20210616, 22:45  #320 
Jan 2012
Toronto, Canada
5×19 Posts 
I'd focus mainly on the lines that are < 3000 digits (roughly SNFS 250 quartic difficulty), as from what I understand partial factorizations of these points don't really help, so these are the only ones we can reasonably break aside from getting lucky with ECM. Don't think there are any more shortcuts as we've found all the algebraic factors @Max0526?

20210617, 11:19  #321  
"Ed Hall"
Dec 2009
Adirondack Mtns
1000001101001_{2} Posts 
Quote:
I'm going to play elsewhere for a while, but I'll keep an eye on this thread. . . 

20210617, 11:25  #322  
"Max"
Jun 2016
Toronto
919 Posts 
Quote:
1) Please read this: https://mersenneforum.org/showpost.p...&postcount=228 If we uncover the explicit expression for a = a(z), we won't need to calculate the GCDs anymore, we will know what they are  they will be algebraic factors of the 6th polynomial on the list (a^4  24*a^3 + 152*a^2  336*a + 196) in terms of z. Doesn't seem to be an extra shortcut yet, but at least we will understand why the GCDs are always there (nobody explained them yet, including the authors), and only for the 6th polynomial. And that expression for a(z) may become the first shortcut I still hope for. If we substitute a(z) into Code:
2*a^4  30*a^3 + 169*a^2  420*a + 392, a^4  12*a^3 + 62*a^2  168*a + 196, a^4  6*a^3 + 17*a^2  84*a + 196 For that to happen, sharing factors with other points (existence of large GCDs) is not a requirement. It just happened so that the factors of the 6th polynomial are shared, some/all others polynomials may have algebraic factorizations in terms of z but not share any numerical factors. 2) There are two connected elliptic curves. The first one produces the first 4 original SNFS polys (for the values a1, a2, a3, a4) to factor any composite by SNFS if we decide to do so. The second one _may_ produce 4 more (for the values a5, a6, a7, a8) for a particular composite we need to factor. Sometimes that composite to factor is shared by the two elliptic curves (and we have 8 original SNFS polys before the spin), and sometimes it is not (only 4 SNFS polys then). The discriminant of the second elliptic curve is much harder to factor completely as it has a polynomial of degree 32 as one of the algebraic factors: Code:
x  15, x  12, x, x^4  96*x^3 + 2232*x^2  17280*x + 32400, x^4  24*x^3 + 288*x^2  4320*x + 32400, x^4 + 192*x^3  5544*x^2 + 34560*x + 32400, x^32  192*x^31  43776*x^30  18800640*x^29 + 5503500288*x^28 + 273683349504*x^27  195095809916928*x^26 + 30785353346101248*x^25  2825577105735545856*x^24 + 176684772621207158784*x^23  8406169261059541893120*x^22 + 310105904650770600493056*x^21  9168826131738256943775744*x^20 + 222352073452169017323945984*x^19  3034551308521081654597386240*x^18 + 31148118165121605968147251200*x^17  232887992784886624221872271226*x^16 + 1184706194482578639874124008896*x^15  3607984676528931780768752872704*x^14 + 6019337787648898873971179520000*x^13  25447638730256460469701181440000*x^12 + 189226164149072887551728025600000*x^11  270078906310708200738914304000000*x^10  3101246803458128590189859098718208*x^9 + 12219551782715019853536631583745024*x^8 + 23360825043808908586558290230722560*x^7  159269714549351610655354977558528000*x^6  138653758460372775633724442812416000*x^5 + 1006003450171503521204261823959040000*x^4 + 1072258009790390526408430768128000000*x^3  1640779783029762601620535197696000000*x^2  1295352901691353352177566433280000000*x + 1214391379708526603038950722542207369 If the GCDs on the second curve exist, they may help us factor our numbers more efficiently, i.e., become the source to derive the second shortcut I am hoping for. And another way to approach the second shortcut might be to ask: What is the expression for a = a(z), e.g., it may or may not be in the form that factors any or all polynomials below in terms of z? Code:
a^4  96*a^3 + 2232*a^2  17280*a + 32400, a^4  24*a^3 + 288*a^2  4320*a + 32400, a^4 + 192*a^3  5544*a^2 + 34560*a + 32400 Last fiddled with by Max0526 on 20210617 at 12:06 

20210617, 12:38  #323 
Apr 2020
253_{16} Posts 
Running t50 on (6,8) c179/snfs229, will continue with t55 if no factors turn up.

20210617, 13:54  #324  
"Ben"
Feb 2007
3586_{10} Posts 
Quote:
Code:
P113 = 48859177299728793320715246065704744864214534488213548350308376738773226206199630803249626926079041962889965804681 P85 = 9056240418333107480731935683874341644262594549554962412771888829244066167328930970849 

20210617, 14:20  #325  
"Max"
Jun 2016
Toronto
919 Posts 
Quote:
Do you need a set of spun polys for c203? We have 8 SNFS polys to start with, so I have more spin options. I should be able to make it SNFS 220, although right now it's SNFS 222/225. Please let me know. 

20210617, 14:23  #326 
"Ben"
Feb 2007
2·11·163 Posts 
Yes please, let's get the line finished. After that I think I will return to more ecm work (probably at a smaller scale).

20210617, 15:33  #327  
"Max"
Jun 2016
Toronto
919 Posts 
Quote:
If a composite that we are trying to factor is cubed, then it is (most probably) not shared with the related second elliptic curve, and only 4 original SNFS polys are available (for a1a4) before we try to spin. If a composite is not cubed, then it is shared with the second elliptic curve, and 8 original SNFS polys are available (for a1a8). Again, bur, you had one tough composite to factor! 

20210617, 16:00  #328  
"Max"
Jun 2016
Toronto
919_{10} Posts 
Quote:
But my question about a = a(z) still stays: What could a(z) be to factor at least one of the following polynomials in terms of z? First/our elliptic curve: Code:
a^2  12*a + 28, a^2  6*a + 7, a^4  24*a^3 + 152*a^2  336*a + 196, 2*a^4  30*a^3 + 169*a^2  420*a + 392, a^4  12*a^3 + 62*a^2  168*a + 196, a^4  6*a^3 + 17*a^2  84*a + 196 Code:
a^4  96*a^3 + 2232*a^2  17280*a + 32400, a^4  24*a^3 + 288*a^2  4320*a + 32400, a^4 + 192*a^3  5544*a^2 + 34560*a + 32400 given, e.g., that and for (8, 6): 1) on our elliptic curve , what are the constant values to produce rational solution(s) for z? 2) on a connected elliptic curve , what are the constant values to produce rational solution(s) for z? Last fiddled with by Max0526 on 20210617 at 16:22 

20210617, 21:26  #329  
"William"
May 2003
New Haven
4501_{8} Posts 
Quote:
a(z) = z^2 z 2 a(z) = 7z^2 5z +4 

20210617, 21:37  #330  
"Max"
Jun 2016
Toronto
919 Posts 
Quote:
The problem here is that we want: 1) create a formula for a=a(z) that factors at least one of our polynomials above in terms of z; _and_ 2) some _rational_ z in that formula to produce our given rational a1 or a5. This rational z requirement slims our choices down almost to zero. I am still not giving up on finding that formula! Last fiddled with by Max0526 on 20210617 at 21:39 

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