## Factorization of 6 consecutive 1803-digit integers

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• Let b=4011209802600 m=16812956160*b x=5000001251617*b y=(2*x^6+3*x^3-148)^2 N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4 then N+k is
Message 1 of 16 , Sep 7, 2007
Let

b=4011209802600
m=16812956160*b
x=5000001251617*b
y=(2*x^6+3*x^3-148)^2
N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4

then N+k is factorized for k in [0,5].

Proof:

Software used:
GGNFS (with SNFS difficulty = 152.1)
GMP-ECM
Msieve
OpenPFGW
Pari-GP
Primo

Neighbouring composites:
c1758=(N-1)/(5*239283371994839*543315317127619*4154925272246149)
c1771=(N+6)/(2*8081*33697745047039*268690037677741)
subjected to 400 ECM curves at B1=250000

7 September 2007
• ... Congratulations on your 10th record in a month! (counting as multiple records when it was record for multiple lengths) I m impressed. The record page is
Message 2 of 16 , Sep 7, 2007
> Let
>
> b=4011209802600
> m=16812956160*b
> x=5000001251617*b
> y=(2*x^6+3*x^3-148)^2
> N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4
>
> then N+k is factorized for k in [0,5].

Congratulations on your 10th record in a month!
(counting as multiple records when it was record for multiple lengths)
I'm impressed.

The record page is updated (people can probably find the link by now after I
have spammed the list with it for a month).

--
Jens Kruse Andersen
• ... It is unlikely that this 72nd order Ansatz will be easily improved upon, since it seems fully to exploit the best known solution to the
Message 3 of 16 , Sep 8, 2007

> b=4011209802600
> m=16812956160*b
> x=5000001251617*b
> y=(2*x^6+3*x^3-148)^2
> N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4
>
> then N+k is factorized for k in [0,5].

It is unlikely that this 72nd order Ansatz will be easily
improved upon, since it seems fully to exploit the best
known solution to the Prouhet-Tarry-Escott problem:

http://euler.free.fr/eslp/TarryPrb.htm
http://mathworld.wolfram.com/Prouhet-Tarry-EscottProblem.html

David
• ... PS: I remark that the search reported in http://portal.acm.org/citation.cfm?id=950414.950443 failed to improve on Nuutti Kuosa s wonderful discovery of 3
Message 4 of 16 , Sep 8, 2007

> It is unlikely that this 72nd order Ansatz will be easily
> improved upon, since it seems fully to exploit the best
> known solution to the Prouhet-Tarry-Escott problem

PS: I remark that the search reported in

http://portal.acm.org/citation.cfm?id=950414.950443

failed to improve on Nuutti Kuosa's wonderful discovery of 3 Sep 1999:

http://euler.free.fr/eslp/k246810.htm

David
• ... I m even more impressed by that fact that Peter Borwein, Petr Lisonek and Colin Percival were unable to improve on Nuutti s polynomial, after expending
Message 5 of 16 , Sep 8, 2007
--- In primeform@yahoogroups.com, "Jens Kruse Andersen"
<jens.k.a@...> wrote:

> Congratulations on your 10th record in a month!
> I'm impressed.

I'm even more impressed by that fact that
Peter Borwein, Petr Lisonek and Colin Percival
were unable to improve on Nuutti's polynomial,
after expending 10^17 floating point operations.

Such Diophantine problems are amongst the weirdest that we know.

David
• ... Their Math.Comp. article currently leaks out of googling 35, 47, 94, 121, 146, 148 even though it /ought/ to be password protected. Verbum sat sapienti?
Message 6 of 16 , Sep 8, 2007

> Peter Borwein, Petr Lisonek and Colin Percival
> were unable to improve on Nuutti's polynomial,
> after expending 10^17 floating point operations.

Their Math.Comp. article currently leaks out of googling

"35, 47, 94, 121, 146, 148"

even though it /ought/ to be password protected.

Verbum sat sapienti?

