20051206, 12:57  #1 
Jun 2005
2×7^{2} Posts 
160 digit factor found of 366 digit (PRP1)
I am trying to prove the following number prime
p=(2^6071)*(2^607169662)+169661 p1 has following prime factors 2 3 ^ 4 7 ^ 2 13 607 2663 117210 040608 611501 and 160 digit composite factor: 2077 722215 701465 989903 905378 105865 831121 690269 515432 941764 890410 612443 142155 675750 703708 696614 154976 284846 472630 448131 063469 226632 172413 531787 662918 734708 751011 and a remaining 178 digit composite factor Is finding a 160 digit composite factor a noteworthy record? Congratulations to Dario Alpern's http://www.alpertron.com.ar/ECM.HTM The Java applet is now stuck on the 160 digit factor so I installed ecm6.0.1. Any pointers for the best command line  never used ecm before. regards Anton Last fiddled with by AntonVrba on 20051206 at 12:58 
20051206, 13:20  #2 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Not if it is an algebraic factor  which it undoubtedly is. Before continuing with ECM, try to find all algebraic factors of your number. It will make your job very much easier.
Alex 
20051206, 13:47  #3  
Nov 2003
2^{2}·5·373 Posts 
Quote:
You are trying to kill an ant with a sledgehammer. ECPP or Cyclotomy methods will prove this prime (if it is) in just a few minutes. Why waste factoring a 160digit composite? 

20051206, 15:28  #4  
Nov 2005
110000_{2} Posts 
Quote:
(2^607  1)*(2^607  169662) + 169661  1 == 2(2^303  1)(2^303 + 1)(2^607  169661) The algebraic factors 2^3031 and 2^303+1 have each been completely factored, as revealed by quick check of the Cunningham project tables. This is enough for a classical proof, so you don't even have to bother with 2^607169661. The algebraic factorization is not a coincidence. How did you choose to study this form? John 

20051206, 15:57  #5 
"Nancy"
Aug 2002
Alexandria
2,467 Posts 
Thanks for the factorisation, John.
2^607  169661 is a fairly easy SNFS number. Anton, you probably won't need this factored any more for your proof, but if want it done, let me know. Alex 
20051206, 16:35  #6  
Jun 2005
2×7^{2} Posts 
Quote:
BTW, Marcel Martin's Primo has done the job of prooving p=(2^6071)*(2^607169662)+169661 prime regards Anton Last fiddled with by AntonVrba on 20051206 at 16:37 

20051206, 21:56  #7 
Aug 2002
Buenos Aires, Argentina
1385_{10} Posts 
My applet also declares that the original number is prime by using the APRTCLE algorithm. So no external programs were needed in this case to prove primality.
Notice that the applet cracked p1 by using Lehman factorization. This is because it is a product of two similar numbers. 
20051206, 22:02  #8 
Aug 2002
Buenos Aires, Argentina
10101101001_{2} Posts 
If you had entered ((2^607  1)*(2^607  169662) + 169661  1)/2/(2^607  169661) in the upper box the applet would have completed the factorization. This is because it search in the Web server known factorizations of Cunningham numbers.

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