Have a prime factor with exponent 1

> > ï¿½

> > p*s+q*r=pi^1*pj^aj*pq^aq.....

> >

> > --- El dom, 3/5/09, Paul Leyland <paul@> escribiï¿½:

> >

> >

> > De: Paul Leyland <paul@>

> > Asunto: Re: [PrimeNumbers] It is never a square

> > Para: "Sebastian Martin Ruiz" <s_m_ruiz@>

> > Fecha: domingo, 3 mayo, 2009 8:21

> >

> >

> > On Thu, 2009-04-30 at 12:07 +0000, Sebastian Martin Ruiz wrote:

> > >

> > >

> > > Hello all:

> > >ï¿½

> > > It is true?

> > >ï¿½

> > > Let p<q<r<sï¿½ four consecutive prime numbers.

> > >ï¿½

> > > p*s+q*r has a square free prime factor.

> > >ï¿½

> > > Then p*s+q*r is never a square.

> > >ï¿½

> > > I have verified it for many values.

> >

> > Isn't this trivial?ï¿½ Any number with a square-free prime factor is not a

> > square.ï¿½ That's what square-free means.

> >

> > I don't think I understand what you are trying to say.

> >

> >

> > Paul

> >

> > - Hi,

I have played with consecutive primes p1,p2,p3,p4 and determined (see link) that any

combinition of pi*pj + pk*pl is possibly square if

3 of them are 4k+3 and 1 is 4k+1 or 3 are 4k+1 and 1 is 4k+3.

Moreover for p1 > 2, this sum is even and a square sum is divisible by 4.

http://groups.google.com/group/sumprimes/web/ps-qr-for-consecutive-primes-p-q-r-s

Using this rule to narrow the search with the above program, I have determined that

p1*p4+p2*p3 is not square for primes < 10^9.

However for,

p1*p2 + p3*p4 we find running the program:

Squares in sum of products of consecutive primes.

Start => 1

Stop => 1000000

Tue 05/05/2009

10:51 AM

n=18242: 203233*203249 + 203279*203293 = 82632101764

Count = 1

Sum = 0

Time in sec = 16.707211

Now

(10:57:02) gp > factor(82632101764)

%2 =

[2 2]

[143729 2]

The 18242th prime, 203233, starts a working combination. This implies squares exist in

the possible combinations. Eg. 203233 = 4m+1, 203249=4m+1, 203279=4m+3,

203293=4m+1 consistent with the rule. This also implies there are possibly other squares

for for larger prime combinations.

Consecutive primes can be arbitrarily far apart. I do not think consecutive primes

play a role in the truth of the assertion since if it did it would imply some underlying

pattern in the prime numbers.

This implies a combination exists such that p1*p4 + p2*p3 is possibly square since there is no

known pattern in the primes.

Consider consecutive primes p1,p2,p3,p4,p5,p6,p7

and test p1*p2 + P5*p6

Then output

n=1257742: 19784717*19784731 + 19784759*19784771 = 782872229861316

gp > factor(782872229861316)

[2 2]

[3 2]

[7 2]

[666187 2]

is a hit

19784717 =4m+1, 19784731,19784759,19784771 = 4m+3

Similarly testing p1*p2 + p6*p7

n=64155: 802751*802759 + 802793*802799 = 1288897007616

Congruent 3,3,1,3 mod 4

gp > factor(1288897007616)

[2 12]

[3 10]

[73 2]

and

p1*p7 + p2*p6

n=2321170: 38038661*38038723 + 38038669*38038709 = 2893883949908224

Congruent 1,3,1,1 mod 4

gp > factor(2893883949908224)

[2 8]

[7 2]

[13 2]

[36947 2]

for p1,p2,p3,p4,p5,p6,p7,p8

test p1*p2 + p7*p8

Start => 1

Stop => 1000000000

Wed 05/06/2009

02:25 AM

n=77: 389*397 + 431*433 = 341056

n=1493: 12503*12511 + 12547*12553 = 313927524

n=169237: 2296009*2296027 + 2296081*2296097 = 10543723352100

My guess is this process can be extended to a random selection of primes taken

k at a time.

Cino

To: primenumbers@yahoogroups.com

From: maximilian.hasler@...

Date: Sun, 3 May 2009 21:57:09 +0000

Subject: [PrimeNumbers] Re: It is never a square

...and I meant:

"

Proposition: If p,q,r,s are consecutive primes,

then valuation( ps+qr, P )=1 for some prime P.

Corollary: If p,q,r,s are consecutive primes,

then ps+qr cannot be a square.

"

(sorry)

Maximilian

> --- In primenumbers@yahoogroups.com, Sebastian Martin Ruiz <s_m_ruiz@> wrote:

[Non-text portions of this message have been removed]

> >

> >

> > Have a prime factor with exponent 1

> > ï¿½

> > p*s+q*r=pi^1*pj^aj*pq^aq.....

> >

> > --- El dom, 3/5/09, Paul Leyland <paul@> escribiï¿½:

> >

> >

> > De: Paul Leyland <paul@>

> > Asunto: Re: [PrimeNumbers] It is never a square

> > Para: "Sebastian Martin Ruiz" <s_m_ruiz@>

> > Fecha: domingo, 3 mayo, 2009 8:21

> >

> >

> > On Thu, 2009-04-30 at 12:07 +0000, Sebastian Martin Ruiz wrote:

> > >

> > >

> > > Hello all:

> > >ï¿½

> > > It is true?

> > >ï¿½

> > > Let p<q<r<sï¿½ four consecutive prime numbers.

> > >ï¿½

> > > p*s+q*r has a square free prime factor.

> > >ï¿½

> > > Then p*s+q*r is never a square.

> > >ï¿½

> > > I have verified it for many values.

> >

> > Isn't this trivial?ï¿½ Any number with a square-free prime factor is not a

> > square.ï¿½ That's what square-free means.

> >

> > I don't think I understand what you are trying to say.

> >

> >

> > Paul

> >

> >