David
• I just need to understand , I found that n^14-n^7+1 is never prime for n integer until more that 6000000 If this is always true , why ? When using pfgw
Message 7 of 16 , Sep 13, 2007
I just need to understand , I found that n^14-n^7+1 is never prime for n
integer until more that 6000000
If this is always true , why ?
When using pfgw -q729^^14-729^^7+1 I have the answer it is PRP
but pfgw -t -q729^^14-729^^7+1 give the answer composite with the
factor found ! What is the bug for PRP answer ?
Thanks for help
Pierre
• Yes, it is always true because you can factor the polynomial n^14-n^7+1 = (n^2-n+1)(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1) S.
Message 8 of 16 , Sep 13, 2007
Yes, it is always true because you can factor the polynomial

n^14-n^7+1 = (n^2-n+1)(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1)

S.

Pierre CAMI wrote:
> I just need to understand , I found that n^14-n^7+1 is never prime for n
> integer until more that 6000000
> If this is always true , why ?
> When using pfgw -q729^^14-729^^7+1 I have the answer it is PRP
> but pfgw -t -q729^^14-729^^7+1 give the answer composite with the
> factor found ! What is the bug for PRP answer ?
> Thanks for help
> Pierre
>
>
>
>
> Post message: primeform@yahoogroups.com
> Web access:
> http://groups.yahoo.com/group/primeform
>
>
>
>
• ... n^14-n^7+1 = (n^2-n+1)*(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1) ... 729^14-729^7+1 is indeed 3-PRP. Note that 729 = 3^6. Some number forms involving powers of b
Message 9 of 16 , Sep 13, 2007
Pierre CAMI wrote:
> I found that n^14-n^7+1 is never prime

n^14-n^7+1 = (n^2-n+1)*(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1)

> When using pfgw -q729^^14-729^^7+1 I have the answer it is PRP
> but pfgw -t -q729^^14-729^^7+1 give the answer composite with the
> factor found ! What is the bug for PRP answer ?

729^14-729^7+1 is indeed 3-PRP. Note that 729 = 3^6.
Some number forms involving powers of b are b-PRP.
Base 3 is default in pfgw. Here is base 2:

C:\Users\Jens>pfgw -b2 -q729^^14-729^^7+1
PFGW Version 1.2.0 for Windows [FFT v23.8]

Switching to Exponentiating using GMP
729^14-729^7+1 is composite: [D66D28B9C614392] (0.0004s+0.0003s)

--
Jens Kruse Andersen
• ... wrote (as Jens had done before) ... The larger factor is Phi(42,n), which is a BPSW probable prime for 28,536 of the first million natural numbers: ct=0;
Message 10 of 16 , Sep 13, 2007
--- In primeform@yahoogroups.com, "Sean A. Irvine" <sairvin@...>
wrote (as Jens had done before)

> n^14-n^7+1 = (n^2-n+1)(n^12+n^11-n^9-n^8+n^6-n^4-n^3+n+1)

The larger factor is Phi(42,n), which is a BPSW probable prime
for 28,536 of the first million natural numbers:

ct=0;
for(n=1,10^6,if(ispseudoprime(subst(polcyclo(42),x,n)),ct=ct+1));
print(ct)

28536

David
• ... Today I made some progress with the PTE problem: http://listserv.nodak.edu/cgi-bin/wa.exeA2=ind0709&L=nmbrthry&P=R994&D ... It may take a good while longer
Message 11 of 16 , Sep 26, 2007

> It is unlikely that this 72nd order Ansatz will be easily
> improved upon, since it seems fully to exploit the best
> known solution to the Prouhet-Tarry-Escott problem:
> http://euler.free.fr/eslp/TarryPrb.htm
> http://mathworld.wolfram.com/Prouhet-Tarry-EscottProblem.html

Today I made some progress with the PTE problem:

http://listserv.nodak.edu/cgi-bin/wa.exeA2=ind0709&L=nmbrthry&P=R994&D
> Perhaps a CRT strategy might lead to a solution with degree s > 12.

It may take a good while longer to see if this feeds into Jens's
consecutive-factorization page, but at least it is pleasing to find
a new PTE solution, rather economically, using 1700-year-old
mathematics.

