Thread: Other Primes View Single Post
2019-04-19, 23:19   #189
paulunderwood

Sep 2002
Database er0rr

22·5·157 Posts

Quote:
 Originally Posted by R. Gerbicz That is just painful. I mean the ecpp proof for the hard number when you have a much cheaper way, searching a better form. Say 2^a divides p-1 3^b divides p+5 ofcourse you want 2^a~3^b (~sqrt(N)) because then you know large p-1,(p+6)-1 factors in the oder of sqrt(N), so a=floor(log(N)/log(2)/2) b=floor(log(N)/log(3)/2) you can search p in the form (because 2^a and 3^b are coprime): p=u*2^a+v*3^b, from divisibilities you can get: Code: v*3^b==1 mod 2^a u*2^a==-5 mod 3^b Don't need to run two variables, fix u, then run v in arithmetic progression. Use sieving.
Somebody, maybe GEN-ERIC with his 16 core Threadripper, should try to beat his record with the above method: The gauntlet has been thrown down!

Last fiddled with by paulunderwood on 2019-04-19 at 23:22