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

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• Have a prime factor with exponent 1   p*s+q*r=pi^1*pj^aj*pq^aq..... ... De: Paul Leyland Asunto: Re: [PrimeNumbers] It is never a
Message 1 of 5 , May 3, 2009
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Have a prime factor with exponent 1

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

--- El dom, 3/5/09, Paul Leyland <paul@...> escribió:

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:

> It is true?

> Let p<q<r<s  four consecutive prime numbers.

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

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

> I have verified it for many values.

Isn't this trivial?  Any number with a square-free prime factor is not a
square.  That's what square-free means.

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

Paul

[Non-text portions of this message have been removed]
• He means : Proposition: If p,q,r,s are consecutive primes, then valuation( ps+rs, P )=1 for some prime P. Corollary: If p,q,r,s are consecutive primes, then
Message 2 of 5 , May 3, 2009
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He means :

"
Proposition: If p,q,r,s are consecutive primes, then valuation( ps+rs, P )=1 for some prime P.

Corollary: If p,q,r,s are consecutive primes, then ps+rs cannot be a square.
"

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
>
>
>
>
>
>
>
> [Non-text portions of this message have been removed]
>
• ...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,
Message 3 of 5 , May 3, 2009
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...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
Message 4 of 5 , May 6, 2009
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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.

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

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:
> >
> >
> > 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
> >
> >

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