David
• ... but either Yahoo or I mangled the link, which should have been http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0709&L=nmbrthry&P=R994 In any case, the
Message 12 of 16 , Sep 26, 2007

> Today I made some progress with the PTE problem:

but either Yahoo or I mangled the link, which should have been

http://listserv.nodak.edu/cgi-bin/wa.exe?A2=ind0709&L=nmbrthry&P=R994

In any case, the sentiment

> It may take a good while longer to see if this feeds into Jens's
> consecutive-factorization page, but at least it is pleasing to find
> a new PTE solution, rather economically, using 1700-year-old
> mathematics.

still stands.

David
• ... The link is http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm but it doesn t work anymore... Andrey
Message 13 of 16 , May 27, 2010
Jens Kruse Andersen wrote:
> > Let
> >
> > b=4011209802600
> > m=16812956160*b
> > x=5000001251617*b
> > y=(2*x^6+3*x^3-148)^2
> > N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4
> >
> > then N+k is factorized for k in [0,5].
>
> Congratulations on your 10th record in a month!
> (counting as multiple records when it was record for multiple lengths)
> I'm impressed.
>
> The record page is updated (people can probably find the link by now after
> I have spammed the list with it for a month).

http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm

but it doesn't work anymore...

Andrey
• And this ? http://users.cybercity.dk/~dsl522332/ ... Von: Andrey Kulsha Betreff: Re: [primeform] Factorization of 6 consecutive 1803-digit
Message 14 of 16 , May 27, 2010
And this ?

http://users.cybercity.dk/~dsl522332/

--- Andrey Kulsha <Andrey_601@...> schrieb am Do, 27.5.2010:

Von: Andrey Kulsha <Andrey_601@...>
Betreff: Re: [primeform] Factorization of 6 consecutive 1803-digit integers
An: primeform@yahoogroups.com
Datum: Donnerstag, 27. Mai, 2010 14:35 Uhr

Jens Kruse Andersen wrote:

> > Let

> >

> > b=4011209802600

> > m=16812956160*b

> > x=5000001251617*b

> > y=(2*x^6+3*x^3-148)^2

> > N=(y-11^4)*(y-35^2)*(y-47^2)*(y-94^2)*(y-146^2)*(y-148^2)/m-4

> >

> > then N+k is factorized for k in [0,5].

>

> Congratulations on your 10th record in a month!

> (counting as multiple records when it was record for multiple lengths)

> I'm impressed.

>

> The record page is updated (people can probably find the link by now after

> I have spammed the list with it for a month).

http://hjem.get2net.dk/jka/math/consecutive_factorizations.htm

but it doesn't work anymore...

Andrey

[Non-text portions of this message have been removed]
• ... Thanks Norman. There s also a record for k=20: http://tech.groups.yahoo.com/group/primeform/message/9982 But it doesn t fit the tables unfortunately. Best
Message 15 of 16 , May 27, 2010
> And this ?
>
> http://users.cybercity.dk/~dsl522332/

Thanks Norman.

There's also a record for k=20:
http://tech.groups.yahoo.com/group/primeform/message/9982

But it doesn't fit the tables unfortunately.

Best regards,

Andrey
• Maybe here ? http://www.math.uni.wroc.pl/~jwr/cons-fac/ ... Von: Andrey Kulsha Betreff: Re: [primeform] Factorization of 6 consecutive
Message 16 of 16 , May 27, 2010
Maybe here ?
http://www.math.uni.wroc.pl/~jwr/cons-fac/

--- Andrey Kulsha <Andrey_601@...> schrieb am Do, 27.5.2010:

Von: Andrey Kulsha <Andrey_601@...>
Betreff: Re: [primeform] Factorization of 6 consecutive 1803-digit integers
An: primeform@yahoogroups.com
Datum: Donnerstag, 27. Mai, 2010 15:18 Uhr

> And this ?

>

> http://users.cybercity.dk/~dsl522332/

Thanks Norman.

There's also a record for k=20:

http://tech.groups.yahoo.com/group/primeform/message/9982

But it doesn't fit the tables unfortunately.

Best regards,

Andrey

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