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 xilman 2008-07-20 16:19

[QUOTE=Raman;138062]So, can you please explain to me up how you derived the 4th degree
polynomial from the 8th degree one for $$10,375-$$
$$x^8-x^7+x^5-x^4+x^3-x+1$$
$$x-10^{25}$$[/quote]

Substitute x = y + 1/y in the octic and see what you get ...

[QUOTE=Raman;138062]
I am starting to sieve for 10,375- now.
10,312+ is in Linear Algebra and will finish up
within about 12 hours or so
(Matrix has less than 20 million rows!)

:exclaim: [SIZE=4]EMERGENCY[/SIZE]
Also that I can't enter the value of [B]m[/B] in the GGNFS
poly file too, because of the fact that
$$\division_{10^{25}}^{(10^{50}+1)}$$
is again not an integer at all[/QUOTE]Solve the equation 10^25x = 1 (mod 10^50) in integers. The solution is the integer you want.

Paul

 frmky 2008-07-20 17:07

[QUOTE=xilman;138066]
Solve the equation 10^25x = 1 (mod 10^50) in integers. The solution is the integer you want.
Paul[/QUOTE]

Although it's certainly a good exercise, actually entering m in the GGNFS poly file causes it to use the rational poly x-m. Enter the rational poly coefficients using Y1 and Y0, and the programs will calculate m.

Greg

 Raman 2008-07-25 09:35

[quote=xilman;138066]
Solve the equation 10[sup]25[/sup]x = 1 (mod 10[sup]50[/sup]) in integers. The solution is the integer you want.
[/quote]
Be careful! There exist no solution to this equation. Since 10[sup]25[/sup] is even, a multiple of it is always even, and on the right hand side, 1 (mod 10[sup]50[/sup]) is always odd. A solution is impossible to exist!

[quote=xilman;138066]
Substitute x = y + 1/y in the octic and see what you get ...
[/quote]
No hopes for degree 4. Substituting x = y + (1/y) in x[sup]8[/sup], so it gives up
$$\sum_{z=0}^8 ^8C_z y^z (1/y)^{8-z}$$
which is clearly being at degree 8.

Other terms will have their appropriate degrees. So, when substituted, the whole algebraic polynomial will be of degree 8 only.

And the linear polynomial becomes more cumbersome, in this form, with
10[sup]25[/sup](y+(1/y)) - (10[sup]50[/sup]+1)

 fivemack 2008-07-25 10:04

Hi Raman.

The calculation of M should be modulo the number you're trying to factor - ie 10^25 N = (10^50+1) mod cofactor. But as xilman pointed out you just fill in the numerator and denominator in the Y0 and Y1 fields.

The idea of substituting y+1/y is to take advantage of the symmetry of the octic; you write {octic} = x^4 * quartic(x+1/x) for some suitably-chosen quartic, and the 10^50+1 and 10^25 are from (x + 1/x) written as (x^2+1)/x.

 R.D. Silverman 2008-07-25 11:16

[QUOTE=fivemack;138325]Hi Raman.

The calculation of M should be modulo the number you're trying to factor - ie 10^25 N = (10^50+1) mod cofactor. But as xilman pointed out you just fill in the numerator and denominator in the Y0 and Y1 fields.

The idea of substituting y+1/y is to take advantage of the symmetry of the octic; you write {octic} = x^4 * quartic(x+1/x) for some suitably-chosen quartic, and the 10^50+1 and 10^25 are from (x + 1/x) written as (x^2+1)/x.[/QUOTE]

The reason this works is that *reversing* the coefficients of any polynomial
results in a homomorphism of its splitting field, sending a root r of the
polynomial to 1/r. Thus, if the coefficients of the polynomial are the same
when reversed, we can replace the polynomial with one whose roots are
r + 1/r and get an isomorphic field.

 bdodson 2008-09-17 17:10

[QUOTE=garo;48284][CODE]Base Index Size 11M(45digits) 43M(50digits) 110M(55digits) 260M(60digits) Decimal
7 271- C214 : 1570202...53660188716054727305891
...
7 301- C189 : 7473377...279834566763898163532521
...
7 393- C217 : 580546345...10110568816475625168427
[/CODE][/QUOTE]

These three cofactors are no longer in the ECMNET input
file, and the indices 271-, 301- and 393- are not in the 7/08
appendix C. That leaves 18, with the NFSNET number 7,319-
sieved. with the matrix running; and 7,313- a C/D number,
also sieved, with matrix running. -Bruce

 bdodson 2009-09-18 18:23

count/recount

[QUOTE=bdodson;142925] ... That leaves 18, with the NFSNET number 7,319-
sieved. with the matrix running; and 7,313- a C/D number,
also sieved, with matrix running. -Bruce[/QUOTE]

OK, the database is now closer to being current than the table
in the first post. There should be 15, with
[code]
7 277- C201 done
7 311- C225 first
7 313- C248 done
7 323- C241 second, &etc. [/code]

If I'm reading the thread activity correctly, out of 18 tables (with
four base-2's and two each for 3, 5, 6, 7, 10, 11 and 12, so
4+2*7 = 18) this one is the one that's gone the longest without
a new factor report? No reserved numbers, either; with 311-
on the more wanted list. -Bruce

Off Topic PS: from the old pages on Sam's site, the cover letter
for page 80 (from 1998) lists a bunch of the tables as having been
extended
[QUOTE]
to insure that every table has at least five holes [/QUOTE]
which explains which tables would be extended, but the trigger
seems to have been an update 2.C. There was also an update 2.E,
followed the the 3rd edition of the tables, Sept 2001. I don't see
any update 3.*'s; so suppose that it's unclear whether dropping one
of the table below five entries would trigger an update and extension,
or we might have some more time to clear an entire table (most likely
3- perhaps).

 10metreh 2009-12-21 07:53

In the DB, someone has entered the (previously unknown) factor of 7,391-:

p57 = 478566296656273815311438559010751123205277732759848440243

with a p187 cofactor. However, it can be found nowhere else - at least the forum and Sam's page don't mention it, and Google doesn't return any results for it. I expect the finder will come forward soon, but anyway, that's one "impossible" out of the way. :smile:

 Raman 2009-12-21 08:05

[quote=10metreh;199499]In the DB, someone has entered the (previously unknown) factor of 7,391-:

p57 = 478566296656273815311438559010751123205277732759848440243

with a p187 cofactor. However, it can be found nowhere else - at least the forum and Sam's page don't mention it, and Google doesn't return any results for it. I expect the finder will come forward soon, but anyway, that's one "impossible" out of the way. :smile:[/quote]
Have you checked up the ECMNET page of Mr. Paul Zimmermann?

 xilman 2009-12-21 08:27

[QUOTE=Raman;199502]Have you checked up the ECMNET page of Mr. Paul Zimmermann?[/QUOTE]Indeed, Paul mailed it out to the usual suspects yesterday evening.

Paul (the other one)

 R.D. Silverman 2010-01-28 20:57

LA Failure?

Did the LA for 7,311- fail?

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