20141219, 01:19  #12 
Tribal Bullet
Oct 2004
3·1,181 Posts 

20141219, 13:51  #13  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
Quote:
so if y is able to be represented as a power that divides by q not allowed with x then y^q is able to be represented with a value of q that doesn't work with x and so the number gets thrown out. that should narrow the field since q only works if it's a factor of one less than a prime factor p of x. Last fiddled with by science_man_88 on 20141219 at 13:52 

20141219, 20:14  #14 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×4,787 Posts 
Hi Risto,
Could you mirror the connection for the client using a plain, CGIlike URL instead of a port? (Or write a cgi whose function would be simply call local service into the :1235 port?) Like http://kasj.tunk.org/getwork?user=User1&lang=julia& to return the same as http://kasj.tunk.org:1235/?user=User1&lang=julia& 
20141221, 15:06  #15  
Mar 2010
Jyvaskyla, Finland
2^{2}·3^{2} Posts 
Quote:
Thanks for the interest :) 

20141221, 15:13  #16  
Mar 2010
Jyvaskyla, Finland
2^{2}·3^{2} Posts 
Quote:
So the number of prime factors of n is reduced from 4 to 3, which is pretty impressive. I didn't find any reference to the minimum factor of 29 at first glance, but I'm guessing it must still apply here. 

20141221, 21:57  #17 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2566_{16} Posts 
Thank you, but a 302 redirect will not solve the problem. If a client host cannot use ports, then:
Code:
HTTP request sent, awaiting response... 302 Moved Temporarily Location: http://kasj.tunk.org:1235/?user=User1&lang=julia& [following] Connecting to kasj.tunk.org91.156.111.210:1235... failed 
20141222, 12:15  #18  
Mar 2010
Jyvaskyla, Finland
24_{16} Posts 
Quote:
Also, I've kept the workspace quite small as there are not that many contributors for now. Of course it will be expanded later as needed, but for future developments it will be easier if a solid rectangle of work has been finished. Moreover, the search is for x >= 1e6 now that I've read some more of the research papers. Last fiddled with by TeknoHog on 20141222 at 12:22 

20141222, 12:31  #19 
Mar 2010
Jyvaskyla, Finland
2^{2}·3^{2} Posts 
Just a general note, I appreciate the alternative approaches proposed here, but my time/focus is limited. I particularly like the idea of using GPUs for the search/match with a list of y's  bignums are a little harder to do on GPUs, so save the precision parts for the CPU. If some of you can code this, great! Maybe you'll find a solution before me...

20141222, 13:17  #20  
Mar 2010
Jyvaskyla, Finland
2^{2}×3^{2} Posts 
Quote:


20141222, 17:57  #21 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2×4,787 Posts 
Right!
Also right on GPUs. 
20141223, 12:17  #22 
"Robert Gerbicz"
Oct 2005
Hungary
5×13×23 Posts 
To see your chances for a new solution:
Rewrite the equation: x^n=(x1)*y^q+1. If the abc conjecture is true for K=1/4 and eps=1, then x^n<0.25*rad(x^n*(x1)*y^q*1)^2=0.25*rad(x*(x1)*y)^2<0.25*(x^2*y)^2, but it is easy to see that y^q<2*x^(n1), from this y<2*x^((n1)/q) is also true. So x^n<0.25*(x^2*2*x^((n1)/q))^2=x^(4+2*(n1)/q) n<4+2*(n1)/q<=4+2*(n1)/3 as there is no new solution for q=2. And from this we get that n<10, what is also proved that this gives no further solution. About the search: it is too slow. If you need to quickly eliminate the q as an exponent for given (x,n) pair I would choose r=k*q+1 prime (where r>x) and check that c=((x^n1)/(x1))^k mod r is 1 or not. If it is not then (x^n1)/(x1) is not a perfect qth power. Or to avoid to compute the multiplicative inverse just test if (x^n1)^k==(x1)^k mod r is true or not. The probability that this holds is roughly O(1/q), use more r primes and perhaps precalculate them (if you fix x then the set of q primes is the same, I would do this way). With this you can almost completely avoid to use bignums. 